section chapter b1 real numbers · 2016-05-21 · real numbers section b chapter 1 topic-1 rational...
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P-1S O L U T I O N S
REAL NUMBERS
SECTION
BSECTIONCHAPTER
1TOPIC-1Rational Numbers
SUMMATIVE ASSESSMENT WORKSHEET-1Solutions
1. 581000
= 0.058 (Decimal point is shifted three places
to the left) 1
2.
256
= 2.04124 ..... . It is non-repeating, so it is not a
rational number.
204
= 2.2360 ..... . It is non-repeating, so it is not a
rational number.
2. 27 = 2.2727 ..... . As, 27 is repeating continously, so it is a rational number.
2 . 3 = 6 = 2.4494 .... . It is non-terminating, non-Repeating So it is not a rational number.
So, 2. 27 is a rational number. 1
3. Let, x = 0.999.... ⇒ 10x = 9.999.... ⇒ 10x–x = (9.999.....) (– 0.999.......) ½ ⇒ 9x = 9 ½ ⇒ x = 1
4.
982
= 982
= 49 = 7
So, it is a rational number. 1 5. Yes, zero is a rational number.
Zero can be expressed as 0 0 0
, ,5 26 100
etc,
which are in the form of, pq
where p and q are
integers and q ≠ 0. 2
6. Let, x = 0. 6 x = 0.6666 ....(i) multiplying 10 on both the sides, we get, 10 x = 6.6666 ....(ii) 1 From (ii) – (i), we get 9 x = 6.0
⇒
x =
69
=
23
.
1
7.
x =
17
0.1428577 10
7 30 28 20 14 60 56 40 35 50 49 1
\ x = 0. 142857 [CBSE Marking Scheme, 2012] 2
8. Since LCM of 7 and 11 is 77,
\ 9
11 =
911
× 77
= 6377
× 33
= 441539
1
and
57
=57
× 1111
= 5577
× 33
= 385539
1
Hence, three rational numbers between57
and
911
are :
386539
,
387539
,
388539
1
9. Any example & verification of example : Let m = 4/5, n = 9/2
Sum = + =4 9 535 2 10
(Rational Number) 1
product =
3610
(Rational Number) 1
diffrence = =
9 4 37–
2 5 10 (Rational Number) 1
division =
9 4 452 5 8
÷ = (Rational Number) 1
P-2 M A T H E M A T I C S - I X T E R M - 1
SUMMATIVE ASSESSMENT WORKSHEET-2Solutions
1.
177413
= 59 × 359 × 7
= 37
.
1
2.
78
= 0.875.
1
3. 1. 011 = 1.011011011....... . 1 4. n – 1. 1
5. 1000x = 237. 237
⇒ x =
237999
[CBSE Marking Scheme, 2012]
Alternative Method : Let
x = 0. 237 = 0.237237237237.......
⇒ 1000x = 237.237237........ ½
⇒ 1000x – x = (237.237237.....) – (0.237237....) ½
⇒ 999x = 237 ½
\
x =
237999
. ½
6.
2157625
= 3.4512 1½
Terminating ½ [CBSE Marking Scheme, 2012]
Alternative Method : 3.4512
625) 2157 1
187528202500
32003125
750 625
1250 1250
×
2157625
= 3.4512 (Terminating) 1
7. Alternative Method : LCM of 5 and 7 is 35
\
35
= 35
× 77
= 2135
½
and
57
=
57
×
55
=
2535
½
so,
2135
< 2235
< 2335
< 2435
< 2535
½
The required three rational numbers are 22
35, 23
35
and
2435
.
½
[CBSE Marking Scheme, 2012] 8. Let a = 3 & b = 4 Here, we find six rational numbers, i.e., n = 6
So d =
– + 1
b an
= 4 – 36 + 1
=
17
1st rational number = a + d = 3 +
17
=227
½
2nd rational number = a + 2d = 3 +
27
=237
½
3rd rational number = a + 3d = 3 +
37
=247
½
4th rational number = a + 4d = 3 +
47
=257
½
5th rational number = a + 5d = 3 +
57
=267
½
6th rational number = a + 6d = 3 +
67
=277
½
So, six rational numbers are
227
,
237
,
247
,
257
,
267
&
277
.
9. LCM of 3, 4, 6, and 12 is 12
1 43 123 12
1 34 124 12
1 2126 6 12
112 1212
2 = 2 = 2 = 16
5 = 5 = 5 = 125
7 =7 =7 = 49
3 = 3 = 3
2
Descending order is
12 12 12 12125, 49 , 16 , 3 1
i.e., 4 3 1265 , 7 , 2 , 3 . 1
[CBSE Marking Scheme, 2012] Alternative Method : Since, LCM of 3, 4, 6, 12 is 12
\1 4 4
12 43 123 4 122 = 2 = 2 = 2 = 16×
1
1 3 3
12 34 124 3 125 = 5 = 5 = 5 = 125×
½
1 2 2
12 2 126 6 2 127 =7 =7 = 7 = 49×
½
12 3 = 12 3
P-3S O L U T I O N S
In Descending order 12 12 12 12125, 49 , 16 , 3 1
i.e., 4 3 1265 , 7 , 2 , 3 . 1
10. 1. 32 = 119
90 1½
0. 35 =
3599
1½
1. 32 + 0. 35 =
1659990
1
[CBSE Marking Scheme, 2012] Alternative Method : Let,
x = 1. 32 = 1.32222.......
⇒ 10x = 13.222..........
⇒ 100x = 132.222......
⇒ 100x – 10x = (132.222.......) – (13.222.....) ½
⇒ 90x = 119.00
⇒
x =
11990 ½
Again, let y = 0. 35 = 0.353535.......
⇒ 100y = 35.3535........ ½
⇒ 100y – y = (35.3535......) – (0.3535.......) ½
⇒ 99y = 35
⇒
y =
3599
½
\ 1. 32 + 0. 35 = x + y =
11990
+ 3599
½
=
119 11 + 35 10990
× ×
=
1309+350990
=
1659990
1
TOPIC-2Irrational Numbers
SUMMATIVE ASSESSMENT WORKSHEET-3Solutions
1. 0.13 is a terminating number. So, it is not an irrational number.
0.13 15 = 0.131515......, 15 is repeating continuously, so it is not an irrational number.
0.1315 = 0.13151315...., 1315 is repeating continuously, so it is not an irrational number.
0.3013001300013...., non-terminating and non-recurring decimal. Hence, it is an irrational number.
So, 0.3013001300013 is an irrational number. 1 2. No, it may be rational or irrational. 1 3. Let, x = 0.777.... ⇒ 10x = 7.777...... ⇒ 10x – x = (7.777......) – (0.777.....) ⇒ 9x = 7
⇒ x =
79
.
1
4. Sum of 2 5 and 3 7 = 2 5 +3 7 . 1
5. Required two irrational number are : 0.1 = 0.10 (i) .10100100010000......... 1 (ii) .1020020002000........ 1
6. 0.5101001000100001....... and 0.502002000200002 ..... . [CBSE Marking Scheme, 2012]
7.
14.5
D
AA O B C E
14.5
4.5
Mark the distance 4.5 units from a fixed point A on a given line to obtain a point B such that AB = 4.5 units. From B, mark a distance of 1 unit and mark the new point as C. Find the mid-point of AC and mark that point as O. Draw a semi-circle with centre O and radius OC. Draw a line perpendicular to AC passing through B and intersecting the semi-circle at D.
Then, BD = 4.5
To represent 4.5 on the number line, let us treat the line BC as the number line, with B as zero, C as 1, and so on. 1
Draw an arc with centre B and radius BD, which intersects the number line at E. 1
\E represents 4.5 1 8. Let two irrational numbers are :
6 and 3 ,
(i) 6 – 3 = Difference is an irrational number.
P-4 M A T H E M A T I C S - I X T E R M - 1
(ii) +6 3 = sum is an irrational number.
(iii) × = =6 3 18 3 2
= product is an irrational number.
(iv) =6 / 3 2 = division is an irrational number. 1 × 4 = 4
SUMMATIVE ASSESSMENT WORKSHEET-4Solutions
1. Since, 5 = 2.236
Hence, the irrational number between 2 and 2.5 is 5 . 1
2. 0.3 0.4+ = (0.333.....) + (0.444.....)
= 0.777...... Let, x = 0.777....... 10x = 7.777....... ⇒ 10 x – x = (7.777.......) – (0.777.......) ⇒ 9x = 7.0
⇒ x = 7
9.
3. 2 is non-terminating, non-recurring. 1
4. 2( 2 + 5) =
2( 2 ) + 2( 5) + 2 × 2 × 5
= 2 + 5 + 2 10 .
= 7 + 2 10 (Irrational Number). 1
5. Let x = 2.218 = 2.218181818........ ⇒ 10x = 22.18181818........ ½ ⇒ 1000x = 2218.181818..... ½ ⇒ 1000x – 10x = (2218.181818......) –
(22.181818.....) ⇒ 990x = 2196.00 ½
⇒ x = 2196990
= 2 3 3 1222 3 3 55× × ×× × ×
= 12255
½
[CBSE Marking Scheme, 2012]
6. Given,
17
= 0.142857142857......... ½
27
= 0.285714285714......
½
Hence, required number can be 0.160160016000... [CBSE Marking Scheme, 2012] 1
7.
–2 –1 –1
2OA
1B
P Q 2 3
3
3
2
D
C1
1
�–�
Let AB = BC =1 unit length
Using Pythagoras theorem, we see that
OC = 2 21 + 1
= 2 ½
Construct CD =1 unit length perpendicular to OC, then using Pythagoras theorem, we see that
OD = 2 2( 2 ) 1+
=3
½
Using a compass with centre O and radius OD, draw an arc which intersects the number line at the point Q, then Q corresponds to 3 . 1
8.
19.3
D
AA O B C E
19.39.3
Mark the distance 9.3 units from a fixed point A on a given line to obtain a point B such that AB = 9.3 units. From B, mark a distance of 1 unit and mark the new point as C. Find the mid-point of AC and mark that point as O. Draw a semi-circle with centre O and radius OC.
Draw a line perpendicular to AC passing through B and intersecting the semi-circle at D. 1
Then, BD = 9.3 1
To represent 9.3 on the number line,let us treat the line BC as the number line, with B as zero, C as 1, and so on.
Draw an arc with centre B and radius BD, which intersects the number line at E. 1
\E represents 9.3
[CBSE Marking Scheme, 2013]
9. 10x = 3.178178 ...(1) 1 ⇒ 10000x = 3178.178178 ...(2) 1
Subtracting (1) from e.q. (2)
9990x = (3178.178 ...... – 3.178....) 1
⇒ x = =3175 6359990 1998
1
P-5S O L U T I O N S
TOPIC-3nth Root of a Real Number
SUMMATIVE ASSESSMENT WORKSHEET-5Solutions
1.
( )5+ 5 ( )5 – 5
=
( ){ }225 – 5
½
= {25 – 5} = 20 ½ [CBSE Marking Scheme, 2014]
2. 12 8× = 2 3 ×2 2 = 4 3 2× = 4 6 . 1
3. 4 28 ÷ 3 7 = 4 ×2 7 ÷ 3 7 = 83
1
4. +8 3 – 2 3 4 3 = +3(8 – 2 4)
= 10 3 2
5. 2 50 × 3 32 × 4 18
⇒10 2 ×12 2 ×12 2
= 2880 2 2
6. x = 3 – 2 2
⇒ x = 1 + 2 – 2 2 ½
⇒ x =
( )21 – 2
⇒ x = 1– 2 ½
⇒
1x
=
11 – 2
×
1 21 2
++
⇒ =
1 21 – 2+
= –
( )1 2+
½
\ x +
1x
= 1 – 2 – 1 – 2
= – 2 2 ½
7. 4 16 =
4 2 2 2 2× × × = 2
3 343 =
3 7 7 7× × = 7
5 243 =
5 3 3 3 3 3× × × × = 3
196 = 14 1
\ 4 16 – 6 3 343 + 18 5 243 – 196
= 2 – 6 × 7 + 18 ×3 – 14 = 2 – 42 + 54 –14 = 56 –56 = 0. 1
8. ( )4 3 3 2+
×( )4 3 – 3 2
=
( )24 3
–
( )23 2
1
= 48 – 18 ½ = 30. ½ [CBSE Marking Scheme, 2013, 2012]
9. 3 45 – 125 200 – 50+
= 9 5 – 5 5 +10 2 – 5 2 1½
= 4 5 5 2+ 1½
10. (i)
( )( )
2 – 1
2 1+
is an irrational number
( )( )
2 – 1
2 1+ =( )( )
( )( )
2 – 1 2 – 1
2 1 2 – 1×
+
= ( )2
2 – 1
2 – 1
=
( )22 – 1
1 = 2 – 1
which is an irrational number. 1 Let, there is a number x such that x3 is an irrational
number but x5 is a rational number.
Let, x = 5 7 be the number
⇒
x3 = ( )35 7 = 35(7) ½
is an irrational number
But, x5 = ( )55 7 = (7)5/5 = 7
= 7, is a rational number ½ (ii) Accepting own mistakes gracefully, co-operative
learning among the classmates. 1
SUMMATIVE ASSESSMENT WORKSHEET-6Solutions
1.
5x
= p 7 ⇒ 5 57
×
= p 7
⇒
p =
257 7×
= 257
. 1
2. b2 = a ⇒
a = b. 1
P-6 M A T H E M A T I C S - I X T E R M - 1
3. ( )a b+ ( )–a b = (a)2 – ( )2
b = a2 – b. 1
4. False 1
Justification :
14775
=
14775
=
4925
= 75
1
which is a rational number. [CBSE Marking Scheme, 2012]
5.
( )24 3 – 3 5
=
( )24 3
+
( )23 5 – 2 × 4 3
× 3 5
[Using (a–b)2 = a2 + b2 –2ab]
= 48 + 45 – 24 15 ½
= 93 – 24 15
= 3(31 – 8 15 ). ½
[CBSE Marking Scheme, 2012]
6. 50 – 98 + 162 1
= 5 5 2× × – 7 7 2× × + 3 3 3 3 2× × × × ½
= 5 2 – 7 2 + 9 2 ½
= 7 2 [CBSE Marking Scheme, 2012]
Alternative Method :
50 – 98 + 162 ½
= 5 5 2× × – 7 7 2× × + 3 3 3 3 2× × × ×
= 5 2 – 7 2 + 3 × 3 2 ½
= –2 2 + 9 2 ½
= 7 2 ½
7. 3 40 =
3 32 5× = 2 3 5 ½
3 320 = 4 3 5 ½
\ 3 3 40 – 4 3 320 –
3 5 = 3 × 2 3 5 – 4 × 4 3 5 –
3 5
= –11 3 5 1
[CBSE Marking Scheme, 2012] Alternative Method :
3 40 =
3 2 2 2 5× × × = 2 3 5 ½
3 320 = 3 2 2 2 2 2 2 5× × × × × ×
= 2 × 2 3 5 = 4 3 5 ½
3 40 – 4
3 320 – 3 5
= 3 × 2 3 5 – 4 × 4 3 5 – 3 5 ½
= 6 3 5 – 16 3 5 – 3 5 = – 11 3 5 . ½
8. LCM of 2 and 3 is 6 1
32 3 3 2× = 2 × 3
6 2 33 2× ½
= 6 6 72 ½
[CBSE Marking Scheme, 2012]
Alternative Method : LCM of 2 and 3 is 6
\ 2 3 3 = 2 ( )
1 23 23 ×
= 2 6 23 ½
and 3 2 = 3 ( )1 32 32 ×
= 3 6 32 ½
Now, 2 3 3 × 3 2 = 26 9 × 3
6 8 ½
= 6 6 72 ½
9. ( )–1 / 6729 = ( )–1 / 663 =
–13 2
=
13
1
[CBSE Marking Scheme, 2012] Alternative Method : 3 729 3 243 3 81 3 27 3 9 3 3 1 729 = 3 × 3 × 3 × 3 × 3 × 3 = 36 2
( )
–16729
= ( )–1
6 63 ½
= ( )–13
=
13
½
10. Justify, 2
5 3+ =
( )( )
5 – 325 3 5 – 3
×+
½
=
( ) ( )2 210 – 6 10 – 6
5 – 35 – 3=
½
=
10 – 62
Again, 28 + 98 + 147
= 2 2 7× × + 2 7 7× × + 3 7 7× × ½
= 2 7 + 7 2 + 7 3 ½
Value : Co-operative learning among classmates without any gender and religious bias . 1
P-7S O L U T I O N S
TOPIC-4Laws of Exponents with Integral Powers
SUMMATIVE ASSESSMENT WORKSHEET-7Solutions
1.
1–2 –2 2 2 4
4 4 4132 143 13 11 13143 132 11 12 12
× = = = ×
=
1213
12
=
132 3 1
2. [(16)½]½
= ( )1
1 22 24
= [ ]124 ½
= 2 ½
3. ( )( )
( – )
( – )
a b cc
b
b a c a
xxxx
÷
= 1 1
–
–
ab ac bc
ba bc ac
x xx x
÷ = (xab–ac–ba+bc) ÷ (xbc–ac) = x0 1
= 1 [CBSE Marking Scheme, 2014]
Alternative Method :
( – )
( – )
ca b c b
b a c a
x xx x
÷
= –
–
ab ac bc
ba bc ac
x xx x
÷
½
= (xab–ac–ba+bc) ÷ (xbc–ac) ½ = (xbc–ac) ÷ (xbc–ac) ½ = 1 ½
4.
30 29 28
31 30 29
3 3 33 3 – 3
+ ++
=
28 2 1
29 2 1
3 (3 3 1)3 (3 3 – 1)
+ ++
1
=
(9 3 1)3(9 3 – 1)
+ ++
½
=
133 11×
=
1333
½
[CBSE Marking Scheme, 2012]
5.
( ) +
1/ 431/3 1/35 8 27
=
( ) +
1/ 433 1/3 3 1/35 (2 ) (3 )
1
= +
1/ 435 (2 3)
1
= ( ) = =
1/ 41/ 43 45 (5) 5 5
1
6. ( )2
2 2 – 5 + ( )2
3 2 3+ – ( )2
2 – 1
= ( )22 2 – 2( 2 )(5) + (5)2 + ( )2
3 2 + 2(3 2 )
( 3 ) + ( 3 )2 – ( )2
2 – (1)2 +2( 2 ) (1) 1
= 8 + 25 – 20 2 + 18 + 3 + 6 6 – 2 – 1 + 2 2 1
= 51 – 18 2 + 6 6 1
[CBSE Marking Scheme, 2012]
7.
2 3 1– – –
3 4 5
4 1 2
(216) (256) (243)
+ +
=
–2 –3 –13 4 53 4 5
4 1 2
(6 ) (4 ) (3 )
+ +
1
=
–2 –3 –14 1 2
6 4 3+ +
1
=
–1 –1 –1
4 1 236 64 3
+ +
1
= 4 × 36 + 1 × 64 + 2 × 3 ½ = 214 ½
8.
–3 3–3
4 281 9 516 25 2
× ÷
=
3 33
4 216 9 281 25 5
× ÷
1
=
3 334 24 2
4 22 3 2
53 5
× ÷
½
P-8 M A T H E M A T I C S - I X T E R M - 1
=
3 34 2 34 22 3 5
3 5 2
× ×
1
=
3 3 32 3 53 5 2
× × ½
=
3 3 3
3 3 32 3 53 5 2
× ×
=
3 3
3 32 33 2
×
½
= 1 ½
SUMMATIVE ASSESSMENT WORKSHEET-8Solutions
1. ( )1
– 4481 81× = ( ) ( )1 1
–4 44 43 3×
= 3–1 × 31 = 30 = 1. 1
2.
( ){ }2–1
–1 4281
=
( ){ }–1
–1 2281
= ( )1481
= ( )1
4 43 = 3. 1
3. = 1
3 45(2 3) + 1
= 1
4 4(5 ) 1
= 5 [CBSE Marking Scheme, 2012] Alternative Method :
13 41 1
3 35 8 27
+
=
13 41 1
3 33 35 (2 ) (3 )
+
½
=
13 45(2 3) +
=
13 45(5) 1
=
14 4(5 ) = 51= 5. ½
4.
3 3– –
4 281 2516 9
× =
3 3– –4 24 23 5
2 3
× ½
=
–3 –33 52 3
× ½
=
3 32 33 5
× ½
=
3
3
25
=
8125
½
[CBSE Marking Scheme, 2012]
5.
3 3– – –3
4 281 25 516 9 2
× ÷
½
=
31 33
4 216 9 281 25 5
× ÷
1
=
3 3 32 3 23 5 5
× ÷ ½
=
38 3 527 5 2
× × =
8 27×
27 8 = 1 1
[CBSE Marking Scheme, 2012, 2013] Alternative Method :
3 3– – –3
4 281 25 516 9 2
× ÷
=
3 3– –4 2 –34 23 5 5
2 3 2
× ÷
1
=
–3 –3 –33 5 52 3 2
× ÷ ½
=
3 3 32 3 23 5 5
× ÷ ½
=
8 27 125× ×
27 125 8
½
=
8 27×
27 8 = 1
½
6.
7 5––1 2 –2 32 2
2 – 4 3 –5
5 7 5 75 7 5 7
× ×× × × ½
=
7 5–4 2 5 32 2
2 1 3 2
7 7 7 75 5 5 5
× ×× × ×
1
=
7 5–6 82 2
3 5
7 75 5
×
=
7 56 52 2
3 8
7 55 7
×
½
P-9S O L U T I O N S
=
1 142 252 2
21 40
7 55 7
×
=
142 25 2
21 40
7 55 7
×
½
= ( )1
2 4 27 5×
= 7 × 52 = 7 × 25 = 175 [CBSE Marking Scheme, 2012]
Alternative Method :
7 5––1 2 –2 32 2
2 – 4 3 –5
5 7 5 75 7 5 7
× ×× × ×
=
7 5–4 2 5 32 2
2 1 3 2
7 7 7 75 5 5 5
× ×× × ×
½
=
7 5–6 82 2
3 5
7 75 5
×
½
=
1 17 –52 26 8
3 5
7 75 5
×
½
=
1 142 252 2
21 40
7 55 7
×
½
=
142 25 2
21 40
7 55 7
× ×
= ( )1
2 4 27 5×
½
= 7 × 52 = 175 ½
7.
( ) 2 2 2( ) ( )
4
. .
( )
a b b c c a
a b c
x x x
x x x
+ + +
=
2 2 2 2 2 2
4 4 4. .
. .
a b b c c a
a b cx x x
x x x
+ + +
2
=
4 4 4
4 4 4
a b c
a b cxx
+ +
+ +
½
= 1 1½ [CBSE Marking Scheme, 2012]
Alternative Method :
( )
( )4
2 2 2a+b (b+c) (c+a)
a b c
x . x . x
x x x
=
2( ) 2( ) 2( )
4 4 4. .
. .
a b b c c a
a b cx x x
x x x
+ + +
1
=
2 2 2 2 2 2
4 4 4. .
. .
a b b c c a
a b cx x x
x x x
+ + +
1
=
2 2 2 2 2 2
4 4 4
a b b c c a
a b cx
x
+ + + + +
+ +
1
= 4 4 4
4 4 4
a b c
a b cxx
+ +
+ + = 1 1
8.
7 5––1 2 –2 32 2
2 – 4 3 –52 3 2 32 3 2 3
× ×× × ×
=
7 5–6 82 2
3 53 32 2
×
1½
=
7 56 8 –
2 2
7 53 5 –
2 2
3 3
2 2
× ×
× ××
1
=
2521 2
21 202
3 23
2
×
= 3 × 22 = 3 × 4 = 12 1½ [CBSE Marking Scheme, 2012]
Alternative Method :
7 5––1 2 –2 32 2
2 – 4 3 –52 3 2 32 3 2 3
× ×× × ×
=
7 5–6 82 2
3 53 32 2
×
1
=
7 56 8 –
2 2
7 53 5 –
2 2
3 3
2 2
× ×
× ××
1
=
21 –20
21 25–
2 2
3 3
2 2
×
1
= 21–20
21 25–
2 2
3
2
= 1
4–
2
3
2
½
= 3 × 22 = 12. ½
SUMMATIVE ASSESSMENT WORKSHEET-9Solutions
1.
15002 15
=
1 15002 15
×
=
1100
2×
=
102
= 5.
1
2.
3 / 4
–1 / 4
1616
= ( )
3 14 416 +
= (16)1 = 16.
1
P-10 M A T H E M A T I C S - I X T E R M - 1
3. (13 + 23 + 33)–3/2 = (1 + 8 + 27)–3/2 ½ = (36)–3/2 ½ = [(6)2]–3/2 = 6–3 ½
=
3
16
=
1216
½
[CBSE Marking Scheme, 2012]
4. Given, a = 2 and b = 3. ab + ba = 23 + 32 ½ = 8 + 9 1 = 17 ½ [CBSE Marking Scheme, 2012]
5. (xa–b)a+b. (xb–c)b+c. (xc–a)c+a
= 2 2 2 2 2 2– – –. .a b b c c ax x x 1
= 2 2 2 2 2 2– – –a b b c c ax + +
1
= x0 = 1 (any number to the power 0 is 1) 1 [CBSE Marking Scheme, 2012]
6. 2–
3
4
(216)
–
3–
4
1
(256) =
–2 –3
3 43 4
4 1–
(6 ) (4 )
1
= 4 × 62 – 43 1 = 144 – 64 = 80 1 [CBSE Marking Scheme, 2011, 2012]
Alternative Method :
2–
3
4
(216)
–
3–
4
1
(256)
= –2
3 3
4
(6)
– –3
4 4
1
(4)
1
=
–2
46
–
–3
14
½
= 4 × 62 – 43 ½ = 144 – 64 ½ = 80 ½
7.
1 1.–
ab aba b a a b a
++
1
= 2 2
2 2
––
b ab b abb a
+ +
1
=
2
2 2
–2–b
a b 1
[CBSE Marking Scheme, 2012] Alternative Method :
–1 –1
–1 –1 –1 –1
a a+
a +b a – b =
1
1 1a
a b+
+
1
1 1–
a
a b
1
=
1a
a bab+
+
1
–a
b aab ½
=
ba b+
+ –
bb a ½
=
( ) ( )( )( )–
–b a b b b a
b a b a+ +
+ ½
=
2 2
2 2
––
b ab b abb a
+ +
½
=
( )2
2 2
2– –
ba b
=
2
2 2
–2–b
a b ½
8. LHS = 1 1 1
1 1 11 1y xy y xy
y xy x
+ ++ + + + + +
1
( xyz = 1 ⇒
1z
= xy) ½
=
11 1 1
y xyxy y xy y xy x
+ ++ + + + + +
1
=
11
y xyxy y
+ ++ +
= 1 = RHS 1½
Hence proved.[CBSE Marking Scheme, 2012]
Alternative Method : (1 + x + y–1)–1 + (1 + y + z–1)–1 + (1 + z + x–1)–1
=1 1 1
1 1 11 1 1x y z
y z x
+ ++ + + + + +
1
=
1 1 11 1 111 1y xyxy xy x
+ ++ ++ + + +
( xyz = 1 ⇒
1z
= xy)
=
1 1 11 11y xy xy yy xy
y xy
+ ++ + + ++ +
1
=
11 1 1
y xyy xy y xy xy y
+ ++ + + + + +
1½
=
( )( )
11
y xyy xy
+ ++ +
= 1.
½
9. 52x–1 – (52)x–1 = 2500 ½ ⇒ 52x–1 – 52x–2 = 2500 ½ ⇒ 52x–2 (5 – 1) = 2500
⇒
52x–2 =
25004
= 625 = 54 1
⇒ 2x – 2 = 4 ⇒ 2x = 6 \ x = 3 1 [CBSE Marking Scheme, 2012]
P-11S O L U T I O N S
Alternative Method : Given, 52x–1 – (25)x–1 = 2500 ⇒ 52x–1 – [(52)]x–1 = 2500 ½ ⇒ 52x–1 – 52x–2 = 2500 ½ ⇒ 52x–2 (5 – 1) = 2500 1
⇒
52x–2 =
25004
= 625 ½
⇒ 52x–2 =54 ½
comparing the powers of both sides, we get 2x – 2 = 4 ½ 2x = 6 ⇒x = 3 ½
TOPIC-5Rationalisatoin of Real Numbers
SUMMATIVE ASSESSMENT WORKSHEET-10Solutions
1.
150
=
15 5 2× ×
=
1 25 2 2
× =
210 1
So, rationalising factor is 2 .
2.
6 – 4 3 6 – 4 36+4 3 6 – 4 3
×
½
=
( )26 – 4 3
36 – 48 ½
=
36 48 – 48 3–12
+
½
=
84 – 48 3–12
= – (7 – 4 3 ) = 4 3 – 7
½
3. x = 3 – 2 2 ⇒
1x
= 3 + 2 2 . ½
21
xx
+ = 8 1
⇒ x +1x
= +2 2 ½
[CBSE Marking Scheme, 2012] Alternative Method :
x = 3 – 2
⇒
1x
=
( )
( )( )3 2 21
3 – 2 2 3 2 2
+×
+
=
( )3 2 2
9 – 8
+
= 3 + 2 2 ½
21
xx
+ =
1 12x x
x x+ + × ×
½
= 3 – 2 2 3 2 2 2+ + + ½
⇒ 21
xx
+ = 8
\ 1
+xx
= +2 2 ½
4. x = 2 + 3
1x
= 2 – 3 1
x +
1x
= 4 1
Squaring, both sides, we get
22
1+x
x = 14 1
5. 3 2
5 2+
+ =
3 2 5 – 25 2 5 – 2
+×
+ 1
= 5 3 5 2 – 6 – 225 – 2
+
1
= 5 3 5 2 – 6 – 223
+ 1
6.
( )
( )( )
2 3 412 3 – 4 2 3 4
+ +×
+ + +
1
=
( ) ( )2
2 3 4 2 3 42 3 2 6 – 42 3 – 4
+ + + +=
+ ++
=
2 3 4 1 – 2 61 2 6 1 – 2 6+ +
×+
1
=
2 22 3 4 – 2 12 – 2 18 – 4 6
1 – (2 6 )+ +
½
=
2 3 4 – 4 3 – 6 2 – 4 61 – 24
+ +
½
P-12 M A T H E M A T I C S - I X T E R M - 1
=
–5 2 – 3 3 2 – 4 6–23
+
½
=
5 2 3 3 4 6 – 223
+ + +
½
7. LHS
=
1 1 1 1 1– –
3 – 8 8 – 7 7 – 6 6 – 5 5 – 2+ +
=
( ) ( )( )
( ) ( )( )
( ) ( )2 2 2 2 22
8 7 7 63 8–
3 – 8 8 – 7 7 – 6
+ +++
–
( )( ) ( )
( )( ) ( )2 2 2 2
6 – 5 5 2
6 – 5 5 – 2
++
1
=
( ) ( ) ( )8 7 7 6 6 53 8– –
9 – 8 8 – 7 7 – 6 6 – 5
+ + +++
1
+
( )5 +2
5 – 4 [ a2 – b2 = (a – b) (a + b)] 1
= 3 + 8 – 8 – 7 + 7 + 6 – 6 – 5 + 5 + 2
= 3 + 2 = 5 = RHS 1
SUMMATIVE ASSESSMENT WORKSHEET-11Solutions
1.
1 ( 2 – 1)( 2 1) ( 2 – 1)
×+
=
2 –1
= 1.414 – 1 1
= 0.414 1 [CBSE Marking Scheme, 2012]
Alternative Method :
12 1+
= 1 ( 2 – 1)
( 2 1) ( 2 – 1)×
+
=
( )( ) ( )2 2
2 – 1
2 – 1
1
=
2 – 12 – 1
= 1.414 – 1
1 = 0.414. 1
2.
12
=
12
×
22
(rationalising)
=
22
= 1.414
2 = 0.707 1
12
+ p = 0.707 + 3.141
= 3.848. 1
[CBSE Marking Scheme, 2012]
3.
5 65 – 6
+
=
5 65 – 6
+
×
5 65 6
++
½
=
( )( )
2
2 2
5 6
5 – ( 6 )
+
½
=
25 6 10 625 – 6
+ +
½
=
31 10 619
+
½
\ a + b 6 =
3119
+
1019
6
½
Comparing the rational and irrational parts of both sides, we get
a =
3119
, b =
1019
½
4. x =
13 – 2 2
×
3 2 23 2 2
++
x =
3 2 29 – 8+
= 3 + 2 2
1
and
y =
13 2 2+
×
3 – 2 23 – 2 2
=
3 2 29 – 8+
= 3 – 2 2 1
x + y+ xy = 3+2 2 + 3 – 2 2 + (3+2 2 )
(3 – 2 2 ) ½
= 6 + 9 – 8
= 7 ½
5. (i) We know that between two rational numbers x and y, such that x < y there is a rational number
+2
x y
.
i.e.,
3 <
72
< 4
Now, a rational number between 3 and
72
< 4 is :
1 7
32 2
+ =
1 6 72 2
+ × = 13
4 ½
A rational number between
72
and 4 is :
P-13S O L U T I O N S
1 7
42 2
+ =
1 7 82 2
+ × = 154
½
\
3 < 134
<
72
< 154
< 4
½
Further a rational between 3 and
134
is :
1 13
32 4
+ =
1 12 132 4
+ × = 25
8
A rational number between
154
and 4 is :
1 15
42 4
+ = 1 15 16
2 4+
× = 31
8 ½
A rational number between
318
and 4 is :
1 31
42 8
+ =
1 31 322 8
+ × = 63
16 ½
\
3 < 25
8 < 13
4<
72
< 154
< 318
< 6316
< 4
Hence, six rational numbers between 3 and 4 are
25 13 7 15 31, , , , ,
8 4 2 4 8 and 63
16 1
(ii) Number system. ½
(iii) Rationality is always welcomed.
6. x =
3 23 – 2
+
and
y =
3 – 23 2+
⇒
x2 + y2 =
23 23 – 2
+
+
23 – 23 2
+
1
=
3 2 2 63 2 – 2 6
+ + +
+ 3 2 – 2 63 2 2 6
+ + +
1
=
5 2 65 – 2 6
+
+
5 – 2 65 2 6+
½
=
( ) ( )( ) ( )
2 2
22
5 2 6 5 – 2 6
5 – 2 6
+ +
½
=
25 24 20 6 25 24 – 20 625 – 24
+ + + +
= 98 1
7. LHS =
( ) ( )( )( )
2 22 5 3 2 5 – 3
2 5 – 3 2 5 3
+ +
+
1
( ) ( )2 24 5 3 2 2 5 3 4 5 3 – 2 2 5 3
2 5 – 3
× + + × × + × + × ×
1
=
20 3 4 15 20 3 – 4 1520 – 3
+ + + +
=
4617
=
4617
+
15 ×
(0)
1
\
4617
+
15 (0) = a + 15 b
= RHS
comparing both sides, we get
a =
4617
, b = 0 1
[CBSE Marking Scheme, 2013]
SUMMATIVE ASSESSMENT WORKSHEET-12Solutions
1.
2 2 8– +
5 +2 10 – 2 2 2
=
( )
2 ( 5 – 2) 2–
( 5 2) ( 10 – 2 2 )5 – 2×
+
×
( 10 2 2 ) 8 2( 10 2 2 ) 2 2
++ ×
+ 1
= 10 – 2 2 2( 10 2 2 ) 8 2–
5 – 4 10 – 8 2+
+
1
=
10 – 2 2 – 2( 10 2 2 )2+
+ 4 2
=
10 – 2 2 – 10 – 2 2 + 4 2 = 0 . 1
2.
12 7 3 3+
=
1(2 7 3 3)
×+
(2 7 – 3 3)(2 7 – 3 3)
=
2 2
2 7 – 3 3(2 7 ) – (3 3)
1
= 2 7 – 3 34 7 – 9 3× ×
1
= 2 7 – 3 3
28 – 27 1
=
2 7 – 3 31
= 2 7 – 3 3 .
P-14 M A T H E M A T I C S - I X T E R M - 1
3. x =
2[ 2 – 2 ]2 – 2
p q p qp q p q+ ++ +
1
=
14q
( )2 22 2 – 4p p q+ 1
⇒ 2qx – p = 2 2– 4p q
⇒ 4q (qx2 – px) = – 4q2 ⇒ qx2 – px + q = 0. 1 [CBSE Marking Scheme, 2012]
Alternative Method :
x =
2 – 2 2 – 22 – – 2 2 – 2
p q p q p q p qp q p q p q p q
+ + + +×
+ + +
½
= ( ) ( )
( ) ( )2 – 2 2 2 – 2
2 – – 2p q p q p q p q
p q p q+ + + × + ×
+
½
= 2 22 2 – 4
2 – 2p p qp q p q
++ +
½
= ( )2 22 – 4
4
p p q
q
+
⇒ 2qx = p + 2 2– 4p q
⇒
2qx – p = 2 2– 4p q Squaring both sides, we get 4q2x2 + p2 – 4pqx = p2 – 4q2 ½ 4q (qx2 – px) = – 4q2 ½ qx2 – px + q = 0. ½
4. (i)
0.7142857 5.000000
– 4910–7
30–28
20–14
60–56
40–355
\
57
=0. 714285 1
and
0.8111 9.00
– 8820
–119
\
911
=0. 81 1
Thus, three different irrational numbers between 57
and
911
are 0.75075007500075000075.......,
0.767076700767000767.... and 0.80800800080000...... (ii) Number system. (iii) Those who are irrational in their approach
fail in their efforts. 1
5. x =
12 – 3
⇒
x =
1 (2 3)2 – 3 (2 3)
+×
+
⇒
=
2 34 – 3+
= 2+ 3 1
⇒ (x – 2) = 3
⇒ ( )2– 2x = ( )23 = 3 1
x2 – 4x + 4 = 3 x2 – 4x + 4 – 3 = 0 ½ \ x2 – 4x + 1 = 0
\ x3 – 2x2 – 7x + 5
x(x2 – 4x + 1) + 2(x2 – 4x + 1) + 3 1
= x × 0 + 2 × 0 + 3 = 3 ½ [CBSE Marking Scheme, 2012]
FORMATIVE ASSESSMENT WORKSHEET-13 & 14 Note : Students should do these activities themselves.
qqq
P-15S O L U T I O N S
TOPIC-1Polynomials
SUMMATIVE ASSESSMENT WORKSHEET-15Solutions
1. p (y) = y2 – y + 1 p (0) = 02 – 0 + 1 = 1 1 2. x + 4 is a factor of x2 + 3x + m ⇒ p (– 4) = 0 ⇒ (– 4)2 + 3(– 4) + m = 0 ⇒ 16 – 12 + m = 0 ⇒ m = – 4 1
3. 3 x2 – x – 1. 1 4. Not defined. 1 5. Linear polynomial. 1 6. Degree of x3 + 5 = 3 Degree of 4 – x5 = 5 Degree of (x3 – 5) (4 – x5) = 3 + 5 = 8. 1 7. Not a polynomial. 1 8. (x + 3) is a factor of x3 + 3x2 – kx – 3 ⇒ p (–3) = 0 ⇒ (–3)3 + 3(–3)2 – k (–3) –3 = 0 1 – 27 + 27 + 3k – 3 = 0 k = 1 1
9. (x + 2) is a factor of p(x) = 2x3 – kx2 + 3x + 10 ½ ⇒ p (–2) = 0 ⇒ 2 (–2)3 – k (–2)2 + 3(–2) +10 = 0 1 ⇒ – 16 – 4k – 6 + 10 = 0 k = – 3 ½ [CBSE Marking Scheme, 2014]
10. p(x) = 2x2 + kx + 2 (x – 1) is a factor of p(x), then p(1) = 0 ½
\ 2 (1)2 + k(1) + 2 = 0 ½
2 + k + 2 = 0 ½
k = –2 – 2 ½ 11. f(x) = 3x + 5 then, f(7) = 3 × 7 + 5 = 26 1 then, f(5) = 3 × 5 + 5 = 20 1 \ f(7) – f(5) = 26 – 20 = 6. 1 12. f(x) = x2 – 5x + 7 Then, f(2) = 22 – 5 × 2 + 7 = 1 1 f(–1) = (– 1)2 – 5 (–1) + 7 = 13 1
and f13
=
213
– 513
+ 7 =
499
1
f(2) – f(–1) + f13
= 1 – 13 + 499
=
–599
1
SUMMATIVE ASSESSMENT WORKSHEET-16Solutions
1. Constant polynomial is 7. 1 2. Binomial. 1
3. Degree of a polynomial 3 is 0. 1 4. Every real number is a zero of the zero polynomial. 1 5. No. of zeroes of cubic polynomial = 3. 1 6. (x + 1)3 – (x + 1) = (x + 1) [(x+1)2 – 1] One factor of (x + 1)3 – (x + 1) is (x + 1). 1 7. Given, p(x) = x2 + 11x + k Since – 4 is a zero of polynomial p(– 4) = 0 ⇒ (– 4)2 + 11 × (– 4) + k = 0
⇒ 16 – 44 + k = 0 \ k = 28. 1
8. 2 x2 + 3x or 3 y2 + 3y. 1
9. Coefficient of x in expression x2 + 2π
x – 7 is 2π
. 1
10. p(x) = x3 – 3x2 – 2x + 6 ½
Then, p ( 2 ) = ( )32 –3 2( 2 ) – 2 ( )2 + 6 ½
= 2 2 – 6 – 2 2 + 6 ½ = 0 ½ 11. Linear polynomial → 1 + x, degree = 1 ½ Quadratic polynomial → x2 + x, degree = 2 ½ Cubic polynomial → x – x3, 7x3, degree = 3 1
POLYNOMIALS
SECTION
BSECTIONCHAPTER
2
P-16 M A T H E M A T I C S - I X T E R M - 1
12. p(x) = x2 – 3x + 6
(i) When x = 2
then, p ( )2 = ( )22 – 3 × 2 + 6 ½
= 2 – 3 2 + 6 ½
= 8 – 3 2 ½ (ii) When x = 3 then, p(3) = 32 – 3 × 3 + 6 ½ = 9 – 9 + 6 ½ = 6 ½ 13. (i) (I) Possible length and breadth of the rectangle
are the factors of its given area. Area = 25a2 – 35a + 12
= 25a2 – 15a – 20a + 12 = 5a(5a – 3) – 4(5a – 3) = (5a – 4)(5a – 3) So, possible length and breadth are (5a – 3) and (5a – 4)
units, respectively. 1 (II) Area = 35y2 + 13y – 12 = 35y2 + 28y – 15y – 12 = 7y(5y + 4) – 3(5y + 4) = (7y – 3)(5y + 4) So, possible length and breadth are (7y – 3) and
(5y + 4) units respectively. 1 (ii) Factorisation of Polynomials. ½ (iii) Expression of one’s desires and news is very
necessary. ½
TOPIC-2Remainder Theorem
SUMMATIVE ASSESSMENT WORKSHEET-17Solutions 1. Let, p(y) = 5y3– 2y2 – 7y + 1, then remainder = p(0) = 0 – 0 – 0 + 1 = 1. 1 2. Put, x + 1 = 0 ⇒ x = – 1 Then remainder is : f(– 1) = (– 1)11 + 101 = –1 + 101 = 100. 1 3. p(x) = 2x3+ 3x2 – 9x + 4 So, x2 + 2x – 3
2x – 1) 2x3 + 3x2 – 9x + 4 2x3 – x2
(–) (+) 4x2 – 9x 4x2 – 2x (–) (+) – 7x + 4 – 6x + 3 (+) (–)
– x + 1 (Remainder) Quotient = x2 + 2x – 3 Remainder = – x + 1 2
4. p(x) = x3 + x2 + x + 1
Put, x – 12
= 0 ⇒x = 12
in p(x) ½
Remainder = p 12
½
= 31
2
+ 21
2
+ 12
+ 1 ½
= 18
+ 14
+ 12
+ 1
= 1+2+4+8
8=
158
. ½
[CBSE Marking Scheme, 2012] 5. Factors of 12 = (±1, ± 2, ± 3, ± 4, ± 6, ± 12 ) ½ p(x) = 6x3 – 25x2 + 32x – 12 p(2) = 6(2)3 – 25(2)2 + 32 × 2 – 12 = 48 – 100 + 64 – 12 = 112 – 112 = 0 ½ \ x = 2 is a zero of p(x) or (x – 2) is a factor of p(x) ½ 6x3 – 25x2 + 32x – 12 = 6x2(x – 2) – 13x(x – 2) + 6(x – 2) ½ = (x – 2) (6x2– 13x + 6) = (x – 2) (6x2– 9x – 4x + 6) ½ = (x – 2) [3x(2x – 3) –2(2x – 3)] = (x – 2) (2x – 3) (3x – 2). ½ 6. 4y2 – 4y + 14 y2 + 4y + 2) 4y4 + 12y3 + 6y2 + 50y + 26 1 4y4 + 16y3 + 8y2
– – – – 4y3 – 2y2 + 50y 1 – 4y3 – 16y2 – 8y + + + 14y2 + 58y + 26 1 14y2 + 56y + 28 – – 2y – 2 1 So, 2y – 2 must be subtracted.
P-17S O L U T I O N S
7. Let p(x) = ax3 + 3x2 – 13 and g(x) = 2x3 – 5x + a R1 = p(2) R2 = g(2) 1½ + 1½ p(2) = a(2)3 + 3(2)2 – 13 R1 = 8a + 12 – 13 = 8a – 1 R2 = 2(2)3 – 5(2) + a = 16 – 10 + a ⇒ R2 = 6 + a R1 = R2 ⇒ 8a – 1 = 6 + a ½ ⇒ 7a = 7 \ a = 1 ½ [CBSE Marking Scheme, 2014]
Alternative Method : Let p(x) = ax3 + 3x2 – 13 and g(x) = 2x3 – 5x + a 1 R1 = p(2) R2 = g(2) p(2) = a(2)3 + 3(2)2 – 13 = 8a + 12 – 13 1 = 8a – 1 So, R1 = 8a – 1 g(2) = 2(2)3 – 5(2) + a = 16 – 10 + a 1 = 6 + a
So, R2 = 6 + a According to the question, R1 = R2 then, 8a – 1 = 6 + a ⇒ 8a – a = 6 + 1 ⇒ 7a = 7 1 \ a = 1 8. (4x2 + 3x – 12) quotient x2 + 2x + 2 ) 4x4 + 11x3 + 2x2 – 11x + 6 ½ 4x4 + 8x3 + 8x2
– – – 3x3 – 6x2 – 11x + 6 ½ 3x3 + 6x2 + 6x – – – – 12x2 – 17x + 6 ½ – 12x2 – 24x – 24 + + + 7x + 30 ½ Thus, quotient = 4x2 + 3x – 12 and remainder = 7x + 30 By Remainder theorem, Divide nd = (Divisor × Quotient) + Remainder ½ \ 4x4 + 11x3 + 2x2 – 11x +6 = [(x2 + 2x + 2) × (4x2 + 3x – 12)] + (7x + 30) ½ = [4x4 + 8x3 + 8x2 + 3x3 – 6x2 + 6x – 12x2 – 24x – 24]
+ 7x + 30 = 4x4 + 11x3 + 2x2 – 11x + 6 1
SUMMATIVE ASSESSMENT WORKSHEET-18Solutions
1. For zeroes, put p(x) = 0 x (x – 2) (x – 3) = 0, then, x = 0, 2, 3. 1 2. Putting, x = 0 p(0) = 0 – 3 × 0 + 2 = 2 Putting, x = 2 p(2) = 22 – 3 × 2 + 2 = 4 – 6 + 2 = 0 Thus, p(0) + p(2) = 2 + 0 = 2. 1 3. Here, p (x) = x4 + x3 – 2x2 + x + 1, and the zero of x – 1 is 1. ½ So, p (1) = (1)4 + (1)3 – 2 (1)2 + 1 + 1 = 2 1 So, by the Remainder Theorem, 2 is the remainder
when x4 + x3 – 2x2 + x + 1 is divided by x – 1. ½ 4. Here, p(x) = x3 – ax2 + 6x – a, and the zero of x – a is a. ½ So, p(a) = a3 – a.a2 + 6a – a = 5a. 1 So, by the remainder theorem 5a is the remainder
when x3 – ax2 + 6x – a, is divided by x – a. ½ 5. Let, p(x) = 3x3 – 5x2 + kx – 2, and q(x) = – x3 – x2 + 7x + k
Put, x + 2 = 0 or x = – 2 in p (x) and q (x) P(– 2) = 3 (– 2)3 – 5(– 2)2 + k(– 2) – 2 = – 24 – 20 – 2k – 2 1 = –2k – 46 q(– 2) = – (– 2)3 – (– 2)2 + 7 (– 2) +k = 8 – 4 – 14 +k 1 = – 10 +k \ p(x) and q(x) leave the same remainder when
divided by x + 2. ⇒ –2k – 46 = k – 10 ⇒ –3k = 36 \ k = – 12 1 6. p(x) = x3 + 8x2 + 17x + a p(x) leave the same remainder when divided by
(x + 2) and (x +1). ½ p(–2) = (–2)3 + 8(–2)2 + 17(–2) + a = – 8 + 32 – 34 + a = – 10 + a 1 p(–1) = (–1)3 + 8(–1)2 + 17(–1) + a(–1) = – 1 + 8 – 17 + a = – 10 – a 1 Remainders are equal
P-18 M A T H E M A T I C S - I X T E R M - 1
So, – 10 + a = – 10 – a ½ ⇒ 2a = 0
\ a = 0 1
7. R1 = f(1) ⇒ 14 – 2(1)3 + 3(1)2 – a(1) + b = 5 ⇒ 1 – 2 + 3 – a + b = 5 ⇒ b – a = 3 ...(1) 1 R2 = f(–1) ⇒ (–1)4 – 2(–1)3 + 3(–1)2 –a(–1) + b = 19 1 + 2 + 3 + a + b = 19 ⇒ a + b = 13 ...(2) 1 Solving (1) and (2), we get b = 8 and a= 5 1 R = f(2) = (2)4 – 2(2)3 + 3(2)2 – 5 × 2 + 8 = 12 – 10 + 8 = 10 1 [CBSE Marking Scheme, 2013]
Alternative Method : f(x) = x4 – 2x3 + 3x2 – ax + b Put, x – 1 = 0 or x = 1 in f(x), we get ⇒ f(1) = (1)4 – 2(1)3 + 3(1)2 – a × 1 + b
⇒ 5 = 1 – 2 + 3 – a + b ⇒ 5 = 2 – a + b ⇒ a – b = 2 – 5 1 ⇒ a – b = – 3 ...(1) Again put, x + 1 = 0 or x = – 1 in f(x), we get f(– 1) = (– 1)4 – 2 (– 1)3 + 3 (– 1)2 – a (– 1) + b ⇒ 19 = 1 + 2 + 3 + a + b ⇒ 19 – 6 = a + b ⇒ a + b = 13 ...(2) 1 Adding equations (1) and (2), 2a = 10 a = 5 By equation (2), 5 + b = 13 \ b = 8 \ f(x) = x4 – 2x3 + 3x2 – 5x + 8 1 Again put, x – 2 = 0 or x = 2 in f(x) f(2) = (2)4 – 2 (2)3 + 3 (2)2 – 5 × 2 + 8 = 16 – 16 + 12 – 10 + 8 = 20 – 10 = 10 1
TOPIC-3Factor Theorem
SUMMATIVE ASSESSMENT WORKSHEET-19Solutions
1. x + 4 is a factor of x2 + 3x + m = p(x) ⇒ p(– 4) = 0 ⇒ 16 – 12 + m = 0 ⇒ m = – 4 1 2. 2x – 1 is a factor of p(x) = 6x2 + kx – 2
⇒
12
p = 0
⇒
1 16. . – 2
4 2+ k
= 0
⇒ k = 1 1 3. 8x3 – (2x – y)3 = (2x)3 – (2x – y)3
= [2x – (2x – y)][(2x)2 + (2x – y)2
+ 2x(2x – y)] Since (a3 – b3) = (a – b) (a2 + b2 – ab) = y[4x2 + 4x2 + y2 – 4xy
+ 4x2 – 2xy] = y[12x2 + y2 – 6xy] 2 4. 9x2 + 6xy + y2 = (3x)2 + 2 × (3x) × y + y2 1 = (3x + y)2 [ a2 + 2ab + b2 = (a + b)2] 1 5. Give Exp = (x – y)2 – 7(x + y) (x – y) + 12(x + y)2
= (x – y)2 – 4(x + y) (x – y) – 3(x + y) (x – y) + 12(x + y)2
= (x – y) [x – y – 4x – 4y] – 3(x + y) [x – y – 4x – 4y] 1 = (x – y) [– 5y – 3x] – 3(x + y) [– 3x – 5y] = (– 5y – 3x) [x – y – 3 (x + y)] 1 = – (5y + 3x) (– 2x – 4y) = (5y + 3x) (2x + 4y) = 2(x + 2y) (5y + 3x) 1
6.
p q3 3 343
729+
=
pq
33 7
( )9
+ 1
=
pq
79
+
pq pq2
2 7 7( )
9 9
+ − × 1
a3 + b3 = (a + b) (a2 + b2 – ab)
=
pq
79
+
pqp q2 2 49 7
81 9 + −
1
7. a6 – b6 = (a3)2 – (b3)2 1 = (a3 – b3) (a3 + b3) 1 = (a – b) (a2 + b2 + ab) (a + b) (a2 + b2 – ab) 1 = (a – b) (a + b) (a2 + b2 + ab) (a2 + b2 – ab). [CBSE Marking Scheme, 2012]
8. p(x) = 2x3 – 3x2 – 11x + 6 p(–2) = – 16 – 12 + 22 + 6 = 0 ⇒ – 2 is a zero of p(x)
P-19S O L U T I O N S
p(3) = 54 – 27 – 33 + 6 = 0 ⇒3 is a zero of p(x) (x + 2) (x – 3) = x2 – x – 6 is a factor of p(x)
2
( )– – 6p x
x x = 2x – 1
\ p(x) = (x + 2) (x – 3) (2x – 1) 4 9. Factor theorem – statement Let p(x) = x3 – 3x2 – x + 3
The factors of the constant term 3 are ±1, ±3 p(1) = 13 – 3(1)2 – 1 + 3 = 0 (x – 1) is a factor p(–1) = (–1)3 – 3(–1)2 – (–1) + 3 = 0 (x + 1) is a factor p(3) = 33 – 3(3)2 – 3 + 3 = 0 (x – 3) is a factor (x – 1) (x + 1) (x – 3) are the factors of p(x). 4
SUMMATIVE ASSESSMENT WORKSHEET-20Solutions
1. x2 – 3x = x(x – 3) 1 2. 12a2b – 6ab2 = 6ab(2a – b) 1 3. 8a3 + 8b3 = (2a)3 + (2b)3
= (2a + 2b) [(2a)2 + (2b)2 – (2a) × (2b)] 1 [ a3 + b3 = (a + b) (a2 + b2 – ab)] = 2(a + b) × 4(a2 + b2 – ab) 1 = 8(a + b) (a2 + b2 – ab) 4. (x + 3y)3 + (3x – y)3 = [x3 + (3y)3 + 3 × x × 3y (x +
3y)] +[(3x)3 – y3 – 3 × 3x × y (3x – y)]
( ) ( )
a b a b ab a b
and a b a b ab a b
3 3 3
3 3 3
+ = + +3 +
( – ) = – – 3 ( – )
= x3 + 27y3 + 9x2y + 27xy2 + 27x3 – y3 – 27x2y + 9xy2 1 = 28x3 + 26y3 – 18x2y + 36xy2. 1
5. To factorise,
a a2 22 1
5 23 3
+ − − ½
a2 – b2 = (a + b) (a – b) 1½
=
a a
2 15 + +2
3 3 −
a a2 1
5 + 2 +3 3
− 1
=
a
17 +
3
(3a + 1)
[CBSE Marking Scheme, 2012] 6. 9x2 + y2 + z2 – 6xy + 2yz – 6xz = (– 3x)2 + (y)2 + (z)2 + 2 × (– 3x)(y) + 2 × (y) (z) +
2 (– 3x) (z) 1 = (– 3x + y + z)2 1 If x = 1, y = 2, z = – 1, then (– 3x + y + z)2 = (– 3 × 1 + 2 – 1)2 ½
= (– 3 + 2 – 1)2
= 4. ½
7. 125x3 – 27y3 + z3 + 45xyz = (5x)3 + (– 3y)3 + (z)3 – 3 × (5x) (– 3y) (z) 1 = (5x – 3y + z) [(5x)2 + (– 3y)2 + (z)2 – (5x)(– 3y) –
(– 3y) (z) – (5x) (z)] 1 a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca ) = (5x – 3y + z) [25x2 + 9y2 + z2 + 15xy + 3yz – 5xz] [CBSE Marking Scheme, 2012] 1
8. We have, x2 – y2 + y2 – z2 + z2 – x2 = 0 (x2 – y2)3 + (y2 – z2)3 +(z2 – x2) = 3(x2 – y2) (y2 – z2)
(z2 – x2) a + b + c = 0 ⇒a3 + b3 + c3 = 3abc (x2 – y2)3 + (y2 + z2)3 + (z2 – x2)3
= 3(x + y) (x – y) (y + z) (y + z) (z – x) (z + x) Similarly, x – y + y – z + z – x = 0 (x – y)3 + (y – z)3 + (z – x)3 = 3(x – y) (y – z) (z – x)
\
2 2 3 2 2 3 2 2 3
3 3 3( – ) ( – ) ( – )
( – ) ( – ) ( – )+ ++ +
x y y z z xx y y x z x
=
3( )( – )( – )( )( )( – )3( – )( – )( – )
+ + +x y x y y z z x y z z xx y y z z x
= (x + y) (y + z) (z + x) 4 9. Let p = m(n2 – p2) + n(p2 – m2 + p(m2 – n2) p(m = n) = n(n2 – p2) + n(p2 – n2) + p(n2 – n2) = n(n2 – p2) – n(n2 – p2) + 0 = 0 m – n is a factor of p Similarly p(n = p) = 0 & p(p = m) n – p is a factor of p. and p – m is a factor of p. 4
SUMMATIVE ASSESSMENT WORKSHEET-21Solutions
1. x – 2 is a factor of p(x), then p(2) = 0 p(2) = 2 × (2)2 + 3 × 2 – k = 0 ⇒ 8 + 6 – k = 0
\ k = 14. 1
2. Since,
1–
3f
= 0
P-20 M A T H E M A T I C S - I X T E R M - 1
\
1–
3 is a zero of polynomial f(x)
So,
x
1+
3 or 3x + 1 is a factor of f(x). 1
3. 64a3 – 27b3 – 144a2b + 108ab2
= (4a)3 – (3b)3 – 3 × (4a)2 × (3b) + 3 × (4a) × (3b)2 ½ = (4a)3 – (3b)3 – 3 × 4a × 3b (4a – 3b) ½ = (4a – 3b)3 1 4. a9 + b9 + 3a6b3 + 3a3b6
= (a3)3 + (b3)3 + 3 × (a3)2 (b3) + 3 (a3) × (b3)2 ½ = (a3)3 + (b3)3 + 3 × a3 b3 × (a3 + b3) ½ = (a3 + b3)3 1 5. Let, p(x) = x6 – ax5 + x4 – ax3 + 3x – a + 2 (x – a) is a factor of the polynomial p(x), then p(a) = 0 1 ⇒ a6 – a × a5 + a4 – a × a3 + 3 × a – a + 2 = 0 1 ⇒ a6 – a6 + a4 – a4 + 3a – a + 2 = 0 ½ ⇒ 2a = – 2 ⇒ a = – 1. ½
6. a7 + ab6 = a(a6 + b6) 1 = a[(a2)3 + (b2)3] = a(a2 + b2) [(a2)2 + (b2)2 – a2 × b2] 1 = a(a2 + b2) (a4 + b4 – a2b2). 1 [CBSE Marking Scheme, 2012]
7. Let, x2 – 4x = a \ Expression = a (a – 1) – 20 ½ = a2 – a – 20 = (a – 5) (a + 4) ½ = (x2 – 4x – 5) (x2 – 4x + 4) = (x – 5) (x + 1) (x – 2)2 2 [CBSE Marking Scheme, 2012]
Alternative Method : (x2 – 4x)(x2 – 4x – 1) – 20 = (x2 – 4x)2 – (x2 – 4x) – 20
½ = (x2 – 4x)2 – 5(x2 – 4x) + 4(x2 – 4x) – 20 ½ = (x2 – 4x)[x2 – 4x – 5] + 4[x2 – 4x – 5] ½ = (x2 – 4x – 5) (x2 – 4x + 4) = (x2 – 5x + x – 5)[(x)2 – 2 × 2 × x + (2)2] ½ = [x(x – 5) + 1(x – 5)][x – 2]2 ½ = (x – 5)(x + 1) (x – 2)(x – 2). ½ 8. p(x) = 3x3 + 5x2 – 11x + 3 p(1) = 3 + 5 – 11 + 3 = 0 \1 is a zero of p(x) p(–3) = – 81 + 45 + 33 + 3 = 0 – 3 is a zero of p(x) (x – 1) (x + 3) = x2 + 2x – 3 is a factor of p(x)
2
( )2 – 3+
p xx x
= 3x – 1, when we divide physically
Hence p(x) = (x – 1) (x + 3) (3x – 1) 4 9. Factor of 60 = (±1, ±2, ±3, ±4, ±5, ±6, ±10, ±12,
±15, ±30, ±60) ½ p(x) = x3 – 12x2 + 47x – 60 p(3) = (3)3 – 12(3)2 + 47(3) – 60 = 27 – 108 + 141 – 60 = 168 – 168 = 0 ½ \ x = 3 is a zero of p(x) or (x – 3) is a factor of p(x) ½ x3 – 12x2 + 47x – 60 = x2(x – 3) – 9x(x – 3) + 20(x – 3) ½ = (x – 3) (x2 – 9x + 20) ½ = (x – 3) (x2 – 5x – 4x + 20) ½ = (x – 3) [x(x – 5) – 4(x – 5)] ½ = (x – 3) (x – 4) (x – 5) ½
SUMMATIVE ASSESSMENT WORKSHEET-22Solutions
1. a3 – 1 = (a – 1) (a2 + 1 + a × 1) Then factors of (a3 – 1) are (a – 1) and (a2 + 1 + a). 1
2. x2 + 5 2x + 12 = x2 + 3 2x + 2 2x + 12
= x
( )x+3 2
+
( )x2 2 +3 2
=
( )( )x x+3 2 +2 2 . 1
Factors are 3 2 and 2 2x x+ + .
3. (x + 2)2 + p2 + 2p(x + 2) = (x + 2)2 + (p)2 + 2 × (x + 2) × p 1 = (x + 2 + p)2, [ a
2 + b2 + 2ab = (a + b)2] 1 4. 8 – 27a3 – 36a + 54a2
= (2)3 – (3a)3 – 18a(2 – 3a) 1
= (2)3 – (3a)3 – 3 × 2 × 3a(2 – 3a) = (2 – 3a)3. 1
5. p(x) = 3x2 – mx – nx If (x – a) is a factor of p(x), then p(a) = 0 1 ⇒ 3(a)2 – m × a – n × a = 0 ½ ⇒ a[3a – m – n] = 0, a ≠ 0 ½ ⇒ 3a – m – n = 0 ½
\ a =
m n3+
. ½
[CBSE Marking Scheme, 2012] 6. 250x3 – 432y3 = 2[125x3 – 216y3] ½ = 2[(5x)3 – (6y)3] ½ = 2(5x – 6y) [(5x)2 + (6y)2 + 5x × 6y]
1½ a3 – b3 = (a – b) (a2 + b2 + ab)
P-21S O L U T I O N S
= 2(5x – 6y) (25x2 + 36y2 + 30xy). ½
7. Let, p(x) = x3 – 3x2 – 9x – 5, p (– 1) = – 1 – 3 × 9 – 5 = 0 Since, (x + 1) is a factor of x3 – 3x2 – 9x – 5 ½ \ (x3 – 3x2 – 9x – 5) = (x + 1) (x2 – 4x – 5) 1 x2 – 4x – 5 = x2 – 5x + x – 5 = x(x – 5) + 1(x + 5) = (x + 1) (x – 5) 1 Factors are : (x + 1) (x + 1) (x – 5) ½ [CBSE Marking Scheme, 2012]
8. p(x) = 2x4 – ax3 + 4x2 – x + 2
If (2x + 1) is a factor of p(x), then 2x + 1 = 0, x =
–12
is a zero of the polynomial p(x) 1
So,
p
–12
= 0
½
p
–12
= 2 ×
4–12
– a
3–12
+
2–14
2
–
–12
2 = 0 1
⇒
2 ×
116
– a
–18
+
14
4
–
–12
+ 2 = 0 ½
⇒
18
+ a8
+ 1 +
12
+ 2 = 0 ½
⇒
298
+ 8
a
= 0
\ a = – 29 ½ 9. (x – 3) is a factor of p(x) = 2x4 + 3x3 – 26x2 – 5x + 6,
then it completely divide p(x) 2x3 + 9x2 + x – 2 (quotient) 1 x – 3) 2x4 + 3x3 – 26x2 – 5x + 6 2x4 – 6x3
– + 9x3 – 26x2
9x2 – 27x2
– + x2 – 5x x2 – 3x – + – 2x + 6 – 2x + 6 + – 0 2 \ Remainder = 0. So, (x – 3) is a factor of p(x). 1
TOPIC-4Algebraic Identities
SUMMATIVE ASSESSMENT WORKSHEET-23Solutions
1.
x12
+ x
32
+ = x2 +
x
32
+
x12
+
34
= x2 + 2x +
34
. 1
2. Given,
x yy x
+ = – 1
⇒
x yyx
2 2+
= – 1
⇒ x2 + y2 = – xy ⇒ x2 + y2 + xy = 0 ⇒ x3 – y3 = (x – y) (x2 + y2 + xy) = (x – y) × 0 = 0. 1
3. (2x – 3y + z)2
= 4x2 + 9y2 + z2 – 12xy – 6yz + 4xz 2[CBSE Marking Scheme, 2014]
Alternative Method : By using the identity, (a + b + c)2
= a2 + b2 + c2 + 2ab + 2bc + 2ca
(2x + (–y) + z)2 = (2x)2 + (–y)2 + z2 + 2(2x)(–y) + 2(–y) (z) + 2(z)(2x) 1
= 4x2 + y2 + z2 – 4xy – 2yz + 4xz 1
4.
x y
31 23 3
−
=
x
313
–
y32
3
– 3 ×
x
13
×
y23
yx
1 23 3
− 1
=
x3
27 –
y3827
–
xy23
x y23 3
−
=
x3
27 –
y3827
–
x y229
+
xy249
1
5. x + y + z = 0 1 ⇒ x + y = – z ⇒ (x + y)3 = (– z)3
⇒ x3 + y3 + 3xy (–z) = – z3
⇒ x3 + y3 – 3xyz = – z3 1 ⇒ x3 + y3 + z3 = 3xyz 1 6. (2a + 3b)3 – (2a – 3b)3 = x3 – y3 where 2a + 3b =x
and 2a – 3b = y
P-22 M A T H E M A T I C S - I X T E R M - 1
= (x –y) (x2 + xy + y2) = [(2a + 3b) – (2a – 3b)][(2a + 3b)2 + (2a + 3b) (2a – 3b) + (2a – 3b)2] 1 = 6b[(4a2 + 12ab + 9b2) + (4a2 – 9b2) + (4a2 – 12ab + 9b2)] = 6b(12a2 + 9b2) 1 = 6b × 3 × (4a2 + 3b2) = 18b(4a2 + 3b2) 1 7. 1113 = (100 + 11)3 1 = (100)3 + 3(100)2 (11) + 3(100) (11)2 + (11)3
1 = 1367631 1
8. 22
1+x
x =
21–
x
x+ 2 1
= 4 + 2 = 6
22
21 +
xx
=
44
1+x
x + 2 1
⇒ (6)2 =
44
1+x
x + 2
36 – 2 = 4
41
+xx
1
44
1+x
x = 34 1
9. RHS. =
12
(x + y + z)[(x – y)2 + (y – z)2 + (z – x)2]
=
12
(x + y + z) [x2 + y2 – 2xy + y2 + z2 – 2yz + z2 + x2 – 2zx]
=
12
(x + y + z) [2x2 + 2y2 + 2z2 – 2xy – 2yz – 2zx] 2
=
12
(x + y + z). 2[x2 + y2 + z2 – xy – yz – zx]
=
12
x3 + y3 + z3 – 3xyz [By identity] 2
= LHS.
SUMMATIVE ASSESSMENT WORKSHEET-24Solutions
1. a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 –ab – bc – ca) a3 + b3 + c3 – 3abc = 0, as a + b + c = 0 a3 + b3 + c3 = 3abc. 1 2. (x – 2)3 = (x)3 – (2)3 – 3 × x × 2 (x – 2) = x3 – 8 – 6x2 + 12x Coefficient of x2 in the expansion of (x – 2)3 = – 6. 1
3. x2 – 9 = (97)2 – (3)2 1 = (97 + 3) (97 – 3) = 100 × 94 ½
= 9400. ½ [CBSE Marking Scheme, 2012]
4. (x + 2y)2 = x2 + 4y2 + 2 × x × 2y 1 = 17 + 4 × 2 ½ = 17 + 8 = 25
⇒ (x + 2y) = ± 25 = ±5. ½
5. + + −2 2( 2 3) ( 5 2 ) = 2 2( 2 ) ( 3) 2 2+ + ×
2 2 3 ( 5) ( 2 ) 2 5 2× + + − × ×
1
= 2 + 3 + 2 6 + 5 + 2 – 2 10
= + −12 2 6 2 10 1
= + −2(6 6 10 ) 1
6. (x + y + 2z) (x2 + y2 + 4z2 – xy – 2yz – 2zx) = (x + y + 2z) [x2 + y2 + (2z)2 – x × y – y × 2z – x × 2z] 1
We know that (a + b + c) (a2 + b2 + c2 – ab – bc – ca)
= a3 + b3 + c3 – 3abc 1 \ (x + y + 2z) (x2 + y2 + (2z)2 – x × y – y × 2z – x × 2z) = (x)3 + (y)3 + (2z)3 – 3 × x × y × 2z = x3 + y3 + 8z3 – 6xyz. 1 7. Given, a + b + c = 3x or 3x – a – b – c = 0 ½ \ (x – a)3 + (x – b)3 + (x – c)3 – 3(x – a) (x – b) (x – c) = [x – a + x – b + x – c][(x – a)2 + (x – b)2 + (x – c)2
– (x – a)(x – b) – (x – b)(x – c) – (x – a)(x – c)] 1½ = [3x – a – b – c][(x – a)2 + (x – b)2 + (x – c)2 – (x – a)
(x – b) – (x – b)(x – c) – (x – a) (x – c)] ½ = 0 × [(x – a)2 + (x – b)2 + (x – c)2 – (x – a)(x – b)
– (x – b)(x – c) – (x – c)(x – a)] = 0. ½
8.
1x
x+
= 5
On squaring both sides, we get 1
21+x
x
= 52
x2 + x
21+ 2 × x + 1
x = 25
[ (a + b)2 = a2 + b2 + 2ab)] 1
x2 +
2
1x
+ 2 = 25 1
x2 +
21x
= 25 – 2
P-23S O L U T I O N S
x2 +
2
1x
= 23 1
9. (i) 103 × 107 = (100 + 3) (100 + 7) = 1002 + (3 + 7) × 100 + 3 × 7 1 = 10000 + 1000 + 21
= 11021 1 (ii) (102)3 = (100 + 2)3
= (100)3 + 23 + 3 × 100 × 2(100 + 2) = 1000000 + 8 + 600 × 102 = 1000000 + 8 + 61200 1 = 1061208 1
SUMMATIVE ASSESSMENT WORKSHEET-25Solutions
1. x2 +x21
=
x
x
21 + – 2 (x)
1x
= (4)2 – 2 = 16 – 2 = 14. 1 2. Area of rectangle = 25a2 – 35a + 12 = 25a2 – 20a – 15a + 12 1 = 5a(5a – 4) – 3(5a – 4) ½ = (5a – 3) (5a – 4) = length × breadth ½ \ Length and breadth are (5a – 3) and (5a – 4)
respectively. 3. 2x + 3y = 8 ⇒ (2x + 3y)3 = 83 ½ ⇒8x3 + 27y3 + 3 × 2x × 3y (2x + 3y) ⇒ = 512 ½ 8x3 + 27y3 + 18 × 2 × 8 = 512 ½ ⇒ 8x3 + 27y3 = 512 – 288 ⇒ 8x3 + 27y3 = 224. ½ 4. Given, x + y = 5, then x3 + y3 + 15xy – 125 = x3 + y3 – 125 + 15xy = (x)3 + (y)3 + (– 5)3 – 3 × x × y × (– 5) ½ = (x + y – 5) [(x)2 + (y)2 + (– 5)2
– x × (– 5) – y × (– 5) – x × y)] 1 = (5 – 5) (x2 + y2 + 25 + 5x + 5y – xy) ½ = 0 (x2 + y2 + 25 + 5x + 5y – xy) ½ = 0 ½
5.
14,x
x+ =
(given)
On squaring both sides, we get
21x
x +
= (4)2 1
⇒
22 1 1
2x xx x
+ + × × = 16 1
⇒
22
12x
x+ +
= 16
⇒
22
1x
x+
= 14. 1
6. Given, a + b + c = 9 ½ ⇒ (a + b + c)2 = 92 ½ ⇒ a2 + b2 + c2 + 2(ab + bc + ca) = 81 ⇒ [as, (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc+ca)] ½ ⇒ 35 + 2(ab + bc + ca) = 81 ½ ⇒ 2(ab + bc + ca) = 81 – 35 = 46 ½ ⇒ ab + bc + ca = 46/2 ⇒ ab + bc + ca = 23. ½ 7. a – b = 7 and a2 + b2 = 85 ½ ⇒ (a – b)2 = (7)2 1 ⇒ a2 – 2ab + b2 = 49 ⇒ 85 – 2ab = 49 ⇒ 2ab = 36 ⇒ ab = 18 1 Then, a3 – b3 = (a – b) (a2 + b2 + ab) ½ = (7) (85 + 18) = (7) (103)
= 721 1
SUMMATIVE ASSESSMENT WORKSHEET-26Solutions
1.
3 3
2 283 17
83 83 17 17+
− × +
=( )
( )2 2
2 2
(83+17) 83 – 83×17+17
83 – 83×17+17
a3 + b3 = (a+ b) (a2 + b2 – ab)
= 83 + 17 = 100. 1
2. 14 = a, 13 = b, – 27 = c
a + b + c = 14 + 13 – 27 = 0 ½
\ a3 + b3 + c3 = 3abc 1
(14)3 + (13)3 + (– 27)3 = 3 × 14 × 13 × (– 27)
= – 14742. ½
3. 249 × 251 = (250 – 1) (250 + 1) 1 = (250)2 – (1)2 = 62500 – 1 ½ = 62499. ½ [CBSE Marking Scheme, 2012]
P-24 M A T H E M A T I C S - I X T E R M - 1
4. x2 – y2 + xy = + −
− − +
2 23 2 3 23 2 3 2
+ −+ × − +
3 2 3 23 2 3 2
½
=
+ −− +
− +5 2 6 5 2 6
15 2 6 5 2 6
½
= ( ) ( )
( )( )+ − −
+− +
2 25 2 6 5 2 6
15 2 6 5 2 6
½
= + + − − +
+−
25 24 20 6 25 24 20 61
25 24 ½
= +40 6 1
= 40 × 2·4 + 1 = 96 + 1 = 97. 1
5.
( ) ( ) ( )( ) ( ) ( )− + − + −
− + − + −
3 3 32 2 2 2 2 2
3 3 3
a b b c c a
a b b c c a
Both Numerator and Denominator are of the form a3 + b3 + c3 ½
We know that when a + b + c = 0 then a3 + b3 + c3 = 3abc ½ For Numerator, a2 – b2 + b2 – c2 + c2 – a2 = 0 For Denominator , a – b + b – c + c – a = 0
\
( ) ( ) ( )( ) ( ) ( )− + − + −
− + − + −
3 3 32 2 2 2 2 2
3 3 3
a b b c c a
a b b c c a
=
( ) ( ) ( )( )
× − − −
− − −
2 2 2 2 2 23
3 ( ) ( )
a b b c c a
a b b c b c 1
=
− + − + − +− − −
( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( )
a b a b b c b c c a c aa b b c c a ½
= (a + b) (b + c) (c + a). ½[CBSE Marking Scheme, 2012]
6. (i) (998)3 = (1000 – 2)3
We know that
(a – b)3 = a3 – b3 – 3ab(a – b) 1
Hence, (998)3 = (1000)3 – (2)3 – (3) (1000) (2) (1000 – 2)
= 1000000000 – 8 – 6000 × 998
= 1000000000 – 5988008
= 994011992. 1
(ii) Algebraic Identities ½
(iii) Satisfaction solves the identity crisis among people. ½
7. Expansion of (x + y + z)2
= x2 + y2 + z2 + 2(xy + yz + zx) 1
⇒ 100 = 40 + 2(xy + yz + zx) 1 ⇒ (xy + yz + zx) = 30 1 x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy
– yz – zx) 1 = 10(40 – 30) = 100
[CBSE Marking Scheme, 2012] Alternative Method :
(x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx) 1
⇒ (10)2 = 40 + 2(xy + yz + zx)
⇒ 100 – 40 = 2(xy + yz + zx) 1
⇒
xy + yz + zx = 60
302
= ½
and x3 + y3 + z3 – 3xyz
= (x + y + z) [x2 + y2 + z2 – (xy + yz + zx)] 1
= 10 [40 – 30]
= 10 × 10 = 100. ½
8. x3 – 8y3 – 36xy – 216
= (x)3 + (– 2y)3 + (– 6)3 – 3(x)(– 2y)(– 6) 1
= [x + (– 2y) + (– 6)][x2 + (– 2y)2 + (– 6)2 – (x)(– 2y) – (– 2y)(– 6) – (x)(– 6)] 1
= (x – 2y – 6) (x2 + 4y2 + 36 + 2xy – 12y + 6x) ½
= 0 × [x2 + 4y2 + 36 + 2xy – 12y + 6x],
( x = 2y + 6 or x – 2y – 6 = 0) 1
\ x3 – 8y3 – 36xy – 216 = 0. ½
FORMATIVE ASSESSMENT WORKSHEET-27 Note : Students should do this activity themselves.
qqq
P-25S O L U T I O N S
TOPIC-1Euclid’s Geometry
SUMMATIVE ASSESSMENT WORKSHEET-28Solutions
1. A system of axioms is called consistent, when it is impossible to deduce from these axioms, a statement that contradicts any axiom or previously proved statement. 1
2. AB + BC = AC ⇒ x + 3 + 2x = 4x – 5 ⇒ 3x + 3 = 4x – 5 ⇒ 4x – 3x = 3 + 5 \ x = 8 1 3. Theorem requires a proof. 1 4. Let, First thing = x Second thing = y then, x = 2y 1
5. A B C D \ AC = BD (given) ⇒ AB + BC = BC + CD 1 ⇒ AB = CD. 1
[CBSE Marking Scheme I, 2012 , 2014]
6. Euclid’s axioms (1) Things which are equal to the same thing are
equal to one another. (2) If equals are added to equals, the wholes are
equal. 1 + 1[CBSE Marking Scheme, 2012, 2014]
7. AB = CD (Given) ⇒ AB + BC = BC + CD ½+½ ⇒ AC = BD Euclid’s axiom used : If equals are added to
equals, the wholes are equal. 1 [CBSE Marking Scheme I, 2012, 2014]
8. AC = DC (Given) CB = CE Adding, AC + CB = DC + CE ½ AB = DE ½ If equals are added to equals, the wholes are
equal. 1 [CBSE Marking Scheme, 2012]
9. Given, AC = BC
A C B 1 So, AC + AC = AC + BC 1 (Equals are added to equals) ⇒ 2AC = AB, (\ AC + CB coincides with AB)
\
AC =
12
AB. 1
10. Given, C is the mid-point of AB. \ AC = CB …(1) 1 If possible, let D be the another mid-point of AB. \ AD = DB …(2) 1
A C BD Now, subtracting (1) from (2), we get AD – AC = DB – CB ⇒ – CD = CD ⇒ 2CD = 0 ⇒ CD = 0 \ C and D coincide. Hence, every line segment has one and only one
mid-point. 1 11. Concermed, Caring. Things equal to same things
are equal to one another. Rehman contributed ` 500. All right angles are, equal to one another (OR) Any
1 postulate of Euclid can be stated.
SUMMATIVE ASSESSMENT WORKSHEET-29Solutions
1. A surface is that which has length and breadth. 1
y
xSurface breadth ( )ylength ( )x
INTRODUCTION TO EUCLID’S GEOMETRY
SECTION
BSECTIONCHAPTER
3
P-26 M A T H E M A T I C S - I X T E R M - 1
2. Dimension of Surface = Length and Breadth (which is 2). 1
3. Lines are parallel if they do not intersect on being extended.
For example :
A
B
or
A B Lines A and B are parallel lines.
1
4.
D
CB
A
AB = AD AC = AD AB = AC 1 Things which are equal to the same thing are
equal to one another. 1 [CBSE Marking Scheme I, 2012]
5. Euclid’s axiom : If C be the mid-point of a line segment AB, then AC = AB.
AC =
12
AB ½
and
AD =
12
AC
½
⇒
AD =
1 12 2
AB
½
⇒
AD =
14
AB. ½
[CBSE Marking Scheme, 2012] 6. Given : Three lines l, m and n in a plane such that
l || m and m || n. 1
l
m
n
To prove : l || n. Proof : If possible, let l be not parallel to n, then l and
n should intersect in a unique point, say A. Thus, through a point A, outside m, there are two
lines l and n, both parallel to m. 1 This contradicts the parallel line axiom. So, our assumption is wrong. Hence, l || n. 1 7. (i) If a straight line l falls on two straight lines m
and n such that the sum of the interior angles on one side of l is equals to two right angles then by Euclid’s fifth postulate the lines will not meet on this side of l. Next, we know that the sum of the interior angles on the other side of line l will also be two right angles. Therefore, they will not meet on the other side also. So, the lines m and n never meet and are therefore, parallel. 2
(ii) Introduction to Euclid’s Geometry. ½ (iii) Universal truth. ½ 8. (i) The terms need to be defined are : Polygon : A simple closed figure made up of three
or more line segments. 1
Line segment : Part of a line with two end points.
Line : Undefined term.
Point : Undefined term. Angle : A figure formed by two rays with a common
initial point.
Ray : Part of a line with one end point.
Right angle : Angle whose measure is 90°. Undefined terms used are : Line, Point. 1 Euclid’s fourth postulate says that ‘‘all right angles
are equal to one another.’’ In a square, all angles are right angles, therefore, all
angles are equal (From Euclid’s fourth postulate). Three line segments are equal to fourth line
segment. (Given) Therefore, all the four sides of a square are equal (by
Euclid’s first axiom ‘‘things which are equal to the same thing are equal to one another.’’) 1
(ii) Introduction to Euclid’s geometry. ½ (iii) Equality leads to democracy. ½
SUMMATIVE ASSESSMENT WORKSHEET-30Solutions
1. ∠1 = ∠2 ...(1) ∠3 = ∠4 ...(2) Adding (1) + (2), we get
∠1 + ∠3 = ∠2 + ∠4 ½ ⇒ ∠ABC = ∠DBC ½ Euclid’s axiom used : If equals are added to
equals, wholes are equal. 1[CBSE Marking Scheme, 2012]
P-27S O L U T I O N S
2. In a circle having centre at P, we have PR = PQ = radius ½ In a circle having centre at Q, we have QR = QP = radius ½ Euclid’s first axiom : Things which are equal to the
same thing are equal to one another. ½ \ PR = PQ = QR. ½
3. x – 15 = 25 x – 15 + 15 = 25 + 15 ⇒ x = 40 1 If equals are added to equals, the wholes are
equal. 1 [CBSE Marking Scheme, 2012]
4. Here, OX =
12
XY, PX =
12
XZ 1
⇒ XY = 2(OX), XZ = 2(PX) Also, OX = PX, (Given)1 XY = XZ, (Because things which are double of the same
things are equal to one another.) 1 5. Here, ∠3 = ∠4 and ∠1 = ∠3 and ∠2 = ∠4. Euclid’s
first axiom says, the things which are equal to same things are equal to one another. 2
1
34
2
So, ∠1 = ∠2. 1
TOPIC-2Euclid’s Postulates
SUMMATIVE ASSESSMENT WORKSHEET-31Solutions
1. If a straight line falling on two straight lines makes the interior angles on the same side of it taken together less than two right angles, then the two straight lines, if produced indefinitely, meet on that side on which the sum of angles is less than two right angles. 1
2. Meeting place of two walls. 1
3. Only one line passes through two distinct points.
A
B
1
4. Playfair’s Axiom (statement) : For every line l and for every point P not lying on l, there exists a unique line m passing through P and parallel to l. It is equivalent to Euclid’s fifth postulate. 2
5. (a) Infinite, if they are collinear.
(b) Only one, if they are non-collinear. 1 + 1
6. AB = 2AE
(E is the mid-point of AB) ½
CD = 2DF
(F is the mid-point of CD) ½
Also, AE = DF (Given)
Therefore, AB = CD (Things which are double of the same thing are equal to one another) 1
7. There are two undefined terms, line and point. They are consistent, because they deal with two different situations.
(i) Says that given two points be A and B, there is a point C, lying on the line which is in between them. 1
(ii) Says that given A and B, we can take C not lying on the line passing through A and B.
These ‘Postulates’ do not follow from Euclid’s postulates.
However, (ii) follow from given postulate (i). 1
8. AB = BC (given) BX = BY (given) 1 If equals are subtracted from equals, then remains
are also equal. AB – BX = BC – BY ⇒ AX = CY 2
9. Since x + y = 10 and x = z, therefore, x + y = z + y 1
⇒ 10 = z + y 1
Hence, z + y = 10. 1
10. Their sales in July will also be equal as things which are double of the same things are equal to one another.
Two other axioms are :
(i) The whole is greater than the Part. (ii) Things which are halves of the same thing are equal
to one another. 4
P-28 M A T H E M A T I C S - I X T E R M - 1
SUMMATIVE ASSESSMENT WORKSHEET-32Solutions
1. Two planes intersect each other to form a straight line. 1
2. M NC P
MN = MC + CN 1
3. A BX Y
Let AB has 2 mid-points, say X and Y,
then,
2AB
= AX and
2AB
= AY ½ + ½
\ AX = AY ½ ⇒X and Y coincides ½ [CBSE Marking Scheme 2013, 12, 11]
4. Let AB be perpendicular to a line l and AP is any other line segment.
In right ∆ABP, ∠B > ∠P, (\ ∠B = 90°) 1 ⇒ AP > AB or AB < AP. 1
5. Given, ∠ABC = ∠ACB ⇒ ∠1 + ∠4 = ∠2 + ∠3 1 ⇒ ∠1 + ∠4 – ∠4 = ∠2 + ∠3 – ∠3 (As, ∠3 = ∠4) 1 ⇒ ∠1 = ∠2. 1 6. Since ∠1 = ∠3 and ∠2 = ∠4, therefore adding both
equations ∠1 + ∠2 = ∠3 + ∠4 1 ⇒ ∠BAD = ∠BCD 1 ⇒ ∠A = ∠C. 1 7. Axiom : If equals be subtracted from equals, the
remainders are equal. Two more axiom are : (i) Things which are halves of the same thing are equal
to one other. (ii) The whole is greater than the part OR any of Eulid’s
Axioms can be stated. 4
FORMATIVE ASSESSMENT WORKSHEET-33Note : Students should do this activity themselves.
qqq
P-29S O L U T I O N S
TOPIC-1Different Types of Angles
SUMMATIVE ASSESSMENT WORKSHEET-34Solutions
1. x + 20° + 2x – 20° + 60° = 180° (\Straight line makes an angle of 180°) ⇒ 3x = 180° – 60° = 120° ⇒ x = 40° Thus, ∠COD = 2x – 20° = 80° – 20° = 60°. 1 2. Let the angle be x, then Angle x = Complement of x ⇒ x = 90° – x ⇒ x = 45°. 1 3. Complementary angle of 65° = 90° – 65° = 25°. 1 (As sum of complemetary angles is 90°) 4. (3x – 15°) + (x + 5°) = 90° ⇒ x = 25° Angles are 60° and 30°. 2
5. Let, ∠a = 2x and ∠b = 3x ½ Then, ∠a + ∠b = 90° ⇒ 2x + 3x = 90° \ x = 18° ½ a = 2 × 18° = 36° b = 3 × 18° = 54° \ c = 180° – b = 180° – 54° = 126°. 1 [CBSE Marking Scheme, 2012]
6. OP ⊥ AB ⇒ ∠POA = 90° ½ Let, ∠POQ = a \ ∠QOR = 3a ∠ROA = 5a ½ ⇒ a + 3a + 5a = 90° ⇒ 9a = 90° ½ ⇒ a = 10° \ x = 10° ½ and y = 3 × 10° = 30° ½ z = 5 × 10° = 50°. ½ [CBSE Marking Scheme, 2012, 2014]
7. 4b + 75° + b = 180° (Linear pair) 1 ⇒ 5b = 180° – 75° = 105°
⇒
b =
1055
°
= 21°
4b = 4 × 21° = 84° 1 ⇒ a = 4b = 84° (V.O.A.) 1 2c + a = 180° (Linear pair) ⇒ 2c = 180 – 84° = 96°
⇒ c = 48°. 1
SUMMATIVE ASSESSMENT WORKSHEET-35Solutions 1. Angles (30° – a) and (125° + 2a) are supplementary
of each other, then 30° – a + 125° + 2a = 180° ⇒ a = 180° – 155° = 25°. 1 2. The complement of (90° – a) = 90° – (90° – a) (As sum of complementary angles is 90°) = 90° – 90° + a = a. 1 3. Let the angle be x, then
By given condition, x =
15
(90° –x) ⇒ 6x = 90°
⇒ x = 15°. 1
4. ∠POC = 2y (Vertically opp. angles)
\ 5y + 2y + 5y = 180° 1 ⇒ 12y = 180° ½ ⇒ y = 15°. ½ [CBSE Marking Scheme, 2012]
5. x + y + z + w = 360° 1 ⇒ (x + y) + (z + w) = 360° ⇒ x + y + x + y = 360°
(Since, z + w = x + y) ½ ⇒ 2(x + y) = 360 °
LINES AND ANGLES
SECTION
BSECTIONCHAPTER
4
P-30 M A T H E M A T I C S - I X T E R M - 1
\ x + y = 180° ⇒ AOB is a straight line, as straight line makes an
angle of 180°. ½ [CBSE Marking Scheme, 2012]
6. Let, ∠AOB = 2x and ∠BOC = 3x ∠AOB + ∠BOC = ∠AOC ⇒ 2x + 3x = 75°
⇒
x =
755
°
= 15°
\ ∠AOB = 2x = 30° 1 and ∠BOC = 3x = 45°. 1 [CBSE Marking Scheme, 2012]
7. Join AC, 1D
C3
1
2
A B
4
In ∆ABC Ext. ∠4 = ∠ACB + ∠CAB ...(i) Again, in ∆ACD Ext. ∠3 = ∠DAC + ∠DCA ...(ii) ½ Adding (i) and (i), we get ∠3 + ∠4 = (∠ACB + ∠DCA)
+ (∠CAB + ∠DAC) 1 = ∠1 + ∠2 \ ∠3 + ∠4 = ∠1 + ∠2. ½ [CBSE Marking Scheme, 2012]
SUMMATIVE ASSESSMENT WORKSHEET-36Solutions 1. Angle (55° + 3a) and (115° – 2a) are supplement of
each other, then 55° + 3a + 115° – 2a = 180° ⇒ a = 180° – 170° = 10°. 1 2. Here, x + 10° + x + x + 20° = 180° ⇒ 3x = 180° – 30° = 150°
\
x =
1503
°
= 50°. 1
3. Here, 5x + 4x = 180° (\Straight line makes an angle of 180°)
⇒ x =
1809
°
= 20°.
1
4. Given, a – b =
16
×180° ½
⇒ a – b = 30° ...(1) a + b = 180° (Linear Pair) ...(2) ½ Adding (1) & (2) 2a = 210° ⇒ a = 105° ½ ⇒ b = 180° – a = 180° – 105° = 75°. ½ [CBSE Marking Scheme, 2012]
5. Let the two supplementary angles are 2x and 3x, then
2x + 3x = 180°
⇒
x =
1805
°
= 36°
1 Hence, the angles are 2x and 3x or 72° and 108°. 1 [CBSE Marking Scheme, 2012]
6.
A C
QP O
BD
12
34
½
OP and OQ are bisectors of ∠AOD and ∠BOC. 1 \ ∠1 = ∠2 and ∠3 = ∠4 ∠AOC = ∠BOD (Vertically opp. angles) \ ∠1 + ∠AOC + ∠3 = ∠2 + ∠BOD + ∠4 But ∠1 + ∠AOC + ∠3 + ∠2 + ∠BOD + ∠4 = 360° 1 \ ∠1 + ∠AOC + ∠3 = ∠2 + ∠BOD + ∠4 = 180° ½ \ OP and OQ are in the same line.
7. Given,
∠POR : ∠ROQ = 5 : 7
Let, ∠POR = 5x
and ∠ROQ = 7x
5x + 7x = 180°
⇒ 12x = 180°
⇒
x =
18012
= 15° 2
∠POR = d = b = 5x = 5 × 15° = 75°
∠ROQ = c = a = 7x = 7 × 15°
= 105°. 1
[CBSE Marking Scheme, 2012]
P-31S O L U T I O N S
TOPIC-2Transversal Line
SUMMATIVE ASSESSMENT WORKSHEET-37Solutions
1. Let, l || AB || CD (by construction)A B
120°
C D
140° ExF
O
O'
I
then, ∠OEF + ∠BOE = 180° ∠O´EF + ∠DOE = 180° [Corresponding interior angles] ∠OEF + ∠O´EF + 120° + 140° = 360° ⇒ ∠OEO´ = 360° – 260° \ x = 100°. 1 2. ∠3 = ∠1 = 60° (corr ∠S) ∠2 + ∠3 = 180° ∠2 + 60° = 180° (∠3 = 60°) ∠2 = 120° = 2 × 60° = 2 ∠1 2 3.
A BD E
100°140°
C21l
Draw l|| AB \ l|| DE as AB || DE
∠1 + 140° = 180 (Co. int. angles) \ ∠1 = 40° ½ Similarly, ∠2 = 80°. ½ ∠1 + ∠BCD + ∠2 = 180 (St. angle) ½ 40° + ∠BCD + 80° = 180°. \ ∠BCD = 60°. ½ [CBSE Marking Scheme, 2012, 2013]
4.
O
D
B
FE
C
5 °y
A5 °y
2 °y
∠COD = 5y° (VOA) 5y + 5y + 2y = 180° ⇒ 12y = 180° ⇒ y = 15° 3
5.
O
QR
SP
∠POR + ∠ROQ = 180° (linear pair) 2 ⇒ 5x + 7x = 180°
⇒ x = 15 \ ∠POR = 75° = ∠QOS, ∠ROQ = 105° = ∠POS 2
SUMMATIVE ASSESSMENT WORKSHEET-38Solutions
1.
m
n
1
2x
∠1 = ∠x = 75° (Alternate angles) ½ ∠2 = 180° – x = 180 – 75° = 105° 1 ∠2 = 105° = 75° + 30°
= 75° + 1
903
× ° . ½
[CBSE Marking Scheme, 2012]
2.
x
yy
l3 l4
l1
l221
x = ∠1 (Corresponding angles) ½ ∠2 = ∠1 (Corresponding angles) ½ ∠2 + 2y = 180° ⇒ x + 2y = 180° ½
P-32 M A T H E M A T I C S - I X T E R M - 1
⇒ 2y = 180° – x
\
y = 90° – 2
x
½
3.
A
D
BC
Oy°
x°
w°
z°
x + y + w + z = 360° ⇒ 2(x + y) = 360° ( x + y = w + z) ⇒ x + y = 180° \ AOB is a straight line. 3
4.
B
O
C
A
E
D
84°
2 °x
z°
y°75°
84° + 2x° = 180° (linear pair) ⇒ 2x = 96° 2 ⇒ x = 48° y + 75 = 2x (VOA) ⇒ y = 2 × 48° – 75° = 96° – 75° = 21°
z = 84° (VOA) 2
SUMMATIVE ASSESSMENT WORKSHEET-39Solutions
1. ∠PQS + ∠QSF = 180° (Angles on the same side of transversal) ⇒ ∠PQS + ∠RFE = 180°, as ∠QSF = ∠EFR
⇒ 60° + ∠RFE = 180° (Corresponding ∠S)
\ ∠RFE = 120°. 1 2. ∠70° + ∠PRS = 180° ⇒ ∠PRS = 110° = ∠TRO (Vertically opposite angles) In ∆TRO, x + 110° + 20° = 180° ⇒ x = 50°. 1
3.
PL
Q
M N ∠M = ∠N (Given) Exterior ∠NLQ = ∠M + ∠N = 2∠N 1 2∠NLP = 2∠N ⇒ ∠NLP = ∠N (Alternate interior angles) \ LP||MN. 1 [CBSE Marking Scheme, 2012]
4. EG is the bisector of ∠AEF \ ∠AEG = ∠GEF = a Similarly, ∠EFH = ∠HFD = b 1
\ ∠GEF = ∠EFH ( a = b) But these are alternate interior angles \ EG || FH ½ Again, ∠AEF = 2a and ∠EFD = 2b \ ∠AEF = ∠EFD = 2a or 2b 1 But these are alternate angles. \ AB || CD. ½
5. AB ||DC ∠CDB = ∠ABD = x = 35° [alternate angles] 1 x + y + 80° = 180° 1 ∠ADB = y = 65° [angle sum property] 1 ∠DCB = z = 180° – [35° + 35°] = 110° 1 [CBSE Marking Scheme, 2012]
Alternative method : AB || DC y + 35° + 80° = 180° (Corresponding interior angles) 1 ⇒ y = 180° – 115° ⇒ y = 65° ½ ∠ABD = ∠CDB ⇒ x = 35°
(Alternate angles) 1 In ∆BCD, 35° + y – 30° + z = 180° 1 ⇒ z = 180° – 5° – y ⇒ z = 175 – 65° ⇒ z = 110°. ½
P-33S O L U T I O N S
TOPIC-3Angle Sum Property of a Triangle
SUMMATIVE ASSESSMENT WORKSHEET-40Solutions
1. We know that x + 60° = 100° (Exterior angle is the sum of the two interior opposite angles) \ x = 40°. 1 2. Since sum of all the exterior angles formed by
producing the sides of a polygon is 360°. \ x° + y° + z° = 360°. 1 3. ∠ACD = ∠A + ∠B 1 [Exterior angle is the sum of
the two interior opposite angles]
= 60 + 70
∠ACD = 130° 1
4. Given, ∠A + ∠B = 65° Given, ∠B + ∠C = 140° ∠A + ∠B + ∠B + ∠C = 65° + 140 = 205° ½ But, ∠A + ∠B + ∠C = 180
(Angle sum prop. of ∆) ⇒ 180° + ∠B = 205° 1 ⇒ ∠B = 25° ½ \ ∠C = 140° – 25° = 115° [CBSE Marking Scheme, 2012, 2013]
Alternative Method : We know that, ∠A + ∠B + ∠C = 180° ...(i)
(Angle Sum prop. of ∆) ⇒ 65° + ∠C = 180° 1 ⇒ ∠C = 115° Again by (i), ∠A + 140° = 180° ⇒ ∠A = 40° ½ Again by (i), 40° + ∠B + 115° = 180° (Angle Sum prop. of ∆) ⇒ ∠B = 25°. ½
5.
E
A G B
H125°
x
C F D
135°
1
Draw EH||AB. Then ∠AGE = ∠GEH and ∠CFE = ∠FEH (Alternate int ∠S) x = ∠GEH + ∠FEH = ∠AGE + ∠CFE ...(i) 1 But ∠AGE + 135° = 180°, ∠CFE + 125° = 180 (linear pair) ∠AGE = 45°, ∠CFE = 55° 1 Hence from (i) x = 45° + 55° = 100°.
SUMMATIVE ASSESSMENT WORKSHEET-41Solutions
1. In ∆ABC,
∠A + ∠B + ∠C = 180°
Also, in ∆DEF,
∠D + ∠E + ∠F = 180°
\ ∠A + ∠B + ∠C + ∠D + ∠E + ∠F
= 360° = 4 × 90°
Hence, k = 4. 1
2. x = ∠APQ = 40° (Alternate angles) ½ x + y = 118° ½ (Exterior angle is the sum of
the two opposite interior angles)
⇒ 40° + y = 118° ½ \ y = 118° – 40° = 78°. ½
3. In ∆ABC, ∠A + ∠B + ∠C = 180° 1
(Angle Sum prop. of ∆) ⇒ 2x – 5° + 5x + 5° + 3x + 50° = 180° 10x = 130° \ x = 13° ∠A = 2x – 5° = 21° ∠B = 5x + 5° = 70° ∠C = 3x + 50° = 89° 1 [CBSE Marking Scheme, 2012]
Alternative Method : In ∆ABC, ∠A + ∠B + ∠C = 180° (Angle Sum prop. of ∆) ⇒ 2x – 5° + 5x + 5° + 3x + 50°
P-34 M A T H E M A T I C S - I X T E R M - 1
= 180° (By given values) ⇒ 10x + 50° = 180° ⇒ 10x = 130° ½ ⇒ x = 13° ∠A = 2x – 5° = 26° – 5° = 21° ½ ∠B = 5x + 5° = 65° + 5° = 70° ½ ∠C = 3x + 50° = 39° + 50° = 89° ½ 4. In ∆BDE, ∠B + ∠D + ∠DEB = 180° (Angle sum property of a triangle) ⇒ 40° + x + 90° = 180° ⇒ x = 50° 1
In ∆DCF, ∠D + ∠FCD = ∠AFD 1 (Exterior angle is the sum of the two interior opposite angles) ⇒ 50° + y = 110° ⇒ y = 60° 1
5. (i) ∠TOP =
12∠TOB
½
∠ORS =
12∠ORD
But, ∠TOB = ∠ORD (l || m and corresponding angles) ½ \ ∠TOP = ∠ORS ½ But, they are corresponding angles w.r.t. transversal
TR and lines OP and RS. Hence, OP || RS. ½
T
P BA
C D
S
R
Ol
m
(ii) Lines and angles. ½ (iii) Bisectoring among human beings gives rise to
deterioration in the society.
SUMMATIVE ASSESSMENT WORKSHEET-42Solutions
1. ∠QPR = 75° (Vertically opposite angles) Again, ∠PQR + ∠QPR = 105° (Exterior angle) ⇒ ∠PQR + 75° = 105° ⇒ ∠PQR = 30°. 1
2. A
CB40°
60°
∠C = 180° – (60° + 40°) = 180° – 100° ½ = 80° ⇒ AC is the smallest side ½ Reason : Side opposite to smaller angle is shortest. 1 [CBSE Marking Scheme, 2012]
3.
P Q
L
y
x
38°
75°R T
∠1
Given : PQ ⊥ PR PQ || RL ∠RQT = 38° ∠QTL = 75° To find : x & y ∠1 = ∠y ...(1) [alt. interior angles] Now, In ∆QRT ∠QTL = ∠TQR + ∠QRT 1 [by exterior property of triangles] ⇒ 75° = 38° + ∠1 ⇒ ∠1 = 37 ° ⇒ ∠y = 37° ...(2) [from eq.(1)] 1 Now, In ∆QTR ∠QPR + ∠x + ∠y = 180° [by ASPT] ⇒ 90° + ∠x + 37° = 180° ∠x = 53° [from eq (2)] 1 4. (i) In ∠ABC, ∠BCD = ∠BAC + ∠ABC [Exterior angle equal to the sum of opposite two interior angles] 6x + 2 = 3x + 15 + 2x – 1 6x + 2 = 5x + 14 6x – 5x = 14 – 2 x = 12° 2
P-35S O L U T I O N S
(i) Exterior angle property of a triangle is used in the above problem. 1
(ii) By doing so, students exhibit the importance of water. 1
5. ∠PBC = x + z ½
⇒ 2∠OBC = x + z ∠QCB = x + y ⇒ 2∠OCB = x + y ½
∠BOC + ∠OCB + ∠OBC = 180° (Angle sum prop. of ∆)
⇒ 2∠BOC + 2∠OCB + 2∠OBC = 360° (Multiply by 2 both sides) ⇒ 2∠BOC + x + y + x + z = 360° ⇒ 2∠BOC + 180° + x = 360° ⇒ 2∠BOC = 180° – x
⇒
2∠BOC = 90° –12
x
3
[CBSE Marking Scheme, 2012]
SUMMATIVE ASSESSMENT WORKSHEET-43Solutions
1. In ∆ABC, ∠A + ∠B + ∠C = 180° (By given conditions) ⇒ ∠A + 2∠A + 6∠A = 180° ⇒ 9∠A = 180° ⇒ ∠A = 20°. 1 2. AB = AC \ ∠C = ∠B Then, ∠A + ∠B + ∠C = 180° (Prop. of isosceles ∆) ⇒ 50° + ∠B + ∠B = 180° ⇒ 2∠B = 130° ⇒ ∠B = 65° 1 3. Exterior angle = Sum of opposite two interior angle
½ other angle = 180 – 110 = 70° 110 = 30 + x 1 ⇒ x = 80°. ½
4.
A
B D C
21
45° 55°
∠1 = ∠2 = x, ∠B = 45° ∠A + ∠B + ∠C = 180°, ∠C = 55° (Angle sum prop. of ∆) ⇒ 2x + 45° + 55° = 180° ⇒ 2x = 80° \ x = 40° ∠ADB = ∠2 + ∠C
(Exterior angle is the sum of the two interior opposite angles) 1 = 40° + 55° = 95° ½ Similarly, ∠ADC = ∠1 + ∠B = 45° + 40° = 85°. ½
5.
A
B C
x
y z
12l
To prove : Sum of all the angles of ∆ABC is 180°. 1 Construction : Draw a line l parallel to BC. Proof : Since l||BC, we have ∠2 = ∠y (Alternate angles are equal) ... (i) 1 Similarly, l||BC ∠1 = ∠z (Alternate angles are equal) ...(ii) Also, sum of angles at a point A on line l is 180°. ½ \ ∠2 + ∠x + ∠1 = 180° i.e., ∠y + ∠x + ∠z = 180°
(from (i) and (ii)) \ ∠x + ∠y + ∠z = 180° ½ ⇒ ∠A + ∠B + ∠C = 180° Sum of all angles of a ∆ is 180° 1 \ 5x + 6x + 7x = 180° ⇒ x = 10° Angles are 50°, 60° and 70° respectively.
Hence proved. [CBSE Marking Scheme, 2012]
FORMATIVE ASSESSMENT WORKSHEET-44, 45 & 46 Note : Students should do these activities themselves.
qqq
P-36 M A T H E M A T I C S - I X T E R M - 1
TOPIC-1Criteria for Congruence of Triangles
SUMMATIVE ASSESSMENT WORKSHEET-47Solutions
1. ASA congruence : Two triangles are congruent, if two angles and the included side of one triangle are equal to two angles and the included side of other triangle. 1
2. Since AB = DE, ∠A = ∠D and ∆ABC ≅ ∆DEF by SAS.
Therefore AC = DF. 1A
B C
D
E F
3.
A
B C
D
Join AD.
In ∆ABC and ∆ACD,
AB = AC (Given)
BD = CD (Given)
AD = AD (Common) 1
By using SSS Congruency Rule,
∆ABD ≅ ∆ACD ½
\ ∠ABD = ∠ACD (By c.p.c.t.) ½
4. In ∆DCB,
∠4 + ∠C + ∠2 = 180° ...(1)
(Angle sum prop. of ∆) ½
In ∆ADB,
∠3 + ∠A + ∠1 = 180° ...(2)
(Angle sum prop. of ∆) ½
Adding eq. (1) & (2),
⇒∠4 + ∠3 + ∠C + ∠A + ∠2 + ∠1 = 360°
∠A + ∠C + ∠B + ∠D = 360°
⇒ ∠A + ∠C + ∠B + ∠D = 360° Hence proved. 1
5. Given : D is the mid-point of side AC
To Prove :
BD =
12
AC
Const : Draw, DE || BC 1
Proof : In ∆ABC,
D is the mid–point of AC & ED || BC
\ By mid point theorem,
AE = EB ...(i)
Now, ED || BC (By const)
∠AED = ∠ABC
(Corresponding ∠’s)
∠AED = 90° 1
\ ∠AED = ∠DEB = 90° ...(2)
Now, In ∆ADE & ∆BDE,
AE = EB (from eq. ...(1))
∠AED = ∠DEB (from eq. ...(2))
ED = ED (Common)
\ ∆ADE ≅ ∆BDE (By SAS rule)
\ AD = BD ...(3) (By cpct)
But, AD = DC =
12
BC ...(4) 1
(\ D is the mid–point of side AC)
From eq. (3) & (4),
BD =
12
AC Hence Proved. 1
TRIANGLES
SECTION
BSECTIONCHAPTER
5
P-37S O L U T I O N S
SUMMATIVE ASSESSMENT WORKSHEET-48Solutions
1. OA = OB (Given) OP = OP (Common) ∠AOP = ∠BOP (Given) ∆OAP ≅ ∠OBP (By SAS) 1 2. In ∆ABY and ∆ACX, AB = AC (Given) AY = AX (Given) 1 ∠A = ∠A (Common) \ By SAS, ∆ABY ≅ ∆ACX. Proved. 1 3. In ∆PAB and ∆PDC, PA = PD (Given) (P is the mid-point of AD) AB = CD (Side of a square) ∠PAB = ∠PDC = 90° By R.H.S., ∆PAB ≅ ∆PDC 1½ \ PB = PC (By c.p.c.t.)
(Angles opp. to equal sides are equal) ⇒ ∠PCB = ∠PBC. Proved. ½
4. Proof : In ∆ABD and ∆BAC, AD = BC (Given) BD = AC (Given) AB = AB (Common side) 2 By SSS congruence axiom, ∆ABD ≅ ∆BAC ⇒ ∠ADB = ∠BCA (By c.p.c.t.) 1 and ∠DAB = ∠CBA. (By c.p.c.t.)
5. Given, ∠DCA = ∠ECB Adding ∠DCE to both sides, we get ∠DCA + ∠DCE = ∠ECB + ∠DCE ⇒ ∠ECA = ∠DCB 1 In ∆ACE and ∆BCD, AC = BC
(Given) ∠ECA = ∠DCB
(Proved) ∠EAC = ∠DBC
(Given) \ ∆ACE ≅ ∆BCD (By AAS cong.) 2 \ BD = AE. Proved. 1 [CBSE Marking Scheme, 2012]
SUMMATIVE ASSESSMENT WORKSHEET-49Solutions
1. AD = BC (Given) ∠BAD = ∠ABC (Given) AB = AB (Common) \ ∆DAB ≅ ∆CBA (By SAS) \ ∠BDA = ∠ACB (By c.p.c.t.) 1 2. OA = OB (O is the mid-point of AB) ∠AOC = ∠BOD
(Vertically opposite angles) ½ OC = OD
(O is the mid-point of CD) ½ ∆AOC ≅ ∆BOC (By SAS) ½ ⇒ AC = BD. (By c.p.c.t.) Proved ½ 3. In ∆ADC and ∆ABC, AC is common ∠DAC = ∠BAC ½ ∠DCA = ∠BCA (Given) ½ Hence, ∆ADC ≅ ∆ABC (By AAS rule) ½ ⇒ CD = BC. (By c.p.c.t.) Proved. ½
4. ∠BDC + ∠CDA = 180° (Linear pair) ⇒ ∠BDC + x = 180° ⇒ ∠BDC = 180° – x ½
Similarly, ∠BEA = 180° – y° ½ Since, x = y (Given) \ ∠BDC = ∠BEA ½ ∠B = ∠B (Common) AB = BC 1 \ ∆BAE ≅ BCD (ASA) ½ \ AE = CD (c.p.c.t.) [CBSE Marking Scheme, 2012]
5. Given, AD = BC
∠1 = ∠2
∠3 = ∠4
Let us consider ∆ABC and ∆ABD
AB = AB (Common side) 1
∠1 = ∠2
∠3 = ∠4
So, ∠1 + ∠3 = ∠2 + ∠4 1
∠DAB = ∠CBA
and, AD = BC 1
Then, from the SAS Congruence Rule
∆ABC ≅ ∆ABD 1
\ BD = AC Hence Proved
P-38 M A T H E M A T I C S - I X T E R M - 1
SUMMATIVE ASSESSMENT WORKSHEET-50Solutions
1. Let the angles of triangle are 5x, 3x and 7x, then 5x + 3x + 7x = 180° ⇒ 15x = 180° Thus, x = 12° \ Angles are 60°, 36°, 84° Each angle is less than 90° \ The triangle is an acute angled triangle. 1 2. ∠BDA = ∠ACB = 90° (Given) AD = BC (Given) AB = AB (Common) \ ∆ABD ≅ ∆BAC (By RHS) 1 3. (i) In ∆AOD and ∆BOC, OA = OB (Given) OD = OC (Given) ∠AOD = ∠BOC
(Vertically opposite angles) So, by SAS criteria, ∆AOD ≅ ∆BOC 1 (ii) ∠CBA = ∠DAB (By c.p.c.t.) AD and BC are two lines intersected by AB such that
∠CBA = ∠DAB and they form a pair of alternate angles.
Hence, AD || BC. 1
4.
C
B
A2 D
134
5
P6
∠1 + ∠5 = 180° = ∠2 + ∠6 (linear pair)
⇒ ∠1 = ∠2 ( ∠5 = ∠6) 1 In ∆CAP and ∆BAP, ∠1 = ∠2 (Proved) ∠3 = ∠4
(AD is the bisector of ∠BAC)
AP = AP \ ∆CAP ≅ ∆BAP (By AAS) 1 ⇒ CP = BP. (By c.p.c.t.) Proved. 1 [CBSE Marking Scheme I, 2012]
5.
P
Q R
O
1 Proof : QO is bisector of ∠PQR
∠OQR =
12∠PQR =
12
∠Q
RO is bisector ∠PRQ
\
∠ORQ = 12∠PRQ
12
∠R
In ∠OQR ∠QOR + ∠OQR + ∠ORD = 180° (Angle sum property)
∠QOR +
12∠Q +
12∠R = 180°
\ ∠QOR = 180° –
12
(∠Q + ∠R)
But in ∠PQR ∠P + ∠Q + ∠R = 180° ∠Q + ∠R = 180° – ∠P
∠QOR = 180° –
12
(180° – ∠P) 2
= 180° – 90° +
12∠P
= 90° +
12
∠P Hence Proved.
These type of rallies spread awareness among people for not to kill girl child and helping in equalising sex ratio. 1
SUMMATIVE ASSESSMENT WORKSHEET-51Solutions
1. ∆CBA ≅ ∆PRQ 1 2. Interior opposite angles. ∠BCD = ∠A + B 1
3.
A
E F
B CD
In ∆BED and ∆CFD,
∠DEB = ∠DFC = 90° ½
BD = DC (D is the mid–point)
ED = FD (Given)
\ ∆BED ≅ ∆CFD (By RHS) 1
⇒ ∠B = ∠C. (By c.p.c.t.) Proved. ½
P-39S O L U T I O N S
4. PQRS is a square. (given)
T
RS
QP
x
x
(i) SRT is an equilateral triangle. (given) \ ∠PSR = 90°, ∠TSR = 60° ½ ⇒ ∠PSR + ∠TSR = 150°. Similarly, ∠QRT = 150° In ∆PST and ∆QRT, we have PS = QR (S) ∠PST = ∠QRT = 150° (A) and ST = RT (S) ½ By SAS, ∆PST ≅ ∆QRT ⇒ PT = QT (By c.p.c.t.) Proved. (ii) In ∆TQR, QR = RT (Square and equilateral ∆ on same base) ½ ⇒ ∠TQR = ∠QTR = x ½ \ x + x + ∠QRT = 180° ⇒ 2x + 150° = 180° ⇒ 2x = 30° \ x = 15°. Proved. 1
5. Proof : We are given two triangles ABC and PQR in which
∠B = ∠Q, ∠C = ∠R and BC = QR We need to prove that ∆ABC ≅ ∠PQR 1 There are three cases. Case I : Let AB = PQ In ∆ABC and ∆PQR, ∠B = ∠Q (Given) BC = QR (Given)
AB = PQ (Assumed) \ ∆ABC ≅ ∆PQR (By SAS rule) 1
A
CB
P
RQ
Case II : Suppose AB ≠ PQ and AB < PQ Take a point S on PQ such that QS = AB
A
CB
P
RQ
S
Join RS. In ∆ABC and ∆SQR, AB = SQ (By construction) BC = QR (Given) ∠B = ∠Q (Given) 1 \ ∆ABC ≅ ∆SQR (By SAS rule) ⇒ ∠ACB = ∠QRS (By c.p.c.t.) But, ∠QRP = ∠ACB ⇒ ∠QRP = ∠QRS Which is impossible unless ray RS coincides with
RP. \ AB must be equal to PQ. ½ So, ∆ABC ≅ ∆PQR Case III : If AB > PQ. We can choose a point T on AB such that TB = PQ
and repeating the arguments as given in Case II, we can conclude that AB = PQ and so,
∆ABC ≅ ∆PQR ½ [CBSE Marking Scheme, 2013, 2012]
TOPIC-2Some Properties of Triangles
SUMMATIVE ASSESSMENT WORKSHEET-52Solutions
1. In ∆PEQ and ∆PER, ∠PEQ = ∠PER = 90° ∠QPE = ∠RPE (Given) PE = PE (Common) \ ∆PEQ ≅ ∆PER (by ASA)
\ PQ = PR. (By c.p.c.t.)
P
RQ E90°90°
1
P-40 M A T H E M A T I C S - I X T E R M - 1
2. AB = AD ⇒ ∠ABD = ∠ADB ...(i) ½ BC = CD ⇒ ∠CBD = ∠CDB ...(ii) ½ Adding eqns. (i) and (ii), we get ∠ABD + ∠CBD = ∠ADB + ∠CDB ½ ⇒ ∠ABC = ∠ADC. Proved. ½ [CBSE Marking Scheme, 2011, 2012]
3.
A
CB P In ∆ABP and ∆ACP, AB = AC (Given) ½ AP = AP (Common) ∠APB = ∠APC = 90°, (AP ⊥ BC) ½ By RHS rule, ∆ABP ≅ ∆ACP ½ ⇒ ∠B = ∠C. (By c.p.c.t.) ½
4.
A
B CY
X
1 AB = BC
⇒
12
AB =
12
BC
But
12
AB = BX
\
12
BC = BX (ii) ½
But
12
BC = BY (i) ½
From (i) and (ii) BX = BY 1
5.
B D C
E
A
O
32
5
61 4
2∠3 + 2 ∠4 = 180° – ∠ABC = 90° \ ∠3 + ∠4 = 45° ∠1 = 90° + ∠3 (External angle) ∠2 = 90° + ∠4 (External angle) Adding, ∠1 + ∠2 = 180° + ∠3 + ∠4 = 180° + 45° (External angle) = 225° 4
SUMMATIVE ASSESSMENT WORKSHEET-53Solutions
1. AB = AC ⇒ ∠C = ∠B By angle sum property,
90°BA
C
∠A + ∠B + ∠C = 180° ⇒ 90° + ∠B + ∠B = 180° ⇒ 2B = 90° \ B = 45°. 1 2. In ∆PQS and ∆PRS, PQ = PR (Given) ½ PS = PS (Common) ∠PSQ = ∠PSR = 90° (PS is altitude) ½ By R.H.S. rule,
∆PQS ≅ ∆PRS ½ ⇒ ∠QPS = ∠RPS (By c.p.c.t.) ½ Hence, S bisects ∠P.
P
RQ S
3. From ∆DCF, we have x + y = ∠AFD = 110° ....(1) Also ∠AFE = 180° – 110° = 70° \ ∠FAE = 90° – 70° = 20° From ∆ACB y = ∠FAE + ∠ABC = 20° + 40° = 60° Then (1) gives x = 50°. 3
P-41S O L U T I O N S
4.
D
CB
A
1 23
4
In ∆ABC, AB = AC ⇒ ∠1 = ∠2 ...(1) Angles opp. to equal sides are equal.
In ∆ADC., AB = AD \ AC = AD ½ In ∆BCD, ∠3 = ∠4 ...(2) ½ ∠1 + (∠2 + ∠3) + ∠4 = 180° ½ ⇒ ∠2 + ∠2 + ∠3 + ∠3 = 180° ½ ⇒ 2(∠2 + ∠3) = 180° ⇒ ∠2 + ∠3 = 90° ½ ⇒ ∠BCD is a right angle. ½ [CBSE Marking Scheme, 2014, 12]
SUMMATIVE ASSESSMENT WORKSHEET-54Solutions
1. Proof : In ∆ABF and ∆ACE, AB = AC (Given) 1 ∠A = ∠A (Common) 1 AF =
AE \ ∆ABF ≅ ∆ACE (by SAS cong.) \ By c.p.c.t., BF = CE. 1 [CBSE Marking Scheme, 2012]
Alternative method :
AB = AC ⇒
AB2
=
AC2
½
⇒ AE = AF, Since E and F are the mid-points of AB and AC. In ∆ABF and ∆ACE, AB = AC (Given) ½ ∠A = ∠A (Common) ½ AF = AE (Proved) ½ \ ∆ABF ≅ ∆ACE (By SSA cong.) ½ \ BF = CE. (By c.p.c.t) ½ 2.
A
CB ½ AB = AC ½ \ ∠B = ∠C = x BA = BC \ ∠A = ∠C = x Also, AC = BC ½ ∠A = ∠B = x But, ∠A + ∠B + ∠C = 180° ½ ⇒ 3x = 180° ⇒ x = 60° \ ∠A = ∠B = ∠C = 60°. 1 [CBSE Marking Scheme, 2012]
Alternative Method :
A
CB
Let ∆ABC be an equilateral triangle, so that AB = AC = BC.
Now, AB = AC ⇒ ∠B = ∠C ....(1) ( Angles opp. to equal sides are equal) CB = CA ½ ⇒ ∠A = ∠B ....(2) ( Angles opp. to equal sides are equal) From (1) and (2), we have ∠A = ∠B = ∠C 1 Also, ∠A + ∠B + ∠C = 180°, (Angle sum property) \ ∠A + ∠A + ∠A = 180° ½ ⇒ 3∠A = 180° ⇒ ∠A = 60° \ ∠A = ∠B = ∠C = 60°. Thus, each angle of an equilateral triangle is 60°. 1
3.
A
C D E B We have here ∠CAE = ∠EAB (given) .....(1) Also ∠CAD + ∠DAE = ∠CAE = ∠EAB
[Use (1)].....(2) Now ∠CAD + ∠C = 90° = ∠DAE + ∠EAB + ∠B ∠CAD + ∠C = ∠DAE + ∠CAD + ∠DAE + ∠B [Use (2)] = 2 ∠DAE + ∠CAD + ∠B \ 2 ∠DAE = ∠C – ∠B
P-42 M A T H E M A T I C S - I X T E R M - 1
⇒ ∠DAE = 12
(∠C – ∠B) 3
4. Proof AB = AC ⇒ ∠ABC = ∠ACB and BD = CD ⇒ ∠DBC = ∠DCB ∠ABC – ∠DBC = ∠ACB – ∠DCB ⇒ ∠ABD = ∠ACD 1 ∆ABD ≅ ∆ACD by SAS 1
⇒ ∠BAP = ∠CAP ∆ABP ≅ ∆ACP by SAS 1
\ BP = PC, ∠APB = ∠APC \ ∠APB = ∠ADC = 90° AP is perpendicular to BC AP is perpendicular bisector of BC 1
D
B P C
A
TOPIC-3Inequalities of a Triangle
SUMMATIVE ASSESSMENT WORKSHEET-55Solutions
1. If D is a point on the side BC of a ∆ABC such that AD bisect ∠BAC, then AB > BD. 1
2. In ∆ABC, AC > AB (Given) \ ∠ABC > ∠ACB (Angles opposite to larger side is greater) ½ \ ∠ABC + ∠1 > ∠ACB + ∠1
(Adding ∠1 on both sides) ½ \ ∠ABC + ∠1 > ∠ACB + ∠2
(AD bisects ∠A, ∠1 = ∠2) ½ \ ∠ADC > ∠ADB.
(Exterior angle property of triangle) ½
3. Proof : In ∆AOB, ∠B > ∠A ∠A < ∠B 1 OB < OA ...(1)
In ∆COD, ∠C > ∠D ∠D > ∠C 1 OC < OD ...(2) Adding (1) + (2), OB + OC > OA + OD BC > AD or, AD < BC [CBSE Marking Scheme, 2012]
4. In ∆PQS, PQ + QS > PS ...(1) 1 (Sum of any two sides is greater than the third side) In ∆PSR, PR + SR > PS ...(2) Adding (1) & (2), 1 PQ + QS + PR + SR > 2PS (QR = QS + SR) PQ + QR + RP > 2PS. 1 [CBSE Marking Scheme, 2012]
5. AD = BD ⇒ ∠ABD = ∠DAB = 59° (Angles opp. to equal sides are equal) 1 In ∆ABD, 59° + 59° + ∠ADB = 180° ⇒ ∠ADB = 180° – 118° = 62° 1 and ∠ACD = 62° – 32° = 30° (Exterior angle is equal to the sum
of interior opposite angles) In ∆ABD, AB > BD
(Side opp. to greatest angle is the longest) 1 Also in ∆ABC, AB < AC
⇒ BD < AC. 1
SUMMATIVE ASSESSMENT WORKSHEET-56Solutions
1. No, Because, 2.3 + 3.1 = 5.4 cm (third side) \ Not possible to construct a triangle. 1 2. Since ∠C > ∠A > ∠B, \ AB > BC > AC.
A
B C
70°
30° 80° 1
P-43S O L U T I O N S
3.
P
TS
Q R
Construction : Produce QS to meet PR in T ½
In ∆PQT, PQ + PT > QT
⇒ PQ + PT > QS + ST ...(1) ½
In ∆SRT, TR + ST > SR ...(2) ½
Adding (1) and (2), we get
PQ + PT + TR + ST > QS + ST + SR 1
⇒ PQ + PR > QS + SR
⇒ QS + SR < PQ + PR. ½
[CBSE Marking Scheme, 2012]
4.
A
C B
In ∆ABC, as AB is the greatest side
⇒ AB > BC ⇒ ∠C > ∠A ...(1)
AB > AC ⇒ ∠C > ∠B ...(2) 1
On adding (1) and (2), we get
2∠C > ∠A + ∠B ⇒ 2∠C + ∠C > ∠A + ∠B + ∠C 1
⇒ 3∠C > 180°
\ ∠C > 60°. 1
[CBSE Marking Scheme, 2012] 5. (i) AB and CD intersect at O \ ∠AOD = ∠BOC (Vertically opp. angles) ...(i) In ∆AOD and ∆BOC, we have ½ ∠AOD = ∠BOC From ...(ii)
∠DAO = ∠CBO = 90° (Given) ½ and AD = BC (Given) \ ∆AOD ≅ ∆BOC (By AAS congruence criterion) ⇒ OA = OB (By c.p.c.t.) i.e., O is the mid-point of AB Hence, CD bisects AB. 1 (ii) Congruency of triangles. 1 (iii) Equality is the sign of democracy. 1 6.
A
B CD
E
Construction : Produce AD to E such that AD = DE. Join EC. ½ In triangles ADB and EDC, AD = DE (Const) ½ BD = DC (Given) ∠ADB = ∠EDC (V.O.A) ½ \ ∆ADB ≅ ∆EDC (SAS congruence axiom) ½ ⇒ AB = EC (By c.p.c.t.) In ∆AEC, AC + EC > AE [Triangle Inequality Property] ½ \ AC + AB > AE (EC = AB)
⇒ AC + AB > AD + DE ⇒ AC + AB > AD + AD ( DE = AD) ½
⇒ AC + AB > 2AD. Proved. 1 [CBSE Marking Scheme, 2013, 12]
FORMATIVE ASSESSMENT WORKSHEET-57 & 58 Note : Students should do these activities themselves.
qqq
P-44 M A T H E M A T I C S - I X T E R M - 1
TOPIC-1Cartesian System
SUMMATIVE ASSESSMENT WORKSHEET-59Solutions
1. P (2, – 3) and Q (–3, 2) lie in IV and II quadrants. 1 2. (3, –2) 1 3. In II quadrant, x < 0 Points = (–1, 0), (–2, 0) 1 4. The distance of a point from the y-axis is called its
x-co-ordinate, or abscissa. 1 5. The P is on x-axis \ y = 0 P is at a distance of 4 units from y-axis to its left. \ In second quadrant, the co-ordinates of the point
P = (– 4, 0). 1 6. II Quadrant. 1
7. (A) P(0, 5) ½ (B) Q(0, – 3) ½ (C) R(5, 0) ½ (D) S(–2, 0) ½
y
xx'
y'
654321
–1–2–3–4
Q(0, – 3)
R(5, 0)(–2, 0)O
P(0, 5)
–1 –5 –4 –3 –2 –1 1 2 3 4 5 6
8. (A) II quadrant ½
(B) III quadrant ½ (C) IV quadrant ½ (D) I quadrant. ½
9.
y
y'
x'x
D A
BC–1
–2
–3
(–1, 3) (1, 3)
0–2–3 2
(–1, –1) (1, – 1)
1
3
2
1
–1
3
Co-ordinates of D are (–1, 3). 10. In a point (– 5, 3), x < 0 and y > 0 \Point (– 5, 3) lies in II quadrant. ½ In a point ( 4, – 3), x > 0 and y < 0 \ Point (4, – 3) lies in IV quadrant ½ In a point (5, 0), x > 0 and y = 0 \ Point (5, 0) lies on x-axis ½ In a point (6, 6), x > 0 and y > 0 \ Point (6, 6) lies in I quadrant ½ In a point (– 5, – 4), x < 0 and y < 0 \Point (– 5, – 4) lies in III quadrant 1
SUMMATIVE ASSESSMENT WORKSHEET-60Solutions
1. (x, y) = (y, x) 1 2. 7 1
3. The co-ordinate of x-axis = (x, 0) 1
4. Negative ordinate i.e., (x, –y) 1
COORDINATE GEOMETRY
SECTION
BSECTIONCHAPTER
6
P-45S O L U T I O N S
5. The point on y-axis has x-co-ordinate 0.
Since it lies at a distance of 4 units in the negative direction of y-axis. 1
\ The point is (0, – 4). 1
6. (a) A point which lies on x and y-axes is (0, 0) i.e., origin ½
(b) A point whose abscissa is 5 and ordinate is 6 i.e., x = 5 and y = 6 is (5, 6) ½
(c) A point whose ordinate is 6 i.e., y = 6 and lies on y-axis is (0, 6) ½
(d) A point whose ordinate is 3 and abscissa is – 7 i.e., y = 3 and x = – 7 is (–7, 3) ½
(e) A point whose abscissa is 3 i.e., x = 3 and lies on x-axis is (3, 0) ½
(f) A point whose abscissa is 4 and ordinate is 4 i.e., x = 4 and y = 4 is (4, 4) ½
7. Firstly, we plot all the points i.e.,
A(3, 0), B(5, 0), C(5, 3) and D(3, 3) on a graph paper and join all these points. ½
–4–5 –3 –2 –1 0 1 2 3 4 5 6
5
4
3
2
1
–1
–2
–3
–4
–5
Y
Y'
XX'
D(3, 3) C(5, 3)
B(5, 0)(3, 0)A
1½
The obtained figure ABCD is a rectangle, since AB = CD and BC = DA and all lines are perpendicular to each other. 1
8. (i) E(– 1, 2) (ii) D(– 4, – 1) (iii) Co-ordinates of A = (2, – 2) Co-ordinates of B = (4, 1) \The abscissa of A – abscissa of B = 2 – 4 = – 2 (iv) Co-ordinates of C = (3, 3) Co-ordinates of F = (– 1, – 4) \ The ordinate of C + ordinate of F = 3 + (– 4) = – 1
1 + 1 + ½+½ + ½+½
SUMMATIVE ASSESSMENT WORKSHEET-61Solutions
1. (0, 0) 1 2. Origin 1 3. Point C(3, a – 3) lie on x-axis \ a – 3 = 0
⇒ a = 3. 1
4. (A) II quadrant ½ (B) III quadrant ½ (C) I quadrant ½ (D) II quadrant. ½ 5. 4, 0, 4, –3. 2 6. The vertices of the rectangle OABC are O(0, 0),
A(– 6, 0), B(– 6, – 3), C(0, – 3) 3 7. (i) Draw X´OX and Y´OY as the co-ordinate axes
and mark their point of intersection O as the origin (0, 0).
In order to plot the points (– 2, 8), we take 2 units on OX´ and then 8 units parallel to OY to obtain the point A(– 2, 8).
Similarly, we plot the point B(– 1, 7). ½ In order to plot (0, – 1.25), we take 1.25 units below
the x-axis on the y-axis to obtain C(0, – 1.25). ½
In order to plot (1, 3), we take 1 unit on OX and then 3 units parallel to OY to obtain the point D(1, 3) ½
X' X
Y'
Y
12
45678
–1–2–3–4–5
–1–2–3–4–5 0 1 2 4 5 6 7 8E(3, –1)
(0, –1.25)C
D(1, 3)
A(–2, 8)
B
3
3
(–1, 7)
1 In order to plot (3, – 1), we take 3 units on OX and
then 1 unit below x-axis parallel to OY´ to obtain the point E(3, – 1) ½
(ii) Co-ordinate geometry. ½ (iii) Co-ordination among people is good for
progress. ½
P-46 M A T H E M A T I C S - I X T E R M - 1
TOPIC-2Plotting a Point in a Plane
SUMMATIVE ASSESSMENT WORKSHEET-62Solutions
1.
3
2
1
1
2
3
3 2 1 1 2 3
(0, 0) (3, 0)
(0, 3) (3, 3)
Y
XX'
Y'
Vertices are (0, 0), (3, 0), (3, 3), & (0, 3). 2 2. For plotting a point A(5, –3), we will take a distance
of – 3 units in the negative direction of y-axis and a distance of 5 units in the positive direction of x-axis, which is shown in the figure given below. Similarly, we plot all the points i.e., B(– 6, 0), C(– 2, – 3) and D(– 4, 3). 1
–6 –5 –4 –3 –1 1 2 3 4 6O
54
21
–1–2
–4–5
–2 5
3
–3
Y
Y'
A(5,–3)
XX'
D(–4, 3)
B(–6, 0)
C(–2,–3)
1
3.
–1
–2
–2 1 2
2
1(–1,1)
–1
(0,2)
(1,0)(–1,0)
A Pentagon
Y
Y'
XX'
(1,1)
4. (i) The co-ordinate of B = (–2, 3) 1 (ii) E is the point which is identified by the coordinates
(–3, –2) 1 (iii) The co-ordinate of the point D is (6, 2) 1 \ Abscissa is 6. (iv) The co-ordinate of the point C is (3, –1) 1 \ Ordinate is –1.
SUMMATIVE ASSESSMENT WORKSHEET-63Solutions
1. Square
Y
XX'
Y'
(0, 2) B C (2, 2)
D (2, 0)(0, 0) A
1 2. The co-ordinate of point Q = (– 3, – 3·5) 1 3. (2 – a + b, b) = (6, 2) \ b = 2 and 2 – a + b = 6 ⇒ 2 – a + 2 = 6 ⇒ – a = 6 – 4 = 2 \ a = – 2. 1
4.
–25 –20 –15 –10 –5 5 10 15 20 25O
Y
XX'
Y'
–5
–10
–15
–20
–25
5
10
15
20
25 B(3,23)
A(23,3)
1
After plotting these two points AB line segment is formed.
P-47S O L U T I O N S
Mid-point =
3+23 23+3,
2 2
= (13, 13) 1
5.
–6 –5 –4 –3 1 2 3 4 5 6 7
1098
654321
–1–2–3–4–5–6–7
XX'
Y'
YA(3,10)
C(–1,–6)
–2 –1
7
O
B(–3,5)
After joining the points, a triangle is formed. 2
6.
y
y'
xx'
54321
–l–2–3
–l–2–3–4–5–6 O1 2 3 4 5 6 7 8
B (0, 4)
A (4, 0)
2
From figure, OA = 4 units OB = 4 units
Area of ∆OAB =
12
× OA × OB
=
12
× 4 × 4 = 8 square unit. 1
7. (i) Plotting of points M(5, – 3) and N(–3, –3) on the graph paper is shown in the diagram.
(ii) Length of MN = 3 + 5 = 8 units (iii) From figure, ½ A(3, – 3), B(1, – 3), C(–1, – 3) ½ + ½ + ½
y
x' x
y'
4321 1 2 3 4 5 6 7 8
–6–5–4
–4–5
B A M(5,–3)N(–3,–3)C
–3–2–1 –1–2–3
2
SUMMATIVE ASSESSMENT WORKSHEET-64Solutions 1. Position of thief = (5, 0) or (– 5, 0)
y
y'
x' x
(0, 5) Police
(5, 0)Thief
(–5, 0)Thief
1
2. The given points are : (–2, – 4) = A (2, 4) = B (0, –2) = C (4, – 6) = D The point A will lie in III quadrant. ½ The point B will lie in I quadrant. ½ The point C will lie O in y-axis (i.e., x = 0) ½ The point D will lie in IV quadrant. ½ Verification :
–6 –5 –4 –3 –1 1 2 3 5 6O
Y
XX'
Y'
6
5
4321
–1
–2
–3
–5
–6
B(2, 4)
C(0, –2)
A(–2, –4)
–2 4
–4
After plotting these points on cartesian plane, we find that point A is lying in III quadrant, B is lying in I quadrant, C is in the negative direction of y-axis and D is in IV quadrant.
Thus, result is verified. 3. According to the question, plot the points A(0, 3),
B(5, 3), C(4, 0) and D(–1, 0) on the graph paper.
P-48 M A T H E M A T I C S - I X T E R M - 1
1 2 3 4 5–1–2–3–4–1
–2
–3
–4
1
2
3
4B(5, 3)
0
D
C(4, 0)
A(0, 3)
1 From figure, ABCD is a parallelogram. The point (2,
2) will lie inside the figure. 1
4.
–1
–2
–3
–4
–5
–6
1 2 3 4 5 6
6
5
4
3
2
1
(–2,5)
(–4, 3)
–6 –5 –4 –3 –2 –1
(0,2.5)
(2,5)
(4,–3)
(6,–1)
TOPIC-3Graph of a Linear Equations
SUMMATIVE ASSESSMENT WORKSHEET-65Solutions
1. Given equation 2x + y – 4 = 0 \ y = – 2x + 4 1 2. Given equation, 3x + 6y – 4 = 0 ... Line intersect the x-axis ... y = 0 ½ 3x + 6 × (0) – 4 = 0
⇒ x =
43
So, the point where line intersect the x-axis is
(43
, 0). ½
3. x-axis. 1 4. Given, y = 3x + 6 Put x = 0, we get y = 3 × 0 + 6 = 6 Similarly at x = 1, y = 9 \ Points (0, 6), (1, 9) lies on the line y = 3x + 6. 1 5. The given equation is : y = x Putting x = 0, we get y = 0 ½ Putting x = 1, we get y = 1 ½ Putting x = 2, we get y = 2 ½ Thus, we have the following table :
x 0 1 2 3
y 0 1 2 3
Now, plot the points R(0, 0), P(1, 1) and a(2, 2) on the graph paper.
–4 –3 –2 –1 4321
–4
–3
–2
–1
4
3
2
1
XX'
Y'
Y
a (2, 2)
P(1, 1)(0, 0) R
½ 6. The given equation is : y = 2x + 1 Putting x = 0, we get y = 0 ½ Putting x = 1, we get y = 3 ½ Putting x = 2, we get y = 5 ½ Thus, we have the following table :
x 0 1 2 3
y 1 3 5 7 Now Let plot the points A(0, 1), B(1, 3), C(2, 5) and
D(3, 7) on the graph paper.
P-49S O L U T I O N S
Y
Y'
X' XA(0, 1)
–1–2–3–4–5 1 2 3 4 5 6 7
7654321
–1
–2
–3
–4
D(3, 7)
C(2, 5)
B(1, 3)
½ 7. The given equation is: y = 3x – 2 Putting x = 0, we get : y = (–2) ½ Putting x = 1, we get y = 1 ½ Putting x = 2, we get y = 4 ½ Thus, we have the following table
½
x 0 1 2 3
y –2 1 4 7 Now plot the points A(0, –2), B(1, 1), C(2, 4) and
D(3, 7) on the graph paper. Join A, B, C, D and extend it in both the directions. Then, line AD is the graph of the equation y = 3x – 2.
y
y'
x' x–5 –4 –3 –2 –1 1 2 3 4 5 6
–1
–2
–3
–4
1234567
0
A(0, –2)
B(1, 1)
C(2, 4)
D(3, 7)
1
FORMATIVE ASSESSMENT WORKSHEET-66 & 67 Note : Students should do these activities themselves.
qqq
P-50 M A T H E M A T I C S - I X T E R M - 1
TOPIC-1Area of Triangle
SUMMATIVE ASSESSMENT WORKSHEET-68Solutions
1.
s =
2a b c+ +
=
13 14 152
+ + = 21 cm ½
Area = ( – )( – )( – )s s a s b s c
= 21 (21 13)(21 14)(21 15)× − − −
= 21 8 7 6 21 4 84× × × = × = cm2. ½
2. Area of equilateral triangle =
2316 3
4a× =
a2 = 4 × 16 ⇒ a = 8 cm
Thus, Perimeter = 3a = 3 × 8 = 24 cm. 1
3.
A
x
B C15
25
x =
2 225 15 40 10− = ×
x = 20 cm. ½
Area of right triangle =
12
× base × height
=
12
× 15 × 20 = 150 cm2. ½
4. Perimeter = 4x + 5x + 6x = 150 15x = 150 ⇒ x = 10 Sides are 40 cm, 50 cm and 60 cm. 1
5.
32
a. 1
6. Perimeter of an equilateral triangle = 3a = 60 m
a = 20 m
Area =
34
a2 =
34
× 20 × 20
= 100 3 m2. 1
7.
A
x
B C6
10
x = 2 210 6 8− = cm ½
Area of right triangle =
12
× base × height
=
12
× 6 × 8 = 24 cm2 ½
8.
A
x
B C
5 2
x
x2 + x2 = ( )25 2
2x2 = 25 × 2 x = 5 cm. 1
9. Perimeter of an equilateral triangle = 3a = 60 ⇒ a = 20 cm ½
Area =
23 320 20
4 4a = × × ½
= 100 3 cm2. 1 [CBSE Marking Scheme, 2012]
10. In right angled ∆ ABC, A
B Ca
a
AREAS
SECTION
BSECTIONCHAPTER
7
P-51S O L U T I O N S
By Pythagoras theorem, AC2 = AB2+ BC2
=a2 + a2 = 2a2 1
⇒ AC = 2 a ½ Perimeter of ∆ABC = AB + BC + CA
= a + a + 2 a
= 2a + 2 a
= a (2 + 2 ) 1½
SUMMATIVE ASSESSMENT WORKSHEET-69Solutions
1.
12 2
xSquare
x
x2 + x2 = ( )212 2
⇒ 2x2 = 144×2 \ x = 12 cm Given, 3a = 4x
(a, be a side of triangle) ⇒ 3a = 4 ×12 \ a = 16 cm
Area of equilateral triangle =
23 316 16
4 4a = × ×
= 64 3 cm2. 1
2. Sides are 3x, 4x, 5x, then 3x + 4x + 5x = 36 ⇒ 12x = 36 ⇒ x = 3 \ Sides a, b, c are 9 cm, 12 cm, 15 cm. 1 3.
A
B C
4
4
90°
Area of ∆ABC = 12
×4×4 = 8 cm2. 1
4. Area of an equilateral triangle
=
234 4 3
4× =
cm2. 1
5. Area of an equilateral triangle
=
234
a
½
\
23
4a
= 81 3
1
⇒ a2 = 81 × 4 a = 9 × 2 = 18 cm Perimeter of equilateral triangle = 3a = 3 × 18 = 54 cm. ½ [CBSE Marking Scheme, 2012]
6. Using Pythagoras theorem, ½ (Hypotenuse)2 = (Base)2 + (Perpendicular)2 ⇒ 102 = (8)2 + (x)2
⇒ 100 – 64 = x2
x = 6 cm 1
\ Area =
12
× base × height ½
=
12
× 8 × 6
= 24 cm2 1 7. (i) Let the sides of the triangle be a, b and c. a : b : c = 12 : 17 : 25
12 17a b=
=
25c
k= (say)
⇒ a = 12k, b = 17k, c = 25k ½ Perimeter = 540 cm ⇒ a + b + c = 540 ⇒ 12k + 17k + 25k = 540 ⇒ 54k = 540
\
k =
54010
54= ½
Hence a = 12 × 10 = 120 cm, b = 17 × 10 = 170 cm, c = 20 × 10 = 250 cm
Now, s =
12
×540 = 270 cm
(s – a) = (270 – 120) cm = 150 cm (s – b) = (270 – 170) cm = 100 cm (s – c) = (270 – 250) cm = 20 cm ½
\ Area of triangle = ( – )( – )( – )s s a s b s c
= 270 150 100 20× × × cm2
= 100 27 15 10 2× × × cm2
= 100 3 3 3 3 5 2 5 2× × × × × × × cm2
= 100 × 3 × 3 × 5 × 2 = 9000 cm2 ½ (ii) Heron’s formula. ½ (iii) A balanced ratio leads to good result. ½
P-52 M A T H E M A T I C S - I X T E R M - 1
SUMMATIVE ASSESSMENT WORKSHEET-70Solutions
1.
A B
CD
aa
a
a
16
For ∆BCD, s =
168
2a a
a+ + = +
\ Area of ∆BCD
= ( 8)( 8 16)( 8 )( 8 )a a a a a a+ + − + − + −
= ( 8)( 8)(8)(8)a a+ −
= 28 64a − ½
\ Area of rhombus = 2 × Area of ∆BCD
⇒ 96 = 2 × 28 64a −
⇒ 6 = 2 64a −
⇒ 36 = a2 – 64 ⇒ a2 = 64 + 36 = 100 \ a = 10 cm. ½ 2.
A
x
B Cx
y
Area of an isosceles right triangle
=
12
× base × height
8 =
12
× x × x
⇒ x = 4 cm
\ y = 2 24 4 32+ = cm.
= 4 2 cm. 1
3. Let ‘b’ be the equal sides length and ‘a’ be the base.A
B CDa
b bx
a/2 a/2
x =
2 2 22 4
2 2a b a
b− − =
Area of triangle =
12
× base × height
=
12
× a × x
=
12
× a × 2 242
b a−
=
2 244a
b a−
1
4. (i) If a, b, c be the lengths of sides BC, CA and AB of
∆ABC and if s =12
(a + b + c) then,
Area of ∆ABC = ( – )( – )( – )s s a s b s c \ Here, a = b = c (= a) and
s =
12
(a + a + a) = 32a
½
s – a =
32 2a a
a− =
s – b =
32 2a a
a− =
s – c =
32 2a a
a− =
½
Area =
32 2 2 2a a a a
=
3
2 2a a
× ×
=
234a
. ...(1) 1
(ii) To find the area, when its perimeter = 180 cm2
Here, a + a + a = 180 ⇒ 3a = 180
⇒ a =
18060
3=
Hence, required area =
23(60)
4×
= 900 3 cm2. 1
(iii) Heron’s formula. ½ (iv) We should follow the traffic rules to save our
lives. ½
P-53S O L U T I O N S
TOPIC-2Heron’s Formula
SUMMATIVE ASSESSMENT WORKSHEET-71Solutions
1. ∆ = ( )( )( )s s a s b s c− − −
where, 2s = a + b + c. 1
2.
34
.a2
1
3. s =
7 24 25
2
+ +
= 28 cm
Area = ( )( )( )28 28 - 7 28 - 24 28 - 25
= 28×21×4×3 = 84 cm2. 1
4. s =
18
2
a b c+ +⇒
=
13 14
2
c+ +
⇒ c = 36 – 27 = 9 cm. 1
5. s
=
8 7 5
2 2
a b c+ + + +=
= 10 cm. 1
6. s =
6+8+10= 12
2 cm.
Area = 12(12 - 6)(12 - 8)(12 - 10)
= 12×6×4×2 = 24 cm2
Cost of painting =
24×9= 2.16
100` . 1
7. a = 8 cm, b = 11 cm, Perimeter = 32 cm 1 \ c = 32 – (8 + 11) = 13 cm, s = 16
\ Area = 16(16 8)(16 11)(16 13)− − −
= 8 30 cm2 1
8. Third side = ( ) ( )2 2125 100− = 75 2
s = 150
Area of ∆ = 150 25 50 75× × × = 3750 m2
9.
5 12 16
20
B13A
D
C
3 For ∆ABD, s = 15
Area of ∆ABD = 15 2 3 10× × × = 30 cm2
For ∆BCD, s = 24
Area of ∆BCD = 24 4 12 8× × ×
= 96 cm2
Area of quadrilateral = 30 + 96 = 126 cm2. 10. Area of cloth required = 10 × Area of cloth for one
piece ½ Area of one piece of cloth is made by sides 60 cm, 60
cm and 20 cm is,
s =
60+60+202
= 70 cm ½
Area = 70(70 – 60)(70 – 60)(70 – 20)
= 70 10 10 50× × ×
= 7 10 10 10 5 10× × × × ×
= 10 × 10 × 35
= 100 35 cm2 1
Area of cloth required
= 10 × 100 × 35
= 1000 35 cm2 1
SUMMATIVE ASSESSMENT WORKSHEET-72Solutions
1. ( )( )( )s s a s b s c− − − 2. Heron’s Formula. 1 3. Sides are 16 cm, 16 cm, 16 cm
Semi, perimeter =
16 16 16
2 2
a b c+ + + +=
=
48= 24 cm
2 ½
\ Area of triangle = ( )( )( )s s a s b s c− − −
P-54 M A T H E M A T I C S - I X T E R M - 1
= 24(24 – 16)(24 – 16)(24 – 16)
= × × ×24 8 8 8
= × × × ×3 8 8 8 8
= 64 3 cm2
A
B C
16 cm 16 cm
16 cm ½
4.
s = 2
a b c+ +,
here a, b, c are the sides of a triangle.
s = 12 16 20 48
24cm2 2
+ += = ½
Area = ( – )( – )( – )s s a s b s c ½
=
24(24 – 12)(24 – 16)(24 – 20)
= 2 3 4 12 8 4× × × × ×
= 2 12 12 2 4 4× × × × ×
= 2×4×12 = 96 cm2 1
5. s =
2a b c+ +
=
70+80+902
= 120 cm ½
Area = ( )( )( )s s a s b s c− − − ½
= 120(120 – 70)(120 – 80)(120 – 90)
= 120 50 40 30× × ×
= 40 3 5 10 40 3 10× × × × × ×
= 40 × 10 × 3 × 5
= 1200 × 2.23
= 2676 cm2 1
6. Sides of a triangle are a = 6 cm, b = 6 cm, c = 4 cm
s =
6 6 4 168
2 2 2a b c+ + + += = = cm
½
A = ( )( )( )s s a s b s c− − − 1
= 8 2 2 4× × ×
= 8 2 cm2 ½
Area of two black triangles
= 8 2 ×2 = 16 2 cm2 ½
Area of two white triangles
= 16 2 cm2. ½
[CBSE Marking Scheme, 2012]
7. Area = ( )( )( )s s a s b s c− − − 1
A = 54 3 17 34 306× × × =
= 306 m2 1
No. of rose beds =
3066
= 51 1
[CBSE Marking Scheme, 2012] Alternative Method :
s =
51 37 20 1082 2
+ + =
= 54 cm ½
Area = ( )( )( )s s a s b s c− − − ½
= 54(54 51) (54 37) (54 20)− − −
= 54 3 17 34× × × ½ = 306 m2 ½ No. of rose beds
=
Total area of triangular fieldArea occupied by each rose bed
½
=
30651
6= ½
SUMMATIVE ASSESSMENT WORKSHEET-73Solutions
1.
s =
120= 60
2
Sides = 5x + 12x + 13x = 120 30x = 120 x = 4 cm 5x = 20 cm, 12x = 48 cm, 13x = 52 cm.
Area of ∆ = ( )( )( )s s a s b s c− − − cm2
= 60 40 12 8 230400× × × =
= 480 cm2 2[CBSE Marking Scheme, 2012]
Alternative Method :
Let the sides are 5x, 12x and 13x, then Perimeter = 5x + 127 + 13x = 120 ⇒ 30x = 120
P-55S O L U T I O N S
⇒ x = 4 cm ½ Then sides are 20 cm, 48 cm, 52 cm
s =
120= 60
2 cm
Area = ( )( )( )s s a s b s c− − − ½
= 60(60 20)(60 48)(60 52)− − −
= 60 40 12 8× × × ½
= 10 6 10 4 6 2 4 2× × × × × × ×
= 10 × 6 × 4 × 2 = 480 cm2. ½
2. a = 15 cm, b = 15 cm, c = 12 cm
s =
15 15 1221
2 2a b c+ + + += = cm
½
Area = ( )( )( )s s a s b s c− − −
= 21 (21 15)(21 15)(21 12)× − − − 1
= 21 6 6 9× × ×
= 18 21 cm2. ½
[CBSE Marking Scheme, 2012] 3. Perimeter of the triangle = 84 cm Ratio of its sides = 13 : 14 : 15 Sum of the ratio = 13 + 14 + 15 = 42 cm 1
\ a =
1342
× 84 = 13 × 2 = 26 cm
b =
1442
× 84 = 14 × 2 = 28 cm
c =
1542
× 84 = 15 × 2 = 30 cm
s =
26 28 302
+ +
s =
842
= 42 cm 1
Using Heron’s formula,
Area of ∆ = ( – )( – )( – )s s a s b s c
=
42(42 – 26) (42 – 28) (42 – 30) cm2 1
= 42 16 14 12× × × cm2
= 2 3 7 16 2 7 3 4× × × × × × × cm2
= 2 × 3 × 7 × 4 × 2 cm2 1
= 336 cm2. 4. For one identical triangular leaf, let a = 28 cm, b = 15 cm and c = 41 cm
Also, s =
28 15 412 2
a b c+ + + +=
=
842
= 42 cm 1
41 cm
28 cm
15 cm
Using Heron’s formula, Area of one triangular leaf
=
( – )( – )( – )s s a s b s c
= 42(42 – 28) (42 – 15) (42 – 41)
= 42 14 27 1× × × 1
= 3 14 14 3 9× × × × = 9 × 14 = 126 cm2 1 There are 6 leaves in a circle. So, total number of leaves in 2 circles = 2 × 6 = 12 \ Area of 12 leaves = (12 × 126) cm2 = 1512 cm2
Hence, total area to be painted red = 1512 cm2. 1
TOPIC-3Application of Heron’s Formula in finding Area of Quadrilaterals
SUMMATIVE ASSESSMENT WORKSHEET-74Solutions
1. In ∆ABC, s = 5 7 8
102
+ + = cm
Area of || gm ABCD
= 2×Area of triangle ABC
= 2 10(10 8)(10 7)(10 5)− − −
= 2 10 2 3 5× × ×
= 20 3 cm2 1
P-56 M A T H E M A T I C S - I X T E R M - 1
2.
50 mD C
BA
50 m50 m
50 m
80 m
Perimeter of rhombus = 4 × sides = 200 ⇒ side = 50 m ½
Area of ∆ABC = ( )( )( )s s a s b s c− − −
=
90 (90 50)(90 50)(90 80)× − − −
= 90 40 40 10× × × 1
= 1200 m2. ½ Area of rhombus = 2 × Area of ∆ABC = 2×1200 = 2400 m2. ½ [CBSE Marking Scheme, 2012]
3.
6 cm
D L C
BA M
O
8 cm
Diagonals bisect each other at O Draw, LOM⊥ AB and CD OL = OM = 3 cm Area of shaded region
OAB =
12
× 8 × 3
= 12 cm2. 2 [CBSE Marking Scheme, 2012]
Alternative Method : Diagonals bisect each other at O. Draw, LOM ⊥AB. OL = OM = 3 cm Area of shaded region OAB
=
12
× 8 × 3 = 12 cm2. 2
4. Rhombus,
D C
BA 13 cm
13 cm
13 cm13 cmO 12 cm
Perimeter = 52 cm 1
Side =
5213
4= cm
Diagonal = 24 cm 1 OB = OD = 12 cm
OA = 2 213 12 169 144 25 5− = − = = ½
Area of rhombus = 4 ×
12
× 5 × 12 ½
Area = 120 cm2 [CBSE Marking Scheme, 2012]
5. Let ABCD be the given field in the form of trapezium in which AB = 35 m, CD = 10 m, BC = 13 m, AD = 14 m and DC || AB ½
35 m
10 m
14 m 13 m
D C
BA E F
Through C, draw CE || DA and let it meet AB at E. Let h metres (CP) be the height of the trapezium. DC || AE and CE || DA \AECD is a parallelogram. AE = DC = 10 m and CE = DA = 14 m In ∆CEB, CB = 13 m, CE = 14 m ½ and BE = AB – AE = 35 – 10 = 25 m Let a = 14 m, b = 13 m and c = 25 m
Then, s =
2a b c+ +
=
14 13 252
+ +
= 26 m ½
Area of ∆CEB = ( – )( – )( – )s s a s b s c
= 26(26 – 14)(26 – 13)(26 – 25)
= 26 12 13 1× × ×
= 13 2 2 6 13× × × ×
= 226 6 m 1
Also, Area of ∆CEB =
12
× base × height
=
12
× BE × h
⇒ 12
× 25 × h = 226 6 m
P-57S O L U T I O N S
\ h =
526
25m ½
Area of trapezium =
12
(DC+AB) × h
=
12
(10 + 35) ×
526
25
=
12
× 45 ×
526
25
=
926 6
5 ×
= 114.66 m2 (approx) 1
SUMMATIVE ASSESSMENT WORKSHEET-75Solutions
1. Area of trapezium ABCD = Area of rectangle AOCD + Area of ∆OBC In ∆OBC, by Phythagoras theorem BC2 = OC2 + OB2
⇒ 172 = OC2 + 82
⇒ OC2 = 289 – 64 = 225 \ OC = 15 cm ½ Area of rectangle AOCD = 6 × 15 = 90 cm2 ½
Area of ∆OBC =
12
× 8 × 15 = 60 cm2 ½
Area of Trapezium ABCD = 90 + 60 = 150 cm2 ½
2.
20 mD C
BA
20 m
20 m
24 m20 m
Semi-perimeter of triangle ABC
s =
2a b c+ +
=
20 20 24 642 2
+ + = = 32 cm ½
Area of ∆ABC = ( )( )( )s s a s b s c− − −
= 32(32 24)(32 20)(32 20)− − −
= 32 8 12 12× × ×
= 8 4 8 12 12× × × × = 8 × 2 × 12 = 192 cm2 1 \ Area of rhombus ABCD = 2 × Area of ∆ABC = 2 × 192 = 384 cm2. ½ 3. Draw a quadrilateral ABCD.
D C
A B
18 m
82 m
50 m
50 m
½ Here, ∠CBD = 90° Area of quadrilateral ABCD = Area of ∆BCD + Area
of ∆ABD ½ In ∆BCD, by phythogoras theorem, BC2 + BD2 = CD2
⇒ 182 + BD2 = 822
⇒ BD2 =822–182
= 6400 BD = 80 m
Area of ∆BCD =
12
× BC × BD ½
=
12
× 18 × 80 = 720 m2 ½
Semi-perimeter of ABD,
s =
50 50 802
+ +
= 90 m
Area = 90(90 – 50)(90 – 50)(90 – 80)
= 9 10 40 40 10× × × ×
= 3×10×40 = 1200 m2 ½ Area of quadrilateral ABCD = 720+1200 = 1920 m2 ½ 4. From Fig; Area of paper of each colour
=
12
× Area of kite.
Area of kite = Area of square + Area of
isosceles triangle. Area of square = (side)2
= (16)2
= 256 cm2
Area of isosceles triangle
= 2 24 –
4a
b a
1
P-58 M A T H E M A T I C S - I X T E R M - 1
Here, b = length of equal side, a = base of triangle.
=
2 244(6) – 4
4
=
4144 – 416
4
= 128
= 64 2 8 2× = cm2 1
Area of paper of each color
=
12
[Area of kite]
=
12
[256 + 8 2 ]
= 128 + 4 2
= 128 + 4×1.41 = 128 + 5.64 = 133.64 cm2 1
FORMATIVE ASSESSMENT WORKSHEET-76 Note : Students should do this activity themselves.
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