11th maths notes of lesson_part 2

8/9/2019 11th Maths Notes of Lesson_Part 2

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1 BHARATHIDHASANAR MATRIC HR.SEC.SCHOOL, ARAKKONAM,MATHEMATICS-PART-2

8. Find the sum of the first

(i) 75 positive integers

1+2+3+……+75 is an arithmetic series

Here a = 1, d = 2-1 = 1 and n = 75

Now, s n = [2a+(n-1)d]

S 75 = [2(1) +(75-1)(1) ]

= [2+74] = [76]

S 75 = 2850

(ii) 125 natural numbers

1+2+3+……+75 is an arithmetic series

Here a = 1, d = 1, n=l=125

S 125 = x [1+125]

= x 126 =125 x 63

S 125 = 7875

9. Find the sum of the first 30 terms of an A.P. whose n th term is 3 + 2 n .

Given that the n th term of an Ap is 3+2n

Now, t n = 3 + 2n = 5 + ((n-1) (2) of the form of a + (m-1)d


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` s 30 = [2(5)+(30-1)(2)]

= 15[10+58] = 15 x 68

s 30 = 1020

10. Find the sum of each arithmetic series

(i) 38 + 35 + 32 +……..+ 2.

38 + 35 + 32 +……..+ 2 is an AP

Here, a = 38, d = 35 – 38 = -3, l = 2

Now, l = a +( n - 1)d → n = + 1 = +1 = 13

Thus S 13 = [2+38] = [40] =260

(ii) 6 + 5 +4 + …….. 25 terms

Here a = 6 d = 5 -6 = and n = 25

S n = [2a + (n - 1)d]

S 25 = [2(6) + (25 - 1) ( )]

= [12 + 24 ( )]

= [12 – 1 8] = 73

11. Find the S n for the following arithmetic series described. (i) a = 5, n = 30 , l = 31


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S n = [a + l]

S 30 = [5 +121] = 15 [126] = 1890

(ii) a = 50 n = 25 d = -4

S n = [2a + (n - 1)d]

S 25 = [2(50) + (25 - 1) (-4)]

= [100+24 + (-4)]

= [100 - 96] = [4]

= 50

12 . In the arithmetic sequence 60, 56, 52, 48…. starting from the first term,how many terms are needed so that their sum is 368?

Given arithmetic sequence is 60, 56, 52 , 48 ,….. Here, a = 60 , d = 56 – 60 = 52 - 56 = -4Also s n = 368

Let us find the number of terms needed

S n = [2a + (n - 1)d]

368 = [2 (60) + (n-1)-4]

[120 – 4n + 4] = 368


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[124-4n] = 368

2n 2 – 62n + 368 = 0(n - 8) (n – 23 ) =0 n = 8 or 23Hence 8 terms or 23terms are needed to get the sum 368.

13. A construction company will be penalised each day for delay inconstruction of a bridge. The penalty will be `4000 for the first day and willincrease by Rs . 1000 for each following day. Based on its budget, the companycan afford to pay a maximum of Rs 1,65,000 towards penalty. Find themaximum number of days by which the completion of work can be delayedPenalty amount to be levied for consecutive days form an arithmetic series witha = 4000, d = 1000Let n be the maximum no. of days for which the work can be delayed

Then, s n = 1, 65, 000

→ [2(4000) + (n-1)(1000)] = 1, 65, 000

n[8000+100n-1000] = 3,30,000

n 2 + 7n – 330 = 0(n+22) (n - 15) = 0

→ n = 15 or n = -22maximum no. of days for which the work can be delayed is 15

14. A sum of `1000 is deposited every year at 8% simple interest. Calculate theinterest at the end of each year. Do these interest amounts form an A.P.? If so,find the total interest at the end of 30 years.Every year Rs.1000 is deposited at 8 % simple interest

Interest for first year, t 1 = 1000 X = 80

Interest for second year, t 2 = 2000 X = 160

Thus, the interest amounts 80, 160, 240, …. At the end of the each year formsan A.P

With a = 80 and d = 80


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= 156 times.

6 MARKS

1. In an arithmetic series, the sum of first 11 terms is 44 and that of the next11 terms is 55. Find the arithmetic series.

We have S n = [2a + (n - 1)d]

Thus S 11 = 44 → [2a + (11 – 1)d] = 44

A + 5d = 4 ……(1) S 22 = S 11 + 55 = 44 + 55 =99

[2a + (22 - 1)d] = 99 → 2a + 21d = 99 …..(2)

Solving (1) and (2) we get a = and d =

The required arithmetic s eries is a + (a + d) +(a + 2d)+…..

= + ( + )+ ( + ) + …….

= + + ….

2. Find the sum of all 3 digit natural numbers, which are divisible by 9.

The sequence of 3 digit number which are divisible by 9 are

108,117, 126 , …..,999 This is an Ap where a = 108 , d = 9 and l = 999

Also , l = a + (n - 1)d → n = + 1

= + 1 + 1 = 99 + 1 = 100

NOW, S N = (a +1) =>s 100 = [999+108]


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= 50(1107) = 55350.

3. Find the sum of first 20 terms of the arithmetic series in which 3rd term is7 and 7th term is 2 more than three times its 3rd term.

Solution :Given that r 3= 7 and 7 7 = 2 = 3 t 3 = 23Now, t n =a +(n- 1) d

Thus a+2d =7A +6d = 23

(2) – (1) => 4d = 16 => d = 4Substituting d =4 in (1) we get a + 2( 4) = 7 => a = -1

Now, s n = ({2a = ( n-1 ) d} =>

EXERCISE : 4.2

Problem 1Calculate the dot-product of two vectors u and v , if theirmagnitudes are | u | = , | v | = 1 and the angle between thevectors is = 45°.

Solution Use the formula of the definition dot-product (see the lessonIntroduction to dot-product under the current topic in this site). You

have

(u ,v ) = |u|*|v|* = *1* = = .

Problem 2

Find the dot-product of two vectors u and v , if their magnitudes areu = 1, v = between the vectors is = 135°.

Solution Use the formula of the definition dot-product (see the lessonIntroduction to dot-product under the current topic in this site). You

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have

(u ,v ) = |u|*|v|* = * *( ) = = .Problem 3

Find the scalar product of the vectors u = (5,4) and v = (3,2).Solution Use the standard formula for scalar product as the sum of products ofthe vectors' components (derived in the lesson Formula for Dot-product of vectors in a plane via the vectors components under thecurrent topic in this site). You have(u ,v ) = 5*3 + 4*2 = 15 + 8 = 23.Problem 4Find the scalar product of the vectors u = ( , ) and v = ( , ).Solution Use the standard formula for scalar product as the sum of products ofthe vectors' components (derived in the lesson Formula for Dot-product of vectors in a plane via the vectors components under thecurrent topic in this site). You have(u ,v ) = * + = = .Problem 5Find the scalar product of the vectors u = ( , ) and v =( , ).Solution Use the standard formula for scalar product as the sum of products ofthe vectors' components (derived in the lesson Formula for Dot-product of vectors in a plane via the vectors components under thecurrent topic in this site). You have

(u ,v ) = ( )*( ) + = = = .Problem 6Find the angle between the vectors u = (8,-1) and v = (4,7).

Solution Let be the angle between the given vectors. According to theformula (4) of the lesson Dot-product of vectors in a coordinate

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plane and the angle between two vectors under the current topic inthis site,

= = = = = = .Hence,

= =~ 1.176 rad = 67.41°.

Answer . = 67.41°.

Problem 7 The unit vector u makes the angle of 45° with the positive direction ofthe x -axis ( Figure 1 ).

The unit vector v makes the angle of 60° with the positive directionof the x -axis.Find cosines of the angle between the vectors u and v , which isequal to 15°.

Solution

Since the vector u is the unit vector, its x - and y -

components are cos(45°) = and sin(45°) = .So, the component form of the vector u is u =

(cos(45°), sin(45°)) = ( , ).Similarly, the component form of the vector v is v =

(cos(60°), sin(60°)) = ( , ).

According to the formula for cosines of the angle

between the vectors u and v (see the lessonDot-product of vectors in a coordinate plane and theangle between two vectors under the current topic inthis site),

Figure 1 . Tothe Problem7

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= .

Take into account that | u | = 1 and | v | = 1, because the vectors u and v are unit vectors, and simplify the numerator. You will get

= .

Answer . cos(15°) = .

Problem 8Prove that the vectors u = (5,2) and v = (2,-5) in a coordinate planeare perpendicular.Solution

The scalar product of these vectors is ( u ,v ) = 5*2 + 2*(-5) = 10 - 10 =0.Since the scalar products of these vectors is equal to zero, the vectorsare perpendicular (see the lesson Perpendicular vectors in acoordinate plane under the current topic in this site).Problem 9Prove that the vectors u = ( ,1) and v = ( , ) in acoordinate plane are perpendicular.Solution

The scalar product of these vectors is ( u ,v ) = ( )*( )

+ = = = .

Since the scalar products of the vectors equal to zero, the vectors areperpendicular (see the lesson Perpendicular vectors in a coordinate

plane under the current topic in this site).Problem 10

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triangle. The proof is completed.Problem 12

Prove that the quadrilateral ABCD with the

vertices in a coordinateplaneA(-3,-4), B(5,-3), C (1,4) and D(-7,3) ( Figure3 ) is a rhombus.

Solution We will check that the vectors AB andDC representing the pair of opposite

sides are equal, as well as the vectors BC and AD representing another pair

of opposite sides are equal.

x - and y - components of the vector AB are5-(-3) = 8 and (-3)-(-4) = 1 respectively.

So, the component form of the vector AB isAB= (8,1).

x - and y - components of the vector BC are1-5 = -4 and 4-(-3) = 7 respectively.

So, the component form of the vector BC isBC = (-4,7).

x - and y - components of the vector DC are1-(-7) = 8 and 4-3 = 1 respectively.

So, the component form of the vector DC isDC = (8,1).

Figure 3 . Tothe Problem 12

x - and y - components of the vector AD are -7-(-3) = -4 and 3-(-4) =7 respectively.

So, the component form of the vector AD is AD= (-4,7).

Thus the vectors AB and DC have identical components. Therefore,these vectors are equivalent, i.e. they are parallel and have the same


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length.

The vectors BC and AD have identical components too. Therefore,these vectors are equivalent too, i.e. they are parallel and have the

equal length also.

Since the quadrilateral ABCD has the opposite sides parallel, thisquadrilateral is a parallelogram.

Now, compare the lengths of the two consecutive sides of theparallelogram ABCD .

The length of the side AB is equal to the length of the vector AB :

| AB | = = = . The length of the side BC is equal to the length of the vector BC :

| BC | = = = .

Thus the parallelogram ABCD has the opposite and the consecutivesides of equal length. In other words, all the sides of theparallelogram ABCD are of equal length.Hence, the parallelogram ABCD is a rhombus. The proof iscompleted.Problem 13

Prove that the quadrilateral ABCD with thevertices in a coordinate planeA(1,2), B(2,-1), C(5,0) and D(4,3) ( Figure 4 ) is asquare.

Solution We will check that the vectors AB and DC are

equal, the vectors BC and AD are equal. Then we will prove that all the sides ofthe quadrilateral are of equal length

Figure 4 . Tothe Problem 13


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and the angle L BAD is the right angle.

x - and y - components of the vector AB are 2-1= 1 and (-1)-2 = -3 respectively.

So, the component form of the vector AB is AB=(1,-3).

x - and y - components of the vector BC are 5-2= 3 and 0-(-1) = 1 respectively.So, the component form of the vector BC is BC =(3,1).

x - and y - components of the vector DC are 4-5= -1 and 3-0 = 3 respectively.

So, the component form of the vector DC is DC =(-1,3).

x - and y - components of the vector AD are 4-1 = 3 and 4-3 = 1respectively.

So, the component form of the vector AD is AD= (3,1).

Thus the vectors AB and DC have identical components. Therefore,these vectors are equivalent, i.e. they are parallel and have the samelength.

The vectors BC and AD have identical components too. Therefore,these vectors are equivalent too, i.e they are parallel and have thesame length.Since the quadrilateral ABCD has the opposite sides parallel, thisquadrilateral is a parallelogram.Now, compare the lengths of the two consecutive sides of theparallelogram ABCD .

The length of the side AB is equal to the length of the vector AB :

| AB | = = = . The length of the side BC is equal to the length of the vector BC :


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| BC | = = = . Thus the parallelogram ABCD has the opposite and the consecutivesides of equal length. In other words, all the sides of the

parallelogram ABCD are of equal length.Now, the scalar product of the vectors AB and AD is 1*3 + (-3)*1 =3 - 3 = 0.Since the scalar product of the vectors AB and AC is zero, thesevectors are perpendicular. Thus the parallelogram ABCD has theright angle.Hence, the quadrilateral ABCD is a square. The proof is completed.

BINOMIAL SERIES:-

Exercise 4.3

1.Expand ( x 2 + 3) 6

Students trying to do this expansion in their heads tend to mess upthe powers. But this isn't the time to worry about that square onthe x . I need to start my answer by plugging the terms and powerinto the Theorem. The first term in the binomial is " x 2", the secondterm in "3", and the power n is 6, so, counting from0 to 6, the

Binomial Theorem gives me:(x 2 +3) 6 = 6C 0 (x 2)6 (3) 0 + 6C 1 (x 2)5 (3) 1 + 6C 2 (x 2)4(3) 2 + 6C 3 (x 2)3(3) 3


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+ 6C 4 (x 2)2(3) 4 + 6C 5 (x 2)1 (3) 5 + 6C 6 (x 2 )0(3) 6

Then simplifying gives me

(1)( x 12 )(1) + (6)( x 10 )(3) + (15)( x 8)(9) + (20)( x 6)(27)

+ (15)( x 4)(81) + (6)( x 2)(243) + (1)(1)(729)

= x 12 + 18 x 10 + 135 x 8 + 540 x 6 + 1215 x 4 + 1458 x 2 +729

2.Expand (2 x – 5 y )7

I'll plug "2 x ", " – 5 y ", and "7" into the Binomial Theorem, counting upfrom zero to seven to get each term. (I mustn't forget the "minus"

sign that goes with the second term in the binomial.)

(2 x – 5 y )7 = 7C 0 (2 x )7( – 5 y )0 + 7C 1 (2 x )6( – 5 y )1 + 7C 2 (2 x )5 ( – 5 y )2

+ 7C 3 (2 x )4( – 5 y )3 + 7C 4 (2 x )3 ( – 5 y )4 + 7C 5 (2 x )2( – 5 y )5

+ 7C 6 (2 x )1( – 5 y )6 + 7C 7 (2 x )0 ( – 5 y )7

Then simplifying gives me: Copyright © Elizabeth Stapel 1999-2009All Rights Reserved

(1)(128 x 7)(1) + (7)(64 x 6)( – 5 y ) + (21)(32 x 5)(25 y 2 ) + (35)(16 x 4)( – 125 y 3)

+ (35)(8 x 3)(625 y 4) + (21)(4 x 2)( – 3125 y 5) + (7)(2 x )(15625 y 6)

+ (1)(1)( – 78125 y 7)

= 128 x 7 – 2240 x 6 y + 16800 x 5 y 2 – 70000 x 4 y 3 +175000 x 3 y 4 – 262500 x 2 y 5

+ 218750 xy 6 – 78125 y 7

You may be asked to find a certain term in an expansion, the idea beingthat the exercise will be way easy if you've memorized the Theorem, but


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will be difficult or impossible if you haven't. So memorize theTheoremand get the easy points.

3.What is the fourth term in the expansion of (3 x – 2) 10 ?

I've already expanded this binomial, so let's take a look:

(3 x – 2) 10 = 10 C 0 (3 x )10 – 0( – 2) 0 + 10 C 1 (3 x )10 – 1( – 2) 1 + 10 C 2 (3 x )10 – 2( – 2) 2

+ 10 C 3 (3 x )10 – 3 ( – 2) 3 + 10 C 4 (3 x )10 – 4( – 2) 4 + 10 C 5 (3 x )10 – 5( – 2) 5

+ 10 C 6 (3 x )10 – 6( – 2) 6 + 10 C 7 (3 x )10 – 7( – 2) 7 + 10 C 8 (3 x )10 – 8( – 2) 8

+ 10 C 9 (3 x )10 – 9( – 2) 9 + 10 C 10 (3 x )10 – 10 ( – 2) 10

So the fourth term is not the one where I've counted up to 4, butthe one where I've counted up just to 3. (This is because, just aswith Javascript, the counting starts with 0, not 1.)

Note that, in any expansion, there is one more term than thenumber in the power. For instance:

(x + y )2 = x 2 + 2 xy + y 2 (second power: three terms)

(x + y )3 = x 3 + 3 x 2y + 3 xy 2 + y 3 (third power: four terms)

(x + y )4 = x 4 + 4 x 3y + 6 x 2y 2 + 4 xy 3 + y 4 (fourth power: fiveterms)

The expansion in this exercise, (3 x – 2) 10 , has power of n = 10, sothe expansion will have eleven terms, and the terms will count up,not from 1 to 10 or from 1 to 11, but from 0 to 10. This is why thefourth term will not the one where I'm using "4" as my counter, butwill be the one where I'm using "3".

10 C 3 (3 x )10 – 3( – 2) 3 = (120)(2187)( x 7)( – 8) = – 2099520 x 7

4.Find the tenth term in the expansion of ( x + 3) 12 .


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To find the tenth term, I plug x , 3, and 12 into the Binomial Theorem, using the number 10 – 1 = 9 as my counter:

12 C 9 (x )12 – 9(3) 9 = (220) x 3 (19683) = 4330260 x 3

5.Find the middle term in the expansion of (4 x – y )8 .

Since this binomial is to the power 8, there will be nine terms inthe expansion, which makes the fifth term the middle one. So I'llplug 4 x , – y , and 8 into the Binomial Theorem, using the number5 – 1 = 4 as my counter.

8C 4 (4 x )8 – 4( – y )4 = (70)(256 x 4 )(y 4 ) = 17920 x 4 y 4

You might be asked to work backwards.

6.Express 1296 x 12 – 4320 x 9 y 2 + 5400 x 6 y 4 – 3000 x 3 y 6 + 625 y 8 inthe form ( a + b )n .

I know that the first term is of the form a n , because, forwhatever n is, the first term is n C 0(which always equals 1)times a n times b 0 (which also equals 1). So 1296 x 12 = a n . By thesame reasoning, the last term is b n , so 625 y 8 = b n . And since thereare alternating "plus" and "minus" signs, I know from experiencethat the sign in the middle has to be a "minus". (If all the signs had

been "plusses", then the middle sign would have been a "plus" also.But in this case, I'm really looking for "( a – b )n ".)

I know that, for any power n , the expansion has n + 1 terms. Sincethis has 5 terms, this tells me that n = 4. So to find a and b , I onlyhave to take the 4th root of the first and last terms of the expandedpolynomial:

Then a = 6 x 3 , b = 5 y 2 , there is a "minus" sign in the middle, and:


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1296 x 12 – 4320 x 9y 2 + 5400 x 6y 4 – 3000 x 3y 6 + 625 y 8 = (6 x 3 – 5 y 2 )4

Don't let the Binomial Theorem scare you. It's just another formula tomemorize. A really complicated and annoying formula, I'll grant you, but

just a formula, nonetheless. Don't overthink the Theorem; there isnothing deep or meaningful here. Just memorize it, and move on.

UNIT-5

ANALYTICAL GEOMETRY

Points A(x1;y1),B(x2;y2) and C(x2;y1) are shown in the diagram below:

Theorem of Pythagoras


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AB2=AC2+BC2

Distance formula Distance between two points:AB=(x2−x1)2+(y2−y1)2−−−−−−−−−−−−−−−−−−√

Notice that (x1−x2)2=(x2−x1)2. Gradient Gradient (m) describes the slope or steepness of the line joining twopoints. The gradient of a line is determined by the ratio of vertical changeto horizontal change.Remember to be consistent: m≠y1−y2x2−x1.

Horizontal lines m=0

Vertical lines m is undefined


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Parallel lines m1=m2

Perpendicular lines m1×m2=−1

Table 1

Mid-point of a line segment


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The coordinates of the mid-point M(x;y) of a line between any twopoints A(x1;y1) and B(x2;y2):M(x;y)=(x1+x22;y1+y22)

Points on a straight line The diagram shows points P(x1;y1), Q(x2;y2) and R(x;y) on a straight line.

We know thatmPR=mQR=mPQ

Using mPR=mPQ, we obtain the following for any point (x;y) on a straightline

y −y1x−x1=y2−y1x2−x1

EX.5.1

1. Given the points P (−5 ;−4) and Q(0;6):1. Determine the length of the line segment PQ.2. Determine the mid-point T(x;y) of the line segment PQ.


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3. Show that the line passing through R(1 ;−34) and T(x;y) isperpendicular to the line PQ.

AnswerDraw a sketch

Assign variables to the coordinates of the given points Let the coordinates of P be (x1;y1) and Q(x2;y2)

x1 =−5;y1 =−4;x2=0;y2=6

Write down the distance formula

PQ=(x2 − x1)2+(y2 − y1)2 −−−−−−−−−−−−−−−−−−√=(0 −(−5))2+(6 −(−4))2 −−−−−−−−−−−−−−−−−−−−−√=25+100 −−−−−−−√=125 −−−√=55 √

The length of the line segment PQ is 55 √ units.Write down the mid-point formula and substitute the values


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T(x;y)=(x1+x22;y1+y22)

xy=x1+x22 =−5+02 =−52=y1+y22 =−4+62=22=1

The mid-point of PQ is T (−52;1).Determine the gradients of PQ and RTm=y2 − y1x2 − x1

mPQmRT=6 −(−4)0 −(−5)=105=2 =−34 −11 −(−52 )=−7472 =−74×27 =−12

Calculate the product of the two gradients:mRT×mPQ =−12×2 =−1

2.Therefore PQ is perpendicular to RT.

Points A (−1;0), B(0;3), C(8;11) and D(x;y) are points on the Cartesianplane. Determine D(x;y) if ABCD is a parallelogram.AnswerDraw a sketch


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The mid-point of AC will be the same as the mid-point of BD. We firstfind the mid-point of AC and then use it to determine the coordinates ofpoint D.Assign values to (x1;y1) and (x2;y2)Let the mid-point of AC be M(x;y)

x1 =−1;y1=0;x2=8;y2=11

Write down the mid-point formula M(x;y)=(x1+x22;y1+y22)

Substitute the values and calculate the coordinates of MM(x;y )=(−1+82;0+112)=(72;112)

Use the coordinates of M to determine DM is also the mid-point of BD so we use M(72;112) and B(0;3) tofind D(x;y)Substitute values and determine x and yM (72;112)=(x1+x22;y1+y22)=(0+x2;3+y2)

727 x=0+x2=0+x=7

11211 y=3+y2=3+y=8

Alternative method: inspection Since we are given that ABCD is a parallelogram, we can use theproperties of a parallelogram and the given points to determine thecoordinates of D.From the sketch we expect that point D will lie below C.Consider the given points A,B and C:

Opposite sides of a parallelogram are parallel, therefore BC must beparallel to AD and their gradients must be equal.

The vertical change from B to C is 8 units up. Therefore the vertical change from A to D is also 8 units up

(y=0+8=8).


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The horizontal change from B to C is 8 units to the right. Therefore the horizontal change from A to D is also 8 units to the

right (x =−1+8=7).

or

Opposite sides of a parallelogram are parallel, therefore AB must beparallel to DC and their gradients must be equal.

The vertical change from A to B is 3 units up. Therefore the vertical change from C to D is 3 units down

(y=11 −3=8). The horizontal change from A to B is 1 unit to the right. Therefore the horizontal change from C to D is 1 unit to the left

(x=8 −1=7).Write the final answer

The coordinates of D are (7;8) .10 marks

1. Find the unknown values in each of the following figures. All lengths are

given in

centimeters. (All measures are not in scale)

In ABC and we have


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2Y + 6 =

Y = 2.4 CM

Again consider and here EG // BC

Thus, by AA criterion of similarity, we have

=

= =

3z = 2z+10 z = 10cm

iii) EFCD is a parallelogram

so, EF = DC = 7CM . Consider

from the figure it is clear that

=

= 12X = 7X + 42

X = 8.4 CM

Consider and

From the figure , it is clear that


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=

DG = X CF

Y = x 6 = 2.5 cm

2. The image of a man of height 1.8 m, is of length 1.5 cm on the film of a

camera. If the

film is 3 cm from the lens of the camera, how far is the man from the camera?

Let AB be the height of the man

CD be the the height of the image of the man of height 1.8 m,

L be the position of the lens of the camera,

LM be the distance between man and lens

LN be the distance between lens and film

Then, AB// CD AB = 1.8m, CD = 1.5m, ln = 3cm

Consider LAB and LCD we have


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LM= x LN = 360cm =3.6m

Hence the distance between the man and the camera is 3.6m

3. A girl of height 120 cm is walking away from the base of a lamp-post at a

speed

of 0.6 m/sec. If the lamp is 3.6 m above the ground level, then find the length

of her

shadow after 4 seconds.

Let AB the height of the lamp post

CD be the height of the girl

CE be the length of the shadow of the girl

Then, AB = 3.6m , CD = 120cm, CM = 1.2m

Given that the walking speed of the girl is 0.6 m/sec.

The distance AC travelled by the girl in 4 second = 4 X 0.6 = 2.4m

Now consider Clearly CD // Ab.


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= =

3EC = 1.2 m

Hence , the length of the shadow of the girl after 4 seconds is 1.2m

5. P and Q are points on sides AB and AC respectively, of D ABC . If AP = 3 cm,

PB = 6 cm, AQ = 5 cm and QC = 10 cm, show that BC = 3 PQ .

Now , = =

From ∆APQ and ∆ABC ,we have

=

AND


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= = ; = =

Thus in ∆BCD and ∆ACB , we have

=

And


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= 72 – 8 = 64cm 2

8. The lengths of three sides of a triangle ABC are 6 cm, 4 cm and 9 cm. 3 PQR

+3 ABC .One of the lengths of sides of 3 PQR is 35cm. What is the greatest perimeter

possible

for ∆PQR ?Given , ∆ PQR~ ∆ABC

= =

Let QR = 35

From (1) , we see that the perimeter of ∆PQR is the greatest only when thecorresponding side to QR must be BC

= =

In the figure, DE || BC and = calculate the value of i)

In ABC DE// BC

ABC

Now, = AD=3k , BD = 5k

Also, = == =

ii)

area of = 9k and area of = 64k

now, area of trapezium BCED

= area of - area of


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= 64k-9k =55k

=

=

10. The government plans to develop a new industrialzone in an unused portion of land in a city.

The shaded portion of the map shown on theright, indicates the area of the new industrial zone.Find the area of the new industrial zone.

Consider ∆EAB, ∆EDC , clearly AB//CD


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We know that if a perpendicular is drawn from the vertex of a right angled

triangle to its hypotenuse then the triangle on the each side of the

perpendicular are similar. So, ∆EAD~EDC. thus we have =

ED 2 = EA X EC = 16 X 81

= 16X81 = 4X9 = 36.

Now, ∆ABD is an isosceles triangle an AE perpendicular to BD so , BE = ED

Thus BD = 2ED = 2x36 = 72cm

12. A student wants to determine the height of a flagpole. He placed a smallmirror on the ground so that he can see the reflection of the top of the flagpole.

The distance of the mirror from him is 0.5 m and the distance of the flagpolefrom the mirror is 3 m. If hiseyes are 1.5 m above the ground level, then find the height of the flagpole.(Thefoot ofstudent, mirror and the foot of flagpole lie along a straight line).

Let AB and ED be the height of a man and the tower respectively

Let c be the point of incidence of the flag pole in the mirror.

In ∆ABC and ∆EDC ,


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II) ∆WYX ~ ∆ YZX iii) ∆WZY ~ ∆ WYX

(ii) Find the height h of the roof.

=

=

H = 4.8 m

EQUATIONS OF A STRAIGHT LINE

Line through two points

The line through two distinct points (x 1 , y 1) and (x 2 , y 2) is given by

(1) y = y 1 + [(y 2 - y 1 ) / (x 2 - x 1)]·(x - x 1),

where x 1 and x 2 are assumed to be different. In case they are equal, theequation is simplified to

x = x 1

and does not require a second point.

Equation (1) can also be written as

y - y 1 = [(y 2 - y 1 ) / (x 2 - x 1)]·(x - x 1 ),

or even as

(x2 - x 1 )·(y - y 1 ) = (y 2 - y 1)·(x - x 1 ),


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where one does not have to worry whether x 1 = x 2 or not. However, thesimplest for me to remember is this

(y - y 1 )/(y 2 - y 1) = (x - x 1)/(x 2 - x 1 )

which is not as universal is the one before.

Ex 5.2

1.Find the equation of the straight line that has slope m = 4and passes through the point ( – 1, – 6).

Okay, they've given me the value of the slope; in this case, m =4. Also, in giving me a point on the line, they have given me an x -value and a y -value for this line: x = – 1 and y = – 6.

In the slope-intercept form of a straight line, I have y , m , x , and b .So the only thing I don't have so far is a value for is b (which givesme the y -intercept). Then all I need to do is plug in what they gaveme for the slope and the x and y from this particular point, andthen solve for b :

y = mx + b( – 6) = (4)( – 1) + b

– 6 = – 4 + b – 2 = b

Then the line equation must be " y = 4 x – 2 ".

What if they don't give you the slope?

2.Find the equation of the line that passes through the points ( – 2,4) and (1, 2).


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Well, if I have two points on a straight line, I can always find theslope; that's what the slope formula is for.

Now I have the slope and two points. I know I can find the equation(by solving first for " b ") if I have a point and the slope. So I need topick one of the points (it doesn't matter which one), and use it tosolve for b. Using the point ( – 2, 4), I get:

y = mx + b4 = ( – 2 / 3)( – 2) + b4 = 4 / 3 + b4 – 4 / 3 = b12/3 – 4 / 3 = bb = 8/3

...so y = ( – 2 / 3 ) x + 8 / 3 .

On the other hand, if I use the point (1, 2), I get:

y = mx + b2 = ( – 2 / 3)(1) + b2 = – 2 / 3 + b2 + 2 / 3 = b6 / 3 + 2 / 3 = bb = 8 / 3

So it doesn't matter which point I choose. Either way, the answer isthe same:

y = ( – 2 / 3 )x + 8 / 3

As you can see, once you have the slope, it doesn't matter which point

you use in order to find the line equation. The answer will work out thesame either way.

http://www.purplemath.com/modules/slope.htmhttp://www.purplemath.com/modules/slope.htmhttp://www.purplemath.com/modules/slope.htmhttp://www.purplemath.com/modules/slope.htm


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In the figure TP is a tangent to a circle. A and B are two pointson

the circle. If < BTP = 72° and < ATB = 43° find < ABT

solution .


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3 . AB and CD are two chords of a circle which intersect each other externallyat P

(i) If AB = 4 cm BP = 5 cm and PD = 3 cm, then find CD .

Since the chord AB and CD meet externally at P , We have

PA X PB = PC x PD

9 X 5 = [3+CD](3)

3+CD = = 15cm

Thus CD = 12CM

(ii) If BP = 3 cm, CP = 6 cm and CD = 2 cm, then find AB

Given that CP = 6cm

CD + PD = 6cm PD = 4cm

PA X PB = PC x PD

i.e., (AB+PB) X PB = PC x PD

(3+AB) X 3 = 6 X 4

3+AB = = 8 cm

AB = 5 cm

10 mark

4. A circle touches the side BC of T ABC at P, AB and AC produced at Q and Rrespectively,prove that AQ = AR = ( perimeter of ∆ABC )

we know that the length of a two tangent drawn to a circle from an externalpoint are equal. Thus we have

BQ = BP [tangent from the point B ]


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CP = CR [tangent from the point C ]

AQ = AR [tangent from the point A ]

Perimeter of ∆ABC = AB + BC+ CA = AB + BP PC CA

= (AB + BP) + (PC+ CA) 1

= (AB + BQ) + (CR + CA) [USING (1) AND (2)]

= AQ +AR

= AR + AR = 2AR (perimeter of ∆ABC )

Thus from (3) we have AR = AQ = ( perimeter of ∆ ABC )

5. If all sides of a parallelogram touch a circle, show that the

parallelogram is a rhombusLet ABCD be the parallelogram . the side AB and ,BC ,CD and DA touch the

circle at the point P,Q,R and S respectively

we know that the length of a two tangent drawn to a circle from an external

point are equal Thus AP = AS ……… (1) BP = BQ ……… (2) CR = CQ ………. (3) DR = DS ………. (4)

Adding (1), (2), (3) and (4), we get

AP + BP +CR+DR = AS + BQ + CQ + DS

→ (AP + BP) + (CR+DR) = (AS + BQ) + (CQ + DS)AB + CD = AD = BC [ABCD is a parallelogram

2AB = 2AD AB = CD & BC = AD ]

AB = AD

Thus we have AB = BC =CD =DA

ABCD is a rhombus.

6. A lotus is 20 cm above the water surface in a pond and its stem is partlybelow the


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water surface. As the wind blew, the stem is pushed aside so that the lotustouched thewater 40 cm away from the original position of the stem. How much of the stemwasbelow the water surface originally?

Let o be bottom of the stem immersed in water . Let B be the lotus . Then Ab be

the length of the stem above the water surface and OA be the length of the

stem below the water surface.

Let OA = x cm

Let c be the point where the lotus touches water surface and OA be the stem

below the water surface

Now, OC = OA + AB = X + 20cmby Pythagoras theorem, we have OC 2 = OA 2 + AC 2

(x + 20) 2 = x 2+ 40 2

X2 + 40x +400 = x 2 +1600

40x = 1200

X = 30cm

Thus the stem is 30 cm below the water surface.

7. A point O in the interior of a rectangle ABCD is joined to each of the verticesA , B , Cand D . Prove that OA 2 + OC 2 = OB 2 + OD 2

Through O , draw EOF \\ AB

Now , ABFE and EFCD are rectangles

In right ∆OEA , by Pythagoras the orem OA 2 = OE 2 + EA 2 ………..(1) In right ∆OFC , by Pythagoras theorem OC 2 = OF 2 + FC 2 ……….. (2)

In right ∆OFB , by Pythagoras theorem OB2 = OF

2 + FB

2 ……….. (3)

In right ∆OED , by Pythagoras theorem OD 2 = OE 2 + ED 2 ………..(4) Adding (3) and (4)


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OB 2 + OD 2 = OF 2 + FB 2 + OE 2 + ED 2

= (OE 2 + FB 2 ) + (OF 2 + ED 2 )

= (OE 2 + EA 2 ) + (OF 2 + FC 2 )

[ ABCD and EFCD are rectangles FB = EA

And ED = FC]

= OA 2 + OC 2

1 . If a straight line intersects the sides AB and AC of a 9 ABC at D and Erespectively andis parallel to BC , then

By Thales theorem =

ANS :2 . In 9 ABC, DE is < to BC , meeting AB and AC at D and E .If AD = 3 cm, DB = 2 cm and AE = 2.7 cm , then AC is equal to

By Thales theorem = → EC =

EC = = 1.8cm

Thus AC = AE + EC = 2.7 + 1.8 = 4.5cm

\ ANS ; 4.5cm\

3. In ∆PQR, RS is the bisector of < R . If PQ = 6 cm, QR = 8 cm, RP = 4 cmthen PS is equal to

Let PS = x cm. SQ = 6-x cm

Rs is a bisector of


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4 . In figure, if =


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Height of the tower = X 40

= 60m

9. The sides of two similar triangles are in the ratio 2:3, then their

areas are in the ratio

2 2 : 3 2 = 4 : 9

10. Triangles ABC and DEF are similar. If their areas are 100 cm2

and 49 cm2 respectively and BC is 8.2 cm then EF =

= → =

=

EF = = 5.74cm

11. The perimeters of two similar triangles are 24 cm and 18 cm

respectively. If one

side of the first triangle is 8 cm, then the corresponding side of the other

triangle is

=


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= = 6cm

12. AB and CD are two chords of a circle which when produced to meet

at a point

P such that AB = 5, AP = 8, and CD = 2 then PD =

Let PD = x cm PA x PB = PC x PD

8 x 3 = (x+2) x

X2 +2x -24 = 0

X = 4 cm

13. In the adjoining figure, chords AB and CD intersect at P such that If

AB = 16 cm, PD = 8 cm, PC = 6 and AP >PB, then AP =

Let Pa = x cm PA x PB = PC x PD

X (16-X) = 6 X 8

X2 -16X + 48 = 0 → (x-4) (x-12) = 0X = 4 or X = 12

But AP > PB AP = 12 cm

14. A point P is 26 cm away from the centre O of a circle and PT is the

tangent

drawn from P to the circle is 10 cm, then OT is equal to

OP 2 = OT 2 +TP 2

OT 2 = 26 2 – 10 2 = 24


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15. In the figure, if < PAB = 120° then < BPT =


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48 BHARATHIDHASANAR MATRIC HR.SEC.SCHOOL, ARAKKONAM,

\

19. The areas of two similar triangles are 16 cm2 and 36 cm2respectively. If the altitudeof the first triangle is 3 cm, then the corresponding altitude of the other

triangle is

=

Altitude of second triangl =

= 4.5 cm

20. The perimeter of two similar triangles ∆ABC and ∆DEF are 36 cmand 24 cm

Respectively. If DE = 10 cm, then AB is

=

AB =

= 15cm

11th maths notes of lesson_part 2

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