cce sample question paper 8...proof : the internal bisector of an angle of a triangle divides the...

CCE SAMPLE QUESTION PAPER 8 FIRST TERM (SA-I)

MATHEMATICS (With Solutions)

I CLASS X rime Allotved : 3 to 3% Hours] IMaximum Marks : 80

General Instructions : (i) All questions are compulsory.

( i i) This question paper consists o f 34 questions divided into four sections A, B, C and D. Section A comprises of 10 questions of 1 mark each, Section B comprises of 8 questions of 2 marks each, Section C comprises of 10 questions of 3 marks each and Section D comprises of 6 questions of 4 marks each.

(iii) Question numbers 1 to 10 in Section A are multiple choice questions where you are to select one correct option out of the given four.

( iv) There is no overall choice. However; internal choice has been provided in 1 question of two marks, 3 questions of three marks each and 2 questions offour marks each. You have to attempt only one of the alternatives in all such questions.

( v ) Use of calculators is not permitted.

Question numbers 1 to 10 are of one mark each.

1. If sin 8 = 3 then the value of (tan 8 + sec 8)2 is 5 '

( a ) 2 (b ) 3 (c) 4 (dl 5

Solution. Choice (c) is correct.

cos o = 41- sin2 e

5 sec 8 = - and z 25 4

sec 8 = - 16

5 z 25 s e c e = - and l + t a n 8 = - 4 16 5 sec 0 = - and 25 4

t a n 2 e = - - I 16

trainer

Rectangle

* 5 sec e = - and 2 9 tan 8 = - . 4 16

5 + sec8= - and 3

tan €I = - 4 4

3 5 2 - - (tan e + see e)2 = (- + -r = ti) = (2)' = 4

4 4

s e c e - t a n 0 2. If 5 sin 0 - 3, then the value of is

sec 8 + t a n 0

1 (a) - 3

4 ' . (b) y

5 Gc) - 3

4 (dl 5

Solution. Choice (a) is correct.

1 sine see 0 -.tan 0 cos 8 cos8 - - s e c € ~ + t a n e 2 sine +-

cos 8 . cos 0

- - (1 - sin 8)Icos 8 (1 + sin €I)/cos 0

- - 1- sin 8 l + s i n e

- 1 - - 4

3. If sec 0 - tan 0 - p, then the value of see I3 + tan

U-Like CCE Sam~le Question Paoer 8 189


I s e c 8 - t a n 8 = p

* (sec 8 -tan 8)(sec 8 +tan 8) = p(sec 8 +tan 8) [Multiplying both sides by (sec 8 + tan 811 a sec2 8 - t a g 8 = p(sec 8 + tan 8) * 1 + tan2 8 - tan2 0 = p(sec 8 +tan 8) 3 1 =p(sec 8 + tan 8)

4. If tan 0 - .h - 1, then the value of sec2 0 is

(a) 2&(& - 1) (b) 2&(& - 1)

(c) 5&(& - 1) (d) 7&(& - 1) Solution. Choice (a) is correct.

. . t a n o = & - 1 * tan2 8 = (& - 1)2

t a n 2 e = 2 + i - 2 @

* l + t a n 2 ~ = 1 + ( 3 - 2 & )

=> . sec28=4-2&

+ sec2 8 = 2(2 - &)

* sec2 e = 2&(& - 1) 5. In figure, AB 11 DE. The length of CD is . .

15 (a) 2.5 cm (b) 2.7 cm (c) 3.6 cm (d) 4.2 cm

Solution. Choice (a) is correct. In figure, AB 11 DE In M C and ALIEC

LA=LE . LB = LD

LACB = LDCE * AABC - ALIEC

[Alternate Lsl [Alternate Lsl

[Vertically oppostite Lsl [AA-similarity]

a BC Considering,. - - -

DE - CD

* 3 x 5 CD = -- = 2.5 cm. 6

6. In figure, the graph of a polynomial p(x) is shown: The n k b e r of zeroes of p(x) is

Y

t

Y (a) 4 (b ) 3

(dl 1 (c) 2 Solution. Choice (e) is correct. The number of zeroes is 2 as the graph intersects the x-axis at two points viz., A and C

7. The decimal expansion-of 23457 will terminate after how many places of 23 54

decimal ? (a) 2 (b) 3 (c) 4 (dl 5


23457 will terminate after 4 places of decimal. Thus, the decimal expansion of - 23 ii4

8. The largest number which exactly divides 280 and 1245 leaving remainders 4 . . and 3 respectively, is

(a ) 138 , (b) 139 (c) 141 (d) 161.

Solution. Choice (a) is correct. 280 - 4 = 276 = 2 x 138 i. 276 is divisible by 138 * HCF of 276 and 1242 is 138.

1245 - 3 = 1242 = 9 x 138 * 1242 is divisible by 138 9. For the distribution

Class-interval 1 0-10 1 10-20 120-30 130-40 140-50 / 50-60 12 18 21 Frequency 15 11 13

U-Like CCE Sample Question Paper 8 191

The sum of lower limits of median class and modal class is (a) 40 ( b ) 50 (c) 60 (dl 70

Solution. Choice (c) is correct. Here, the maximum frequency is 21 and the class co~esponding to this frequency is 30 -40,

therefore 30 - 40 is the modal class. The lower limit of the class 30 - 40 is 30.

n 90 Since 30 - 40 is the class whose cumulative frequency 60 is greater than - = - = 45,

2 2

Class interval 1 0-10 1 10-20 ( 20-30 1 30-40 1 40-50 1 50-60

therefore 30 - 40 is the median class. The lower limit of the class 30 - 40 is 30. The sum of lower limits of median class and modal class is 30 + 30, i.e., = 60. 10. The linear pair of equations (3k + 1)x + 3y - 5 = 0 and 22 - 3y + 5 = 0 have infinite

solutions. Then the value of k is (a) 1. (b) 0 (c) 2 (d ) , - 1

Solution. Choice (d) is correct. The given pair of linear equations will have infinite. solutions if

3k + 1 * -=-I 2

* 3 k + 1 = - 2 * 3 k = - 3 * k = - 1

Question numbers 11 to 18 carry 2 marks each. 11. Show that 4n caqnot end with the digit 0 for any natural number. Solution. We know that any positive integer ending with the digit 0 is divisible by 5 and ,

so its prime factorisation must contain the prime 5. We have 4n = (2 x 2In = (22)n = 22n

The only prime in the factorisation of 4n is 2. * There is no other prime in the factorisation of 4n = 22n. . =, 5 does not occur in the prime factorisation of 4" for any n. [By uniqueness of the Fundamentid Theorem of Arithmetic]. Hence, 4n does not end with the digit zero for any natural number n. . . '

12. Find the zeroes of thq quadratic polynomial 4a? - h - 3 and verify the relation between zeroes and its coefficients.

Solution. We have k 2 - k - 3 = k 2 - 6 x + 2 x - 3

78 1 90 Cumulative freauencv I 11 I 26 I 39 60

192 U-Like Mathematics-X

a 4 x 2 - 4 x - 3 = 2 ~ ( % - 3 ) ' + ( 2 ~ - 3 ) => 4x2-42-3=(2.-3)(%+1) So, the value of 4x2 - 4x - 3 is zero when 22 - 3 = 0,2x + 1 = 0

3 1 i.e.,whenx= - andx =--. 2 2

3 1 Therefore, the zeroes of 4x2 - 4x - 3 and x = - and - - 2 2

3 1 Now sum of zero& = - - - - 2 2

= - (Coefficient of x) * Coefficient of x2

3 1 Product of zeroes = - x -- . - ~ \ ~ . 2 2

.. ~ . - 3 = -- 4 Constant term - -

Coefficient of x2 13. Find the value(s) of k for whicli the pair of linear equations ks + 3y- k - 2 and

122 + ky - k has no solution. . Solution. We know that : for apair of linear equations

alx + bly = cl a@ + b g = cz

has no solution if a1 bl cl - = - + - a~ bz C z

.- So, the given system of equations will have no solution. if

=> k 3 -- - - 3 k - 2 12 k

and - - k k

= k 2 = 36 and 3 i k - 2 - k = t 6 and 3 i k - 2 Clearly, for k = t 6, we have, 3 t k - 2 Hence, the given pair of linear equations will have no solution if k - t 6.

1 Solution. Given, sin (A - B) = - 2


a sin@-B)=sin3O0 + A-B=3O0 ... (1)

1 Also, given cos (A + B) = - = cos 60' 2

* cos (A + B) = cos 60° - A+B=60° Adding (1) and (2), we get

2.4 = 30" + 60" A = 9 0 ° + 2 = 4 5 "

Subtracting (1) from (21, we get 2B = 60" - 30" = 30"

z Hence, A - 45" and B - 15".

Or If A, B and C are interior angles of a triangle ABC then show that

Solution. We know A+B+C=18O0

==i. A + B = 180"-C [Sum of angles of aA = 18O0]

[Dividing both sides by 21

- tan (9) = tan (W - g)

C - tan - = cot - iAIBi 2 ['; tan (90" - 8) = cot 81

Hence, the result. 15. In figure, altitudes AD and CE of AABC intersect each other at the point P.

Show that . . i M P - ACDP (ii) AABD - ACRE

194 U-Like MathernaticsX

Solution. We are given that in AABC, AD I BC.and CE I AB. . . LADB = LADC or LPDC [Each = 90" as AD I BCl and LCEB = LCEA or LPEA [Each = 90" as CE I BCl (i) In AAEP and ACDP, we have

LAPE = LCPD [vertically opposite Lsl LPEA = LPDC [Each = 90"l

So, biAA-criterion of similarity, we have AAEP - ACDP

(ii) In AABD and ACBE, we have Q=LB [coimnonl

' LADB = LCEB [Each = 90°1 .So, by AA-criterion of similarity, we have

AABD - ACBE 16. 0 is a point inside a M C . The bisectors of LAOB, DOC and LCOA meet the

sides AB, BC and CA at the points D, E and F respectively. Prove that AD.BE.CF = BD.EC.FA

~oluti6n. Given : 0 is a point inside a AABC, OD, OE and OF are the bisectors of angles AOB, BOC'and COA meeting AB, BC and CA at D, E and F respectively.

To prove : AD.BE.CF = BD.EC.FA -

A

Proof : The internal bisector of an angle of a triangle divides the opposite side in the ratio of the sides containing the angle.

As in MOB, OD is the internal bisector of LAOB

Similarly

and CF OC - = - FA OA

Multiplying the corresponding sides of (I), (2) and (31, we get AD BE CF OA OB OC - . - . - - - . - . - BD EC FA - OB OC OA

= 1 Hence, AD.BE.CF = BD.EC.FA

U-L~ke CCE Sample Questlon Paper 8 195

17. The following is the cumulative frequency distribution of marks obtained by 85 students.

Marks Below 10 Below 20 Below 30 Below 40 Below 50 Below 60 Below 70 Below SO Below 90 Below 100

Mode = 1 + fl - fo h 2 f l - fo - fi

Number of Students 5 9 17 29 43 60 70 78 83 85

Frequency

5 4 (9 - 5) 8 0 7 - 9 ) ,

Below 40 Below 50 Below 60 Below 70 Below 80 Below 90 Below 100

Write the above cumulative frequency distribution as frequency distribution. Solution.

Marks

0-10 10 - 20 20 - 30.

Marks

Below 10 .Below 20 Below 30

Number of students (Cumulatiue frequency)

5 9

. 17

18. In a continuous frequency distribution, if lower limit of the modal class - 20, frequency of modal class - 20, frequency of the class preceeding the mod4 class = 14, frequency of the class succeding the modal class - 16 and size of each class is 5. Find the mode of the distribution.

Solution. Given, 1 = 20, fi = 20, fo = 14, fi = 16, h = 5 Using the formula :

29 43 60 70 78 83 , 85

30 - 40 40 - 50 50 - 60 60 - 70 70 - 80 80 - 90 90 - 100

12 (29 - 17) 14 (43 - 29) 17 (60 - 43) 10 (70 - 60) 8 (78 - 70) 5 (83 - 78) 2 (85 - 83)

= 2 0 + 3 - = 23

p z G q Questzon numbers 19 to 28 cany 3 marks each. 19. Roohi travels 300 km to her home partly by train and partly by bus. She takes

4 hours if she travels 60 km by train and then remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.

Solution. Let the speed of the train be x kmlh and the speed of the bus bey km/h. . Total distance covered = 300 km. When Roohi travels 60 km by train and the remaining by bus. If she covers 60 krn by train then the distance covered by bus is (300 - 60) km = 240 km

60 Time taken to cover 60 km by train = - hours X

240 and time taken to cover 240 km by bus = - hours Y

[: hours Thus, the total time taken to cover a distance of 300 km = - t -

It is given that the total time of travelling is 4 hours.

" When Roohi travels 100 km by train and the remaining by bus If she covers 100 km by train then the distance covered by bus is (300 - 100) km = 200 km

100 ' Time taken to cover 100 km by train = - hours

X

I 200 and time taken to cover 200 km by bus = -hours Y

I (IT t 7) hours Thus, the total time taken to cover a distance of 300 km = -

It is given that the total time of travelling is 4 hours 10 minutes.

. . - loo + -- 200 - 4 hours.10 minutes x Y

U-Like CCE Sample Question Paper 8 - . . 197

LIultiplying (1) by 25 and (2) by 15, we have .

1500 + 3000 i25 and ---- - = - X. Y 2

Subtracting (2a)from (la), we get

a y = 80 km/h Substituting y = 80 in (11, we get

8

60 * - = I X

+ x=6Okm/h Hence, speed of train = 60 km/h and speed of bus ='SO km/h.

Or Solve for x and y :

6(ax t b y ) - 3 a t 2 b . .

6(bx - ay) = 3b - 2a Solution. The given equations are : . .

6(az+by)=3a+2b 6(bx - ay) = 3b - 2a

Equations (1) and (2) can be re-written as

3b-2a bx-ay=- . 6

Multiplying (la) by a and (2a) by b and adding them, we have

(a2x + aby) + (b2x - aby) = 3a2 + 2ab + 3b2 - 2ab

6 6

... (la)

... (1)

... (2) .

... (la)

. . . (2a)

1 . Substituting z-= - in (la), we get 2

1 1 Hence, the solution of the given system of equations is x - - and y - -

'2 3' - - 20. Find the LCM and HCF of 12, 15 and 21, by applying the prime factorisation

method. Solution. We have

1 2 = 2 x 2 x 3 1 5 = 3 x 5 2 1 = 3 x 7

Here. 3' is the smallest Dower of the common factor 3 respectively. So, HCF (12,15,21) = 3<= 3 Also, Z2, 3', 5 l ahd 7' are the greatest powers of the prime factors 2,3,5 and 7 respectively

involved in the three numbers. So, LCM (12,15,21) = 2' x 3' x 5' x 7'

=12x35=420

21. Prove that 6 + ,h is irrational.

Solution. Let us assume, to the contrary, that 6 + & is rational. That is, we can find coprime p and q (q + 0) such that

Therefore: 2 - 6 = & 4


* 4

Since, p and q are integers, we get

P - 6 is rational, and so .& is rational. 4

But this contradicts the fad that & is irrational.

This contradiction has arisen because of our incorrect assumption that 6 + & is rational. . .

So, we conclude that 6 t & is irrational. Or

Prove that 3 - & is an irrational number. . Solution. Let us assume, to the contrary, that 3 - & is rational, we can find co-prime a

and b (b it 0) such that '

+ a 3-- =& b

Rearranging this equation, we have

Since a and b are integers,,we get, 3 - is rational, and so & is rational. b

But this contradicts the fact that & is irrational.

This contradiction has arisen because of our'incorrect assumption that 3 - & is rational.

So, we conclude that 3 - & is irrational. 22. If one zero of the polynomial (a2 t 9)x2 t 132 t 6a is reciprocal of the other,

find the value of a. Solution. The zeroes of the quadratic polynomial ax2 + bx + c and the roots of the

quadratic equation ax2 + bx + c = 0 are the same. Let one zero of the given polynomial be a, then one root of the equation (a2 + 9)x2 + 13x t 6a = 0

I will be a. It is given that the other zero of the given polynomial be .: Reciprocal of a is - a a I

I 1 i.e., the other root.of the quadratic equation (a2 + 9)x2 + 13x t 6a = 0 will be -.

a 1

:: Product of the roots = a.- = 1 ... (1) a I

1 Let a and - be the roots of the quadratic equation (a2 + 9)x2 + 13x + 6a = 0, then a

Constant term Product of the roots =

Coefficient of x2 6a =- . . .(2)

a2 +9

From (1) and (2), we obtain

. 6 a = 1 ' a 2 + 9 .

=) ' a 2 + 9 = 6 a =+ a 2 - 6 a + 9 = 0 * (a - 3)' = 0 =+ a - 3 = 0 a a = 3 Thus, the value of a is 3. 23. Without &ing trigonometric tablgs, evaluate the following :

(sinz 25" + sin2 65") + &(tan 5" tan 15" tan 30" tan 75" tan 85"). Solution. We have

(sin2 25" + sin2 6508) + &(tan 5" tan 15' tan 30" tan 75" t an 85") 1

= [sin2 25O + sin2 (90" - 25')] + & [tan 5" tan 15" tan 30" tan (90' - 15") tan (90° - 5'11

= (sin2 25 + cos2 25') + &[tan 5" tan 15' tan 30" cot 15' cot 5"l , [.: sin (90" - 8) = cos 8, tan (90" - 8) = cot 81

= 1+ &(tan 5" cot 5")(tan 15" cot 15") tan 30"

24. Rove the following :

Solution. We have L.H.S. = tan2 8 + cot2 8 + 2

= t&8+co t28+2 tan8co te I. : tan e.cot 8 = 11 = (tan 8 + cot 8)2


25. In figure, M is mid-point of side CD of a parallelogram ABCD. The line EM is drawn intersecting AC at L and AD produced at E. Prove that EL = 2BL.

: W . . ~ . : B C

Solution. , In ADME and ACMB, we have fDME = LCMB Wertically opposite Lsl LEDM= LBCM . ' [Alternate Lsl

DM = MC [.; M is the mid-point of CD] So, by AAS-criterion of congruence, we have

ADME = ACMB a DE=BC ,

Also, AD = BC [.: Opposite sides of a parallelogram ABCDI . . DE+AD=2BC => AE =, 2BC . ... (1) . Now, in AALE and ACLB, we have

A L E = LCLB [Vertically opposite Lsl LLAE = LLCB [Alternate Lsl

So, by AA-criterion of similarity of triangles, we have AALE - ACLB

* EL AE - = - ' BL BC

EL 2BC - = - * BL BC

[using (111

* EL -2

=> BL . EL = 2BL

26. Prove that the ratio of the area of two similar triangles is equal to the square of the ratio of their corresponding medians.

Solution. Given : AABC - ADEF and AP, DQ are their medians.

To prove : ar(AABC) f12 =-

~~(ADEF) D Q ~ Proof: Since the ratio of the areas of

two similar triangles is equal to the ratio of the square of .their corresponding sides.

. ar(hABC) A B ~ =- ... (1) B P C E Q F

" ~~(ALIEF) D E ~

Now, AABC - ADEF

* AB BC - DE EF

Thus, in triangles APB and DQE, we have .

and L B = L E [.; AABC - ADEN So, by AA-criteridn of similarity of triangles, we have

M B - ADQE

From (2) and (31, we get

AB2 AP2 * -=- . ..(4) D E ~ D Q ~

From (1) and (4), we get

27. Find the mean of the following distribution using step-deviation method.

Income (in f ) 1 0 - 5 0 150-100 1100-150(150-2001200-2501250-300

I I I

Total 1 n=Zfi=500 I I Zfiui = 425

Using the formula :

Solution. Let the assumed mean be a = 175 and h = 50:.

10 70

fiui

- 75 - 150 - 125

0 225 ..

. 550

Frequency

xi - 175 ui = 50

- 3 - 2 - 1

0 1 2

100

Class-mark (xi)

25 75

125 175 225 275

Incomes (in 7)

0-50' 50 - 100'

loo - 150 150 -200 . 200 - 250 250 - 300

80 90

Frequency (fi)

90 150 100 80 70 10

150


= 175 + 42.5 = 217.50

or The mean of 10 numbers is 12.5. The mean of first six numbers is 15 and the mean

of last five numbers is 10. Find the sixth number. . ....., ...., Solution. Let the ten numbers be xl, xz, x5, x6, X I , xlO, then

.... X 1 + X z + + X g + X g Mean of first six numbers =

6

2 6 + X 7 + X g + X g + X I 0 Mean of last five numbers =

5 - - 10x5=x6+x l+xg+xg+x10 * ~ ~ = x ~ + x ~ + x ~ + x ~ + x ~ ~ .. .(2)

X I + x2 + X 3 + X 4 + X g + X g + X I + X g + X g + X I 0 Mean of 10 numbers =

10

=) 1 2 . 5 ~ 10=90+50-xg + - 125 = 140 -x6 = xg=140-125 * x6 = 15 Hence the sixth number is 15. 28. Find the median of the following data :

[using (1) and (211

Classes I 0-10 110-20120-30130-40140-50150-60160-70170-80 Frequency 1 13 1 11 1 16 1 30 1 14 1 6 I 7 l 3

n 100 Hence, - = - = 50 2 2 .

Solution. The cumulative frequency distribution with the given frequency becqmes :

Now, 30 - 40 is the class, whose cumulative frequency 70 is greater than = 50. 2

Cumulatiue frequency 13 24 40 (cfJ 70 84 90 97

Classes 0-10

10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70

Frequency 13 11 16 30 (f) 14 6 7


Therefore, 30 - 40 is the median class. Thus the lower-limit (1) of the median class is 30. Using the formula :

5 0 - 4 0 x' lo = 3 0 + - 30

10 = 3 0 + - x 1 0 30 10

= 30 + - -

3 = 30 + 3.33 = 33.33

Question numbers 29 to 34 carry 4 marks each.

I 29. Obtain all other zeroes of & + 623 - %? - 10% - 5, if two of its zeroes are - E

I Solution. Since two zeroes are - and - - , therefore ( x - ' ~ ] [ ~ + t ) = x z - i i s a

factor of the givenpolynomial. 5 Now, we divide the given polynomial by x2 - ;

3x4 First term of quotient is = 3x2 x I

Second term of quotient is = 6x x

6x3 I -.. - 3x2 - 5 - +

X Third term of quotient is

, 3 So, 3x4 + 6x3 - 2x2 - lox - 5 = x - - (3x2 + 6x + 3)


So, its zeroes are given by x = - 1 and x = - 1. Therefore, the zeroes of the givbn polynomial

are -,- - , - I and-1. C C 30. Prove that in a triangle, if a line is drawn parallel to one side of a triangle to

intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

Solution. Given : A triangle ABC in which a line parallel to BC intersects other two sides AB and AC at D and E respectively. A

AD A E Toprove: -=-. DB EC

Construction : Join BE, CD and draw DM I AC and EN I AB. Proof : Since EN is perpendicular to AB, therefore,

EN is the height of triangles ADE and BDE. . -. 1 . . a r ( M E ) = -(base x height) '

2 1 =-(AD xEN) ... (1) 2

and 1 ar(AEDE) = -(base x height) 2

A(AI~ x EN) a a r ( M E ) - - 2

ar(ABDE) EN) 2

I J ( A E ~ D M ) AE Similarly, a r ( m E ) - - 2 -- -

ar(mEC) DM) EC 2

[using (1) and (2)1

Note that ABDE and ADEC are on the same base DE and between the same parallels BC and DE.

. . ar(ABDE) = ar(ADEC) ... (5) From (4) and (51, we have.

a r ( M E ) - - ... (6) ar(ABDE) EC

Again from (3) and (6), we have AD AE - DB EC

Hence, AD AE - = - DB EC

Or Prove that the ratio of the areas of two similar triangles is equal to the ratio of

the squares on their corresponding sides. Solution. Given : AABC and APQR such that AABC - APQR.

To prove : ~ ~ ( A A B C ) A B ~ B C ~ C A ~

Construction : Draw AD I BC and PS I QR. P

~ X B C X A ~ ar(AABC) - 2 Proof: -

(QQR) 1. QRx pS 0

1 [Area of A = -(base) 2 k height]

A

=). ar (AABC) - BC x AD - ... (1) ar(APQR) QRx PS ,

. Now, in AALlB and APS'Q, we have LB = LQ [AS AABC - APQRI

LADB = LPSQ [Each = 9Oo1 3rd BAD = 3rd LQPS

Thus, AALIB and APSQ are equiangular and hence, they are similar. AD AB Consequently, - =- ... (2) PS PQ

[If A's are similar, the ratio of their corresponding sides is same1

But AB BC - = -

PQ QR

* AD BC PS=d2R . ..(3) [using (2)l

Now, from (1) and (3), we get

ar(AABC) BC AD =- x- ar(APQR) QR PS

=z. ar(AABC) BC BC ar(APQR) = m X a

=). ar (AABC) B C ~

=- ar(APQR) Q R ~

[using (311

... (4)

U-Like CCE Sample Question Paper~8 207

As AABC - APQR, therefore

Hence, A B ~ B C ~ C A ~ a r ( M C ) - -

ar(APQR) PQ' QR' RP2 31. Prove that :

tan e - cot e cos e + sin 8 - - 1- t an0 1-co te cos0-sine

Solution. We have

tan 8 - cote L.H.S. = 1 1-tan0 1

tan 8

- - tan 0 cot @.tan 0 - 1-tan8 t a n @ - 1

- - [tan e + 11 1- tan 0.

[From (4) and (511

sin 0 + .-

= cos e

sin e I--

cos e

= ~ i n e + c o s e cos 8 - sin 0

= R.H.S. Or

Evaluate :

cos (90" - 8) sec (90" - 8) tan 0 +

sin2 75" + sin2 15"

cosec (900 - 0) sin (90" - 8) cot (90" - 8) tan 20" tan 40" tan 45" tan 50" tan 70" + 2 sin2 38".sec2 52" - sin2 45".

Solution. We have

- - sin8.cosec8.tane + sin2 (90" - 15") + sinz 15" sec 0 . ~ 0 s 8.tan 0 tan 20" tan 40" tan 45" tan (90" - 40") tan (90" - 20")

(sin 0.cosec 0) cosZ 15" + sin2 15" . .. - ..- - + + 2 sin2 38" cosec2 38" - sinz 45O " ., -.~,(sec 8 . ~ 0 s 8) tan 20" tan 40" (1) cot 40" cot 20" .

= A + ‘ 1 . + 2 sin2 38" cosec2 38' - sin2 45" 1 (tan 20".cot 20"Xtan 40°.cot 40")

7 = - 2

32. Prove that : 1 1 -- 1 =-- 1

cosec 0 - cot 0 sin 0 . sin 0 cosec 0 + cot 0 Solution. We have

. L.H.S. = 1 1 -- cosec 0 - cot 0 sin 0

- - 1 1 -- 1 cos 0 s ine

sin 0 sin 0

1 1 - - -- 1- C O ~ 0 ~ i n 8

sin 0

- - sin 0 1 -- 1- cos 8 sin0

- - sin 8(1+ cos 0) 1 -- (1 - cos 0)(1+ cos 0) sin 0

- - sin 8(1+ cos 8) 1 -- 1-cos2e sin0


- - sih e( i + cos e) 1 -- sin2 e sin e

-- - 1+ cos 8 1 -- sin 0 sin €I

1 cos 0 1 =----+---

sine s m e sine

- 1 1 cos e ---.--+- sin tl sine sine

1 cos e =-- s i n [ s 0 sin 01

=-- .I [cosec 0 - cot 01 sin 0

1 (cosec 0 - cot e)(cosec e + cot 8) =-- sin 8 cosec @ +cot 0

1 cosec2 0 - cot2 0 =-- sin 0 cosec I3 + cot 0

- - 1+cot2 0-cot20 . cosec 0 + cot 0 I

1 =-- 1 sin e cosec 8 + cot 0

= R.H.S. 33. Represent the following system of linear equations graphically. From the

graph, find the points where the lines intersect y-axis. 3 x + y - 5 = 0 ; ? a - y - 5 = 0

Solution. Given equations are : 3 x + y - 5 = o * y = 5 - 3 x ... (1)

and 2x-y -5=0*y=2z-5 . . .(2) Let us draw the graphs of the equations (1) and (2) by finding two solutions for each of these

equations. They are given in tables :

y = 5 - 3 x y = 2 x - 5

A


. Plot the points A(O,5), B(2, - I ) , C(0, -5) and D ( 3 , l ) on graph paper and draw the lines AB and CD representing given equations, as shown in figure.

In figure, we observe that the two lines representing the two equations are intersecting at the point B(2, - 1).

I Hence, x - 2 and y - - 1. The line AB cuts the y-axis at the point A(O,5) and the line CD cuts the y-axis at the point

C(0, - 5). 34. During the medical check-up of 35 students of a class their weights were

40 - 42 42-44 44 - 46 46 - 48 48 - 50 50 - 52

2 4

' 5 14 4 3

Draw n less than type and a more than type ogive from the given data. Hence ledian weight from the graph.

U-Like CCE Sample Question Paper 8 21 1

Solution. Less Than Method : We prepare the-c.umulative frequency table by less than tme method as given below :

Weight (in kg)' 38 - 40 40 - 42 42 - 44 44 - 46 46-48 48 - 50 50 - 52

Frequency - 3

-

Here 40, 42, U,......, 52 are the upper limits o'f the' respective class intervals less than 38 - 40,40 - 42, ...., 50 - 52. To represent the data in the table graphically, we mark the upper limits of the class intervals on the horizontal axis (x-axis) and their corresponding cumulative frequencies on the vertical axis (y-axis), choosing a. convenient scale other than the class intervals, we assume a class interval 36 - 38 prior to the first class intervals 38 - 40 with zero frequency.

Now plot the points (38,0), (40,3), (42,5), (44, B), (46,141, (48,281, (56,321 and (52;35) on a graph paper and join them by ti free'hand smooth curve. The curve we get is called an ogive of less than type (see figure).

More Than Method : We prepare the &dative frequency table by more than type method as given below :

2 4 5

14 4 3

Weight (in kg) 38 - 40 40 - 42 42 - 44 44 - 46 46 - 48 48 - 50 50 - 52

Weight less than

40

Cumulative . Frequency

3 42 44 46 48 50 52

Frequency

Other than the given class intervals, we assume the class interval 52 - 54 with zero frequency.

To represent the data in the table graphically, we mark the lower limits of the class intervals on the horizontal axis (x-axis) and their corresponding cumulative frequencies on the vertical axis (y-axis), choosing the same scale as in less than method. Now, plot the points (38,351, (40, 321, (42,301, (44,26), (46, 21), (48, 7), (50,3) and (52,O) on the same graph paper and join them by a free hand smooth curve. The curve we get is called an ogive of more than t y p e (see figure). We h d that the two types of cumulative frequency curves intersect at point P. From P perpendicular PM is drawn on x-axis. The value of weight corresponding to M is 46.5 kg. Hence, the median is 46.5 kg.

3 2 4 5

14 4 3

5 9

14 28 32 35

Weight more than

38 40 42 44 46 48 50

Cumulative frequency

35 . 32

30 26 21 7 3

Weights (in kg) L

cce sample question paper 8...proof : the internal bisector of an angle of a triangle divides the...

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