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Answer

In Δ ABC,∠B = 90º
By Applying Pythagoras theorem , we get

AC² = AB² + BC² = (24)² + 7² = (576+49) cm² = 625 cm²
⇒ AC = 25

(i) sin A = BC/AC = 7/25    cos A = AB/AC = 24/25
(ii) sin C = AB/AC = 24/25
     cos C = BC/AC = 7/25

2.  In Fig. 8.13, find tan P – cot R.

Answer

By Applying Pythagoras theorem in ΔPQR , we get
PR² = PQ² + QR² = (13)² = (12)² + QR² = 169 = 144  + QR²
⇒  QR² = 25 ⇒  QR = 5 cm

Let BC be 3k and AC will be 4k where k is a positive real number.

By Pythagoras theorem we get,
AC² = AB² + BC² 
(4k)² = AB² + (3k)²
16k² – 9k² = AB²
AB² = 7k²
AB = √7 k

4. Given 15 cot A = 8, find sin A and sec A.

By Pythagoras theorem we get,
OP² = OM² + MP² 
(13k)² = (12k)² + MP² 
169k² – 144k² = MP²
MP² = 25k²

MP = 5

Answer

Let ΔABC in which CD ⊥ AB.

A/q,

By applying Pythagoras theorem in ΔCAD and ΔCBD we get,
CD² = AC² – AD² …. (iii)
and also CD² = BC² – BD² …. (iv)
From equations (iii) and (iv) we get,
AC² – AD² = BC² – BD²
⇒ (kBC)² – (k BD)² = BC² – BD²
⇒ k² (BC² – BD²) = BC² – BD²
⇒ k² = 1
⇒ k = 1
Putting this value in equation (ii), we obtain
AC = BC
⇒ ∠A = ∠B  (Angles opposite to equal sides of a triangle are equal-isosceles triangle)

Let ΔABC in which ∠B = 90º and ∠C = θ

Let BC = 7k and AB = 8k, where k is a positive real number.
By Pythagoras theorem in ΔABC we get.
AC² = AB² + BC² 
AC² = (8k)² + (7k)²
AC² = 64k² + 49k²
AC² = 113k²
AC = √113 k

8.  If 3cot A = 4/3 , check whether (1-tan²A)/(1+tan²A) = cos²A – sin²A or not.

Answer


Let ΔABC in which ∠B = 90º,
A/q,
cot A = AB/BC = 4/3
Let AB = 4k and BC = 3k, where k is a positive real number.
By Pythagoras theorem in ΔABC we get.
AC² = AB² + BC² 
AC² = (4k)² + (3k)²
AC² = 16k² + 9k²
AC² = 25k²
AC = 5k
tan A = BC/AB = 3/4
sin A = BC/AC = 3/5
cos A = AB/AC = 4/5
L.H.S. = (1-tan²A)/(1+tan²A) = 1- (3/4)²/1+ (3/4)² = (1- 9/16)/(1+ 9/16) = (16-9)/(16+9) = 7/25
R.H.S. = cos²A – sin²A = (4/5)² – (3/4)² = (16/25) – (9/25) = 7/25
R.H.S. = L.H.S.
Hence,  (1-tan²A)/(1+tan²A) = cos²A – sin²A

9. In triangle ABC, right-angled at B, if tan A =1/√3 find the value of:
(i) sin A cos C + cos A sin C
(ii) cos A cos C – sin A sin C

Answer

Let ΔABC in which ∠B = 90º,
A/q,
tan A = BC/AB = 1/√3
Let AB = √3 k and BC = k, where k is a positive real number.
By Pythagoras theorem in ΔABC we get.
AC² = AB² + BC² 
AC² = (√3 k)² + (k)²
AC² = 3k² + k²
AC² = 4k²
AC = 2k
sin A = BC/AC = 1/2                   cos A = AB/AC = √3/2 ,
sin C = AB/AC = √3/2                   cos A = BC/AC = 1/2
(i) sin A cos C + cos A sin C = (1/2×1/2) + (√3/2×√3/2) = 1/4+3/4 = 4/4 = 1
(ii) cos A cos C – sin A sin C = (√3/2×1/2) – (1/2×√3/2) = √3/4 – √3/4 = 0

10. In Δ PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.

Answer

Given that, PR + QR = 25 , PQ = 5
Let PR be x.  ∴ QR = 25 – x

By Pythagoras theorem ,
PR² = PQ² + QR²
x² = (5)² + (25 – x)²
x² = 25 + 625 + x² – 50x
50x = 650
x = 13
∴ PR = 13 cm
QR = (25 – 13) cm = 12 cm

(i) False.

AC² = AB² + BC² 
5² = 3² + 4²
25 = 9 + 16
25 = 25

(ii) True.
Let a ΔABC in which ∠B = 90º,AC be 12k and AB be 5k, where k is a positive real number.
By Pythagoras theorem we get,
AC² = AB² + BC² 
(12k)² = (5k)² + BC² 
BC² + 25k² = 144k²
BC² = 119k²

Such a triangle is possible as it will follow the Pythagoras theorem.

(i) tan 48° tan 23° tan 42° tan 67°
    = tan (90° – 42°) tan (90° – 67°) tan 42° tan 67°
    = cot 42° cot 67° tan 42° tan 67°
    = (cot 42° tan 42°) (cot 67° tan 67°) = 1×1 = 1

(ii) cos 38° cos 52° – sin 38° sin 52°
    = cos (90° – 52°) cos (90°-38°) – sin 38° sin 52°
    = sin 52° sin 38° – sin 38° sin 52° = 0

A/q,
tan 2A = cot (A- 18°)
⇒ cot (90° – 2A) = cot (A -18°)
Equating angles,
⇒ 90° – 2A = A- 18° ⇒ 108° = 3A
⇒ A = 36°

tan A = cot B
⇒ tan A = tan (90° – B)
⇒ A = 90° – B
⇒ A + B = 90°

A/q,
sec 4A = cosec (A – 20°)
⇒ cosec (90° – 4A) = cosec (A – 20°)

sin 67° + cos 75°
= sin (90° – 23°) + cos (90° – 15°)
= cos 23° + sin 15°

cosec²A – cot²A = 1
⇒ cosec²A = 1 + cot²A
⇒ 1/sin²A = 1 + cot²A
⇒sin²A = 1/(1+cot²A)

Now,
sin²A = 1/(1+cot²A)
⇒ 1 – cos²A = 1/(1+cot²A)
⇒cos²A = 1 – 1/(1+cot²A)
⇒cos²A = (1-1+cot²A)/(1+cot²A)
⇒ 1/sec²A = cot²A/(1+cot²A)
⇒ secA = (1+cot²A)/cot²A

also,
tan A = sin A/cos A and cot A = cos A/sin A
⇒ tan A = 1/cot A

2. Write all the other trigonometric ratios of ∠A in terms of sec A.

Answer

We know that,
sec A = 1/cos A
⇒ cos A = 1/sec A
also,
cos²A + sin²A = 1
⇒  sin²A = 1 – cos²A
⇒  sin²A = 1 – (1/sec²A)
⇒  sin²A = (sec²A-1)/sec²A

also,
sin A = 1/cosec A
⇒cosec A = 1/sin A

Now,
sec²A – tan²A = 1
⇒ tan²A = sec²A + 1

also,
tan A = 1/cot A
⇒ cot A = 1/tan A


3. Evaluate :
(i) (sin²63° + sin²27°)/(cos²17° + cos²73°)
(ii)  sin 25° cos 65° + cos 25° sin 65°

Answer

(i) (sin²63° + sin²27°)/(cos²17° + cos²73°)
   = [sin²(90°-27°) + sin²27°]/[cos²(90°-73°) + cos²73°)]
   = (cos²27° + sin²27°)/(sin²27° + cos²73°)
   = 1/1 =1                       (∵ sin²A + cos²A = 1)

(ii) sin 25° cos 65° + cos 25° sin 65°
   = sin(90°-25°) cos 65° + cos(90°-65°) sin 65°
   = cos 65° cos 65° + sin 65° sin 65°
   = cos²65° + sin²65° = 1

4. Choose the correct option. Justify your choice.
(i) 9 sec²A – 9 tan²A =
      (A) 1                 (B) 9              (C) 8                (D) 0
(ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ)
      (A) 0                 (B) 1              (C) 2                (D) – 1
(iii) (secA + tanA) (1 – sinA) =
      (A) secA           (B) sinA        (C) cosecA      (D) cosA

= 9 (sec²A – tan²A)
= 9×1 = 9             (∵ sec2 A – tan2 A = 1)

(vii) (sin θ – 2sin³θ)/(2cos³θ-cos θ) = tan θ
(viii) (sin A + cosec A)² + (cos A + sec A)² = 7+tan²A+cot²A
(ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA)
     [Hint : Simplify LHS and RHS separately]
(x) (1+tan²A/1+cot²A) = (1-tan A/1-cot A)² = tan²A

Answer

(i) (cosec θ – cot θ)² = (1-cos θ)/(1+cos θ)
L.H.S. =  (cosec θ – cot θ)²
           = (cosec²θ + cot²θ – 2cosec θ cot θ)
           = (1/sin²θ + cos²θ/sin²θ – 2cos θ/sin²θ)
           = (1 + cos²θ – 2cos θ)/(1 – cos²θ)
           = (1-cos θ)²/(1 – cosθ)(1+cos θ)
           = (1-cos θ)/(1+cos θ) = R.H.S.

(ii)  cos A/(1+sin A) + (1+sin A)/cos A = 2 sec A
 L.H.S. = cos A/(1+sin A) + (1+sin A)/cos A
            = [cos²A + (1+sin A)²]/(1+sin A)cos A
            = (cos²A + sin²A + 1 + 2sin A)/(1+sin A)cos A
            = (1 + 1 + 2sin A)/(1+sin A)cos A
            = (2+ 2sin A)/(1+sin A)cos A
            = 2(1+sin A)/(1+sin A)cos A
            = 2/cos A = 2 sec A = R.H.S.

(iii) tan θ/(1-cot θ) + cot θ/(1-tan θ) = 1 + sec θ cosec θ
L.H.S. = tan θ/(1-cot θ) + cot θ/(1-tan θ)
           = [(sin θ/cos θ)/1-(cos θ/sin θ)] + [(cos θ/sin θ)/1-(sin θ/cos θ)]
           = [(sin θ/cos θ)/(sin θ-cos θ)/sin θ] + [(cos θ/sin θ)/(cos θ-sin θ)/cos θ]
           = sin²θ/[cos θ(sin θ-cos θ)] + cos²θ/[sin θ(cos θ-sin θ)]
           = sin²θ/[cos θ(sin θ-cos θ)] – cos²θ/[sin θ(sin θ-cos θ)]
           = 1/(sin θ-cos θ) [(sin²θ/cos θ) – (cos²θ/sin θ)]
           = 1/(sin θ-cos θ) × [(sin³θ – cos³θ)/sin θ cos θ]
           = [(sin θ-cos θ)(sin²θ+cos²θ+sin θ cos θ)]/[(sin θ-cos θ)sin θ cos θ]
           = (1 + sin θ cos θ)/sin θ cos θ
           = 1/sin θ cos θ + 1
           = 1 + sec θ cosec θ = R.H.S.

(iv)  (1 + sec A)/sec A = sin²A/(1-cos A)
L.H.S. = (1 + sec A)/sec A
           = (1 + 1/cos A)/1/cos A
           = (cos A + 1)/cos A/1/cos A
           = cos A + 1
R.H.S. = sin²A/(1-cos A)
            = (1 – cos²A)/(1-cos A)
            = (1-cos A)(1+cos A)/(1-cos A)
            = cos A + 1
L.H.S. = R.H.S.

(v) (cos A–sin A+1)/(cos A+sin A–1) = cosec A + cot A,using the identity cosec²A = 1+cot²A.
L.H.S. = (cos A–sin A+1)/(cos A+sin A–1)
           Dividing Numerator and Denominator by sin A,
           = (cos A–sin A+1)/sin A/(cos A+sin A–1)/sin A
           = (cot A – 1 + cosec A)/(cot A+ 1 – cosec A)
           = (cot A – cosec²A + cot²A + cosec A)/(cot A+ 1 – cosec A) (using cosec²A – cot²A = 1)
           = [(cot A + cosec A) – (cosec²A – cot²A)]/(cot A+ 1 – cosec A)
           = [(cot A + cosec A) – (cosec A + cot A)(cosec A – cot A)]/(1 – cosec A + cot A)
           =  (cot A + cosec A)(1 – cosec A + cot A)/(1 – cosec A + cot A)
           =  cot A + cosec A = R.H.S.


Dividing Numerator and Denominator of L.H.S. by cos A,

= sec A + tan A = R.H.S.

(vii) (sin θ – 2sin³θ)/(2cos³θ-cos θ) = tan θ
L.H.S. = (sin θ – 2sin³θ)/(2cos³θ – cos θ)
           = [sin θ(1 – 2sin²θ)]/[cos θ(2cos²θ- 1)]
           = sin θ[1 – 2(1-cos²θ)]/[cos θ(2cos²θ -1)]
          = [sin θ(2cos²θ -1)]/[cos θ(2cos²θ -1)]
          = tan θ = R.H.S.

(viii) (sin A + cosec A)² + (cos A + sec A)² = 7+tan²A+cot²A
L.H.S. = (sin A + cosec A)² + (cos A + sec A)²
               = (sin²A + cosec²A + 2 sin A cosec A) + (cos²A + sec²A + 2 cos A sec A)
           = (sin²A + cos²A) + 2 sin A(1/sin A) + 2 cos A(1/cos A) + 1 + tan²A + 1 + cot²A
           = 1 + 2 + 2 + 2 + tan²A + cot²A
           = 7+tan²A+cot²A = R.H.S.

(ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA)
L.H.S. = (cosec A – sin A)(sec A – cos A)
           = (1/sin A – sin A)(1/cos A – cos A)
           = [(1-sin²A)/sin A][(1-cos²A)/cos A]
           = (cos²A/sin A)×(sin²A/cos A)
           = cos A sin A
R.H.S. = 1/(tan A+cotA)
            = 1/(sin A/cos A +cos A/sin A)
            = 1/[(sin²A+cos²A)/sin A cos A]
            = cos A sin A
L.H.S. = R.H.S.

(x)  (1+tan²A/1+cot²A) = (1-tan A/1-cot A)² = tan²A
L.H.S. = (1+tan²A/1+cot²A)
           = (1+tan²A/1+1/tan²A)
           = 1+tan²A/[(1+tan²A)/tan²A]
           = tan²A

Exercise 7.1

1. Find the distance between the following pairs of points:
(i) (2, 3), (4, 1) (ii) (−5, 7), (−1, 3) (iii) (a, b), (− a, − b)

Answer

(i) Distance between the points is given by


(ii) Distance between (−5, 7) and (−1, 3) is given by


(iii) Distance between (a, b) and (− a, − b) is given by


2. Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2.

Distance between points (0, 0) and (36, 15)

Yes, Assume town A at origin point (0, 0).
Therefore, town B will be at point (36, 15) with respect to town A.
And hence, as calculated above, the distance between town A and B will be 39 km.

Let the points (1, 5), (2, 3), and (- 2,-11) be representing the vertices A, B, and C of the given triangle respectively.Let A = (1, 5), B = (2, 3) and C = (- 2,-11)

Since AB + BC ≠ CA
Therefore, the points (1, 5), (2, 3), and ( – 2, – 11) are not collinear.

Let the points (5, – 2), (6, 4), and (7, – 2) are representing the vertices A, B, and C of the given triangle respectively.

Therefore, AB = BC
As two sides are equal in length, therefore, ABC is an isosceles triangle.

By using distance formula, we get

It can be observed that all sides of this quadrilateral ABCD are of the same length and also the diagonals are of the same length.
Therefore, ABCD is a square and hence, Champa was correct

6. Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(i) (- 1, – 2), (1, 0), (- 1, 2), (- 3, 0)
(ii) (- 3, 5), (3, 1), (0, 3), (- 1, – 4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2)

 

Answer

 

Let the points ( – 1, – 2), (1, 0), ( – 1, 2), and ( – 3, 0) be representing the vertices A, B, C, and D of the given quadrilateral respectively.
It can be observed that all sides of this quadrilateral are of the same length and also, the diagonals are of the same length. Therefore, the given points are the vertices of a square.

(ii) Let the points ( – 3, 5), (3, 1), (0, 3), and ( – 1, – 4) be representing the vertices A, B, C, and D of the given quadrilateral respectively.
It can be observed that all sides of this quadrilateral are of different lengths. Therefore, it can be said that it is only a general quadrilateral, and not specific such as square, rectangle, etc.

(iii) Let the points (4, 5), (7, 6), (4, 3), and (1, 2) be representing the vertices A, B, C, and D of the given quadrilateral respectively.
It can be observed that opposite sides of this quadrilateral are of the same length. However, the diagonals are of different lengths. Therefore, the given points are the vertices of a parallelogram.

7. Find the point on the x-axis which is equidistant from (2, – 5) and (- 2, 9).

Answer

We have to find a point on x-axis. Therefore, its y-coordinate will be 0.
Let the point on x-axis be (x,0)
(x – 2)² + 25 = (x – 2)² + 81
x² + 4 – 4x + 25 = x² + 4 + 4x + 81
8x = 25 -81
8x = -56
x = -7
Therefore, the point is ( – 7, 0).

8. Find the values of y for which the distance between the points P (2, – 3) and Q (10, y) is 10 units.

Answer

It is given that the distance between (2, – 3) and (10, y) is 10.
64 + (y + 3)² = 100
(y +3)² = 36
y + 3 = ±6
y + 3 = +6 or y + 3 = -6
Therefore, y = 3 or -9

9. If Q (0, 1) is equidistant from P (5, – 3) and R (x, 6), find the values of x. Also find the distance QR and PR.

Answer

PQ = QR

41 = x² + 25
16 = x²
x = ±4
Therefore, point R is (4, 6) or ( – 4, 6).
When point R is (4, 6),


10. Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (- 3, 4).

Answer

Point (x, y) is equidistant from (3, 6) and ( – 3, 4).
x² + 9 – 6x + y² + 36 – 12y = x² + 9 + 6x + y² + 16 – 8y
36 – 16 = 6x + 6x + 12y – 8y
20 = 12x + 4y
3x + y = 5
3x + y – 5 = 0

Exercises 7.3

1. Find the area of the triangle whose vertices are:
(i) (2, 3), (-1, 0), (2, -4)

Answer

(i) For collinear points, area of triangle formed by them is zero.
Therefore, for points (7, -2) (5, 1), and (3, k), area = 0
1/2 [7 { 1- k} + 5(k-(-2)) + 3{(-2) + 1}] = 0
7 – 7k + 5k +10 -9 = 0
-2k + 8 = 0
k = 4

(ii) For collinear points, area of triangle formed by them is zero.
Therefore, for points (8, 1), (k, – 4), and (2, – 5), area = 0
1/2 [8 { -4- (-5)} + k{(-5)-(1)} + 2{1 -(-4)}] = 0
8 – 6k + 10 = 0
6k = 18
k = 3

3. Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, -1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.

Answer

Let the vertices of the triangle be A (0, -1), B (2, 1), C (0, 3).
Let D, E, F be the mid-points of the sides of this triangle. Coordinates of D, E, and F are given by

Answer

Let the vertices of the quadrilateral be A ( – 4, – 2), B ( – 3, – 5), C (3, – 2), and D (2, 3). Join AC to form two triangles ΔABC and ΔACD.

Answer

Let the vertices of the triangle be A (4, -6), B (3, -2), and C (5, 2).
Let D be the mid-point of side BC of ΔABC. Therefore, AD is the median in ΔABC.

                       = 1/2 (-8+18-16)
                       = -3 square units
However, area cannot be negative. Therefore, area of ΔABD is 3 square units.
Area of ΔABD = 1/2 [(4) {0 – (2)} + 4{(2) – (-6)} + (5) {(-6) – (0)}]
                         = 1/2 (-8+32-30)
                         = -3 square units
However, area cannot be negative. Therefore, area of ΔABD is 3 square units.
The area of both sides is same. Thus, median AD has divided ΔABC in two triangles of equal areas.

Exercise 6.1

1. Fill in the blanks using correct word given in the brackets:-

(i) All circles are __________. (congruent, similar)
► Similar

(ii) All squares are __________. (similar, congruent)

► (a) Equal, (b) Proportional

2. Give two different examples of pair of
(i) Similar figures
(ii) Non-similar figures

Answer

(i) Two twenty-rupee notes, Two two rupees coins.
(ii) One rupee coin and five rupees coin, One rupee not and ten rupees note.

3. State whether the following quadrilaterals are similar or not:

1. In figure.6.17. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

⇒ 1.5/3 = 1/EC

EC = 3×10/15 = 2 cm
Hence, EC = 2 cm.

(ii) In △ ABC, DE∥BC (Given)
∴ AD/DB = AE/EC [By using Basic proportionality theorem]
⇒ AD/7.2 = 1.8/5.4
⇒ AD = 1.8×7.2/5.4 = 18/10 × 72/10 × 10/54 = 24/10
⇒ AD = 2.4
Hence, AD = 2.4 cm.

2. E and F are points on the sides PQ and PR respectively of a ΔPQR. For each of the following cases, state whether EF || QR.
(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

Answer

3. In the fig 6.18, if LM || CB and LN || CD, prove that AM/MB = AN/AD

By using basic proportionality theorem, we get,

From (i) and (ii), we get
AM/MB = AN/AD

4. In the fig 6.19, DE||AC and DF||AE. Prove that

In ΔABC, DE || AC (Given)

∴ BD/DA = BF/FE …(ii) [By using Basic Proportionality Theorem]

From equation (i) and (ii), we get
BE/EC = BF/FE

5. In the fig 6.20, DE||OQ and DF||OR, show that EF||QR.

6. In the fig 6.21, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.

7. Using Basic proportionality theorem, prove that a line drawn through the mid-points of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).

Answer

⇒ AD/BD = 1 … (i)

Therefore, AD/DB = AE/EC [By using Basic Proportionality Theorem]

⇒1 = AE/EC [From equation (i)]
∴ AE =EC
Hence, E is the mid point of AC.

8. Using Converse of basic proportionality theorem, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).

Answer

Given: ΔABC in which D and E are the mid points of AB and AC respectively such that AD=BD and AE=EC.

To Prove: DE || BC

Proof: D is the mid point of AB (Given)

⇒ AD/BD = 1 … (i)

⇒AE/EC = 1 [From equation (i)]

From equation (i) and (ii), we get
AD/BD = AE/EC
Hence, DE || BC [By converse of Basic Proportionality Theorem]

9. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO/BO = CO/DO.

Answer

Given: ABCD is a trapezium in which AB || DC in which diagonals AC and BD intersect each other at O.

To Prove: AO/BO = CO/DO

Construction: Through O, draw EO || DC || AB

Proof: In ΔADC, we have
OE || DC (By Construction)

∴ AE/ED = AO/CO  …(i) [By using Basic Proportionality Theorem]

In ΔABD, we have
OE || AB (By Construction)

∴ DE/EA = DO/BO …(ii) [By using Basic Proportionality Theorem]

From equation (i) and (ii), we get
AO/CO = BO/DO
⇒  AO/BO = CO/DO

10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/BO = CO/DO. Show that ABCD is a trapezium.

Answer

1. State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:

∠B = ∠Q = 80° (Given)
∠C = ∠R = 40° (Given)
∴ ΔABC ~ ΔPQR (AAA similarity criterion)

(ii) In  ΔABC and ΔPQR, we have
AB/QR = BC/RP = CA/PQ
∴  ΔABC ~ ΔQRP (SSS similarity criterion)

(iii) In ΔLMP and ΔDEF, we have
LM = 2.7, MP = 2, LP = 3, EF = 5, DE = 4, DF = 6
MP/DE = 2/4 = 1/2
PL/DF = 3/6 = 1/2
LM/EF= 2.7/5 = 27/50
Here, MP/DE = PL/DF ≠ LM/EF
Hence, ΔLMP and ΔDEF are not similar.

(iv) In ΔMNL and ΔQPR, we have
MN/QP = ML/QR = 1/2
∠M = ∠Q = 70°
∴ ΔMNL ~ ΔQPR (SAS similarity criterion)

(v) In ΔABC and ΔDEF, we have
AB = 2.5, BC = 3, ∠A = 80°, EF = 6, DF = 5, ∠F = 80°
Here, AB/DF = 2.5/5 = 1/2
And, BC/EF = 3/6 = 1/2
⇒ ∠B ≠ ∠F
Hence, ΔABC and ΔDEF are not similar.

(vi) In ΔDEF,we have
∠D + ∠E + ∠F = 180° (sum of angles of a triangle)
⇒ 70° + 80° + ∠F = 180°
⇒ ∠F = 180° – 70° – 80°
⇒ ∠F = 30°
In PQR, we have
∠P + ∠Q + ∠R = 180 (Sum of angles of Δ)
⇒ ∠P + 80° + 30° = 180°
⇒ ∠P = 180° – 80° -30°
⇒ ∠P = 70°
In ΔDEF and ΔPQR, we have
∠D = ∠P = 70°
∠F = ∠Q = 80°
∠F = ∠R = 30°
Hence, ΔDEF ~ ΔPQR (AAA similarity criterion)

Page No: 139

2.  In the fig 6.35, ΔODC ∝ ¼ ΔOBA, ∠ BOC = 125° and ∠ CDO = 70°. Find ∠ DOC, ∠ DCO and ∠ OAB.

Answer

DOB is a straight line.
Therefore, ∠DOC + ∠ COB = 180°
⇒ ∠DOC = 180° – 125°
= 55°

∴ ∠OAB = ∠OCD [Corresponding angles are equal in similar triangles.]
⇒ ∠ OAB = 55°
∴ ∠OAB = ∠OCD [Corresponding angles are equal in similar triangles.]

3. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that AO/OC = OB/OD

Answer

In ΔDOC and ΔBOA,
∠CDO = ∠ABO [Alternate interior angles as AB || CD]
∠DCO = ∠BAO [Alternate interior angles as AB || CD]
∠DOC = ∠BOA [Vertically opposite angles]
∴ ΔDOC ~ ΔBOA [AAA similarity criterion]

Page No: 140

Answer

In ΔPQR, ∠PQR = ∠PRQ
∴ PQ = PR …(i)
Given,QR/QS = QT/PR
Using (i), we get
QR/QS = QT/QP …(ii)
In ΔPQS and ΔTQR,
QR/QS = QT/QP [using (ii)]
∠Q = ∠Q
∴ ΔPQS ~ ΔTQR [SAS similarity criterion]

5. S and T are point on sides PR and QR of ΔPQR such that ∠P = ∠RTS. Show that ΔRPQ ~ ΔRTS.

Answer

In ΔRPQ and ΔRST,
∠RTS = ∠QPS (Given)
∠R = ∠R (Common angle)
∴ ΔRPQ ~ ΔRTS (By AA similarity criterion)

It is given that ΔABE ≅ ΔACD.
∴ AB = AC [By cpct] …(i)
And, AD = AE [By cpct] …(ii)
In ΔADE and ΔABC,

∠A = ∠A [Common angle]
∴ ΔADE ~ ΔABC [By SAS similarity criterion]

(i) ΔAEP ~ ΔCDP
(ii) ΔABD ~ ΔCBE
(iii) ΔAEP ~ ΔADB
(iv) ΔPDC ~ ΔBEC

(i) In ΔAEP and ΔCDP,
∠AEP = ∠CDP (Each 90°)
∠APE = ∠CPD (Vertically opposite angles)
Hence, by using AA similarity criterion,
ΔAEP ~ ΔCDP

(ii) In ΔABD and ΔCBE,
∠ADB = ∠CEB (Each 90°)

(iii) In ΔAEP and ΔADB,
∠AEP = ∠ADB (Each 90°)

(iv) In ΔPDC and ΔBEC,
∠PDC = ∠BEC (Each 90°)
∠PCD = ∠BCE (Common angle)
Hence, by using AA similarity criterion,
ΔPDC ~ ΔBEC

8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ~ ΔCFB.

Answer

In ΔABE and ΔCFB,
∠A = ∠C (Opposite angles of a parallelogram)
∠AEB = ∠CBF (Alternate interior angles as AE || BC)
∴ ΔABE ~ ΔCFB (By AA similarity criterion)

9. In the fig 6.39, ABC and AMP are two right triangles, right angled at B and M respectively, prove that:

(i) ΔABC ~ ΔAMP

(ii) CA/PA = BC/MP

 

Answer

 

(i) In ΔABC and ΔAMP, we have

∠A = ∠A (common angle)

∠ABC = ∠AMP = 90° (each 90°)

∴  ΔABC ~ ΔAMP (By AA similarity criterion)

 

(ii) As, ΔABC ~ ΔAMP (By AA similarity criterion)

If two triangles are similar then the corresponding sides are equal,

Hence, CA/PA = BC/MP

 

10. CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively. If ΔABC ~ ΔFEG, Show that:

(i) CD/GH = AC/FG
(ii) ΔDCB ~ ΔHGE
(iii) ΔDCA ~ ΔHGF

Answer

(i) It is given that ΔABC ~ ΔFEG.
∴ ∠A = ∠F, ∠B = ∠E, and ∠ACB = ∠FGE
∠ACB = ∠FGE
∴ ∠ACD = ∠FGH (Angle bisector)
And, ∠DCB = ∠HGE (Angle bisector)
In ΔACD and ΔFGH,
∠A = ∠F (Proved above)
∠ACD = ∠FGH (Proved above)
∴ ΔACD ~ ΔFGH (By AA similarity criterion)

Page No: 141

11. In the following figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ΔABD ~ ΔECF.

Answer

It is given that ABC is an isosceles triangle.
∴ AB = AC
⇒ ∠ABD = ∠ECF
In ΔABD and ΔECF,
∠ADB = ∠EFC (Each 90°)
∠BAD = ∠CEF (Proved above)
∴ ΔABD ~ ΔECF (By using AA similarity criterion)

13. D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA² = CB.CD

Answer

In ΔADC and ΔBAC,
∠ADC = ∠BAC (Given)
∠ACD = ∠BCA (Common angle)
∴ ΔADC ~ ΔBAC (By AA similarity criterion)
We know that corresponding sides of similar triangles are in proportion.

14. Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ΔABC ~ ΔPQR.

Answer

15. A vertical pole of a length 6 m casts a shadow 4m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

Answer

16. If AD and PM are medians of triangles ABC and PQR, respectively where ΔABC ~ ΔPQR prove that AB/PQ = AD/PM.

It is given that ΔABC ~ ΔPQR
We know that the corresponding sides of similar triangles are in proportion.∴ AB/PQ = AC/PR = BC/QR …(i)
Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R …(ii)
Since AD and PM are medians, they will divide their opposite sides.∴ BD = BC/2 and QM = QR/2 …(iii)
From equations (i) and (iii), we get
AB/PQ = BD/QM …(iv)
In ΔABD and ΔPQM,
∠B = ∠Q [Using equation (ii)]
AB/PQ = BD/QM [Using equation (iv)]
∴ ΔABD ~ ΔPQM (By SAS similarity criterion)⇒ AB/PQ = BD/QM = AD/PM.

1.  Sides of triangles are given below. Determine which of them are right triangles? In case of a right triangle, write the length of its hypotenuse.
(i) 7 cm, 24 cm, 25 cm
(ii) 3 cm, 8 cm, 6 cm
(iii) 50 cm, 80 cm, 100 cm
(iv) 13 cm, 12 cm, 5 cm

Answer

(i) Given that the sides of the triangle are 7 cm, 24 cm, and 25 cm.
Squaring the lengths of these sides, we will get 49, 576, and 625.
49 + 576 = 625
(7)² + (24)² = (25)²
The sides of the given triangle are satisfying Pythagoras theorem.Hence, it is right angled triangle.
Length of Hypotenuse = 25 cm

(ii) Given that the sides of the triangle are 3 cm, 8 cm, and 6 cm.
Squaring the lengths of these sides, we will get 9, 64, and 36.
However, 9 + 36 ≠ 64
Or, 3² + 6² ≠ 8²
Clearly, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side.
Therefore, the given triangle is not satisfying Pythagoras theorem.

(iii) Given that sides are 50 cm, 80 cm, and 100 cm.
Squaring the lengths of these sides, we will get 2500, 6400, and 10000.
However, 2500 + 6400 ≠ 10000
Or, 50² + 80² ≠ 100²
Clearly, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side.
Therefore, the given triangle is not satisfying Pythagoras theorem.
Hence, it is not a right triangle.

(iv) Given that sides are 13 cm, 12 cm, and 5 cm.
Squaring the lengths of these sides, we will get 169, 144, and 25.
Clearly, 144 +25 = 169

2. PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM² = QM × MR.

Answer

3. In Fig. 6.53, ABD is a triangle right angled at A and AC ⊥ BD. Show that
(i) AB² = BC × BD
(ii) AC² = BC × DC
(iii) AD² = BD × CD

Answer

(i) In ΔADB and ΔCAB, we have
∠DAB = ∠ACB (Each equals to 90°)
∠ABD = ∠CBA (Common angle)
∴ ΔADB ~ ΔCAB [AA similarity criterion]
⇒ AB/CB = BD/AB
⇒ AB² = CB × BD

(ii) Let ∠CAB = x
In ΔCBA,
∠CBA = 180° – 90° – x
∠CBA = 90° – x
Similarly, in ΔCAD
∠CAD = 90° – ∠CBA = 90° – x
∠CDA = 180° – 90° – (90° – x)
∠CDA = x
In ΔCBA and ΔCAD, we have
∠CBA = ∠CAD
∠CAB = ∠CDA
∠ACB = ∠DCA (Each equals to 90°)
∴ ΔCBA ~ ΔCAD [By AAA similarity criterion]
⇒ AC/DC = BC/AC
⇒ AC² =  DC × BC

(iii) In ΔDCA and ΔDAB, we have
∠DCA = ∠DAB (Each equals to 90°)
∠CDA = ∠ADB (common angle)
∴ ΔDCA ~ ΔDAB [By AA similarity criterion]
⇒ DC/DA = DA/DA
⇒ AD² = BD × CD

4. ABC is an isosceles triangle right angled at C. Prove that AB² = 2AC² .

Answer

5. ABC is an isosceles triangle with AC = BC. If AB² = 2AC², prove that ABC is a right triangle.

Answer

Since, the diagonals of a rhombus bisect each other at right angles.

= 4AO² + 4BO² [Since, AO = CO and BO =DO]
= (2AO)² + (2BO)² = AC² + BD²

Page No: 151

8. In Fig. 6.54, O is a point in the interior of a triangle

ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that
(i) OA² + OB² + OC² – OD² – OE² – OF² = AF² + BD² + CE² ,
(ii) AF² + BD² + CE² = AE² + CD² + BF².

Answer

Join OA, OB and OC

Similarly, in ΔBOD
OB² = OD² + BD²
Similarly, in ΔCOE
OC² = OE² + EC²
Adding these equations,
OA² + OB² + OC² = OF² + AF² + OD² + BD² + OE² + EC²
OA² + OB² + OC² – OD² – OE² – OF² = AF² + BD² + CE².

(ii) AF² + BD² + EC² = (OA² – OE²) + (OC² – OD²) + (OB² – OF²)
∴ AF² + BD² + CE² = AE² + CD² + BF².

9. A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.

Answer

Therefore, the distance of the foot of the ladder from the base of the wall is 6 m.

10. A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut ?

Answer

Let AB be the pole and AC be the wire.
By Pythagoras theorem,

11. An aeroplane leaves an airport and flies due north at a speed of 1,000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1,200 km per hour. How far apart will be the two planes after hours?

Answer

Distance covered by first aeroplane due north in  hours (OA) = 100 × 3/2 km = 1500 km

Distance covered by second aeroplane due west in hours (OB) = 1200 × 3/2 km = 1800 km

12. Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.

Answer

Let CD and AB be the poles of height 11 m and 6 m.
Therefore, CP = 11 – 6 = 5 m
From the figure, it can be observed that AP = 12m
Applying Pythagoras theorem for ΔAPC, we get

13. D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE² + BD² = AB² + DE².

Answer

Putting these values in equation (iii), we get
DE² + AB² = AE² + BD².

14. The perpendicular from A on side BC of a Δ ABC intersects BC at D such that DB = 3CD (see Fig. 6.55). Prove that 2AB² = 2AC² + BC².

= 9CD² – CD²  [∴ BD = 3CD]               
= 9CD² = 8(BC/4)² [Since, BC = DB + CD = 3CD + CD = 4CD]
Therefore, AB² – AC² = BC²/2
⇒ 2(AB² – AC²) = BC²
⇒ 2AB² – 2AC² = BC²
∴ 2AB² = 2AC² + BC².

15.  In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3BC. Prove that 9AD² = 7AB².

Answer

Let the side of the equilateral triangle be a, and AE be the altitude of ΔABC.

∴ BD = a/3

Applying Pythagoras theorem in ΔADE, we get
AD² = AE² + DE² 

⇒ 9 AD² = 7 AB²


16. In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.

Answer

Let the side of the equilateral triangle be a, and AE be the altitude of ΔABC.
∴ BE = EC = BC/2 = a/2
Applying Pythagoras theorem in ΔABE, we get
AB² = AE² + BE²

4AE² = 3a²
⇒ 4 × (Square of altitude) = 3 × (Square of one side)


17. Tick the correct answer and justify: In ΔABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm.
The angle B is:
(A) 120°

Answer

Given that, AB = 6√3 cm, AC = 12 cm, and BC = 6 cm
We can observe that
AB² = 108
AC² = 144
And, BC² = 36
AB² + BC² = AC²
The given triangle, ΔABC, is satisfying Pythagoras theorem.
Therefore, the triangle is a right triangle, right-angled at B.
∴ ∠B = 90°
Hence, the correct option is (C).

Exercise 6.6

1. In Fig. 6.56, PS is the bisector of ∠QPR of ΔPQR. Prove that QS/SR = PQ/PR

2. In Fig. 6.57, D is a point on hypotenuse AC of ΔABC, such that BD ⊥ AC, DM ⊥ BC and
DN ⊥ AB. Prove that :
(i) DM² = DN.MC
(ii) DN² = DM.AN

Answer

Given that, D is a point on hypotenuse AC of ∆ABC, DM ⊥ BC and DN ⊥ AB.
Now, join NM.
Let BD and NM intersect at O.

3. In Fig. 6.58, ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced. Prove that AC² = AB² + BC² + 2 BC.BD.

4. In Fig. 6.59, ABC is a triangle in which ∠ABC < 90° and AD ⊥ BC. Prove that AC² = AB² + BC² – 2BC.BD.

Answer

Given that, in figure, ABC is a triangle in which ∠ABC < 90° and AD ⊥ BC.

Proof:

In right △ADC,
∠D = 90°

= AD² + (BC – BD)²  [∵BC = BD + DC]
= AD² + (BC – BD)²  (BC = BD + DC)
= AD² + BC² + BD² – 2BC.BD [∵ (a + b)² = a² + b² + 2ab]
= (AD² + BD²) + BC² – 2BC . BD
= AB² + BC² – 2BC . BD
{In right △ADB with ∠D = 90°, AB² = AD² + BD²} (By Pythagoras theorem) 

5. In Fig. 6.60, AD is a median of a triangle ABC and AM ⊥ BC. Prove that :
(i) AC² = AD² + BC.DM + (BC/2)²
(ii) AB² = AD² – BC.DM + (BC/2)²
(iii) AC² + AB² = 2 AD² + 1/2 BC²


Answer

Given that, in figure, AD is a median of a ∆ABC and AM ⊥ BC.
Proof:
(i) In right ∆AMC,

Answer

Given that, ABCD is a parallelogram whose diagonals are AC and BD.

AD = BC (Opposite sides of a parallelogram)
AM = BN (Both are altitudes of the same parallelogram to the same base) ,
△AMD ⩭ △BNC (RHS congruence criterion)
MD = NC (CPCT) —(i)

In right △BND,
∠N = 90°
BD² = BN² + DN² (By Pythagoras theorem)
= BN² + (DC + CN)² (∵ DN = DC + CN) 
= BN² + DC² + CN² + 2DC.CN [∵ (a + b)² = a² + b² + 2ab]
= (BN² + CN²) + DC² + 2DC.CN
= BC² + DC² + 2DC.CN — (ii) (∵In right △BNC with ∠N = 90°)
BN² + CN² = BC²  (By Pythagoras theorem)

In right △AMC,
∠M = 90°
AC² = AM² + MC² (∵MC = DC – DM)
= AM² + (DC – DM)² [∵ (a + b)² = a² + b² + 2ab]
= AM² + DC² + DM² – 2DC.DM
(AM² + DM²) + DC² – 2DC.DM
= AD² + DC² – 2DC.DM
[∵ In  right triangle AMD with ∠M = 90°, AD² = AM² + DM² (By Pythagoras theorem)]
= AD² + AB² = 2DC.CN  — (iii)
[∵ DC = AB, opposite sides of parallelogram and BM = CN from eq (i)]
Now, on adding Eqs. (iii) and (ii), we get
AC² + BD² = (AD² + AB²) + (BC² + DC²)
= AB² + BC² + CD² + DA²

7. In Fig. 6.61, two chords AB and CD intersect each other at the point P. Prove that :
(i) Δ APC ~ Δ DPB 
(ii) AP . PB = CP . DP

Answer

Given that, in figure, two chords AB and CD intersects each other at the point P.

Proof:

(i) ∆APC and ∆DPB
∠APC = ∠DPB (Vertically opposite angles)
∠CAP = ∠BDP (Angles in the same segment)
∴ ∆APC ~ ∆DPB (AA similarity criterion)

(ii) ∆APC ~ ∆DPB [Proved in (i)]
∴ AP/DP = CP/BP
(∴ Corresponding sides of two similar triangles are proportional)
⇒ AP.BP = CP.DP
⇒ AP.PB = CP.DP

8. In Fig. 6.62, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that
(i) Δ PAC ~ Δ PDB
(ii) PA.PB = PC.PD

Answer

Given that, in figure, two chords AB and CD of a circle intersect each other at the point P (when produced) out the circle.

Proof:

(i) We know that, in a cyclic quadrilaterals, the exterior angle is equal to the interior opposite angle.
Therefore, ∠PAC = ∠PDB …(i)
and ∠PCA = ∠PBD …(ii)
In view of Eqs. (i) and (ii), we get
∆PAC ~ ∆PDB  (∵ AA similarity criterion)

(ii) ∆PAC ~ ∆PDB  [Proved in (i)]
∴ PA/PD = PC/PB
(∵ Corresponding sides of the similar triangles are proportional)
⇒ PA.PB = PC.PD

9. In Fig. 6.63, D is a point on side BC of ΔABC such that BD/CD = AB/AC. Prove that AD is the bisector of ∠BAC.

10. Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out (see Fig. 6.64)? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds?

Answer

Length of the string that she has out

Exercise 5.1

1. In which of the following situations, does the list of numbers involved make as arithmetic progression and why?

(i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km.

Answer

It can be observed that
Taxi fare for 1st km = 15
Taxi fare for first 2 km = 15 + 8 = 23
Taxi fare for first 3 km = 23 + 8 = 31
Taxi fare for first 4 km = 31 + 8 = 39

Clearly 15, 23, 31, 39 … forms an A.P. because every term is 8 more than the preceding term.

(ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.

Answer

Let the initial volume of air in a cylinder be V litres. In each stroke, the vacuum pump removes 1/4 of air remaining in the cylinder at a time. In other words, after every stroke, only 1 – 1/4 = 3/4th part of air will remain.
Therefore, volumes will be V, 3V/4 , (3V/4)² , (3V/4)³…
Clearly, it can be observed that the adjacent terms of this series do not have the same difference between them. Therefore, this is not an A.P.

(iii) The cost of digging a well after every metre of digging, when it costs Rs 150 for the first metre and rises by Rs 50 for each subsequent metre.

Answer

Cost of digging for first metre = 150
Cost of digging for first 2 metres = 150 + 50 = 200
Cost of digging for first 3 metres = 200 + 50 = 250
Cost of digging for first 4 metres = 250 + 50 = 300
Clearly, 150, 200, 250, 300 … forms an A.P. because every term is 50 more than the preceding term.

(iv) The amount of money in the account every year, when Rs 10000 is deposited at compound interest at 8% per annum.

Answer

We know that if Rs P is deposited at r% compound interest per annum for n years, our money will be

Clearly, adjacent terms of this series do not have the same difference between them. Therefore, this is not an A.P.

2. Write first four terms of the A.P. when the first term a and the common differenced are given as follows
(i) a = 10, d = 10
(ii) a = -2, d = 0
(iii) a = 4, d = – 3
(iv) a = -1 d = 1/2
(v) a = – 1.25, d = – 0.25

Answer

(i) a = 10, d = 10
Let the series be a₁, a₂, a₃, a₄, a₅ …
a₁ = a = 10
a₂ = a₁ + d = 10 + 10 = 20
a₃ = a₂ + d = 20 + 10 = 30
a₄ = a₃ + d = 30 + 10 = 40
a₅ = a₄ + d = 40 + 10 = 50
Therefore, the series will be 10, 20, 30, 40, 50 …
First four terms of this A.P. will be 10, 20, 30, and 40.

(ii) a = – 2, d = 0
Let the series be a₁, a₂, a₃, a₄ …
a₁ = a = -2
a₂ = a₁ + d = – 2 + 0 = – 2
a₃ = a₂ + d = – 2 + 0 = – 2
a₄ = a₃ + d = – 2 + 0 = – 2
Therefore, the series will be – 2, – 2, – 2, – 2 …
First four terms of this A.P. will be – 2, – 2, – 2 and – 2.

(iii) a = 4, d = – 3
Let the series be a₁, a₂, a₃, a₄ …
a₁ = a = 4
a₂ = a₁ + d = 4 – 3 = 1
a₃ = a₂ + d = 1 – 3 = – 2
a₄ = a₃ + d = – 2 – 3 = – 5
Therefore, the series will be 4, 1, – 2 – 5 …
First four terms of this A.P. will be 4, 1, – 2 and – 5.

(iv) a = – 1, d = 1/2
Let the series be a₁, a₂, a₃, a₄ …a₁ = a = -1
a₂ = a₁ + d = -1 + 1/2 = -1/2
a₃ = a₂ + d = -1/2 + 1/2 = 0
a₄ = a₃ + d = 0 + 1/2 = 1/2
Clearly, the series will be-1, -1/2, 0, 1/2
First four terms of this A.P. will be -1, -1/2, 0 and 1/2.

(v) a = – 1.25, d = – 0.25
Let the series be a₁, a₂, a₃, a₄ …
a₁ = a = – 1.25
a₂ = a₁ + d = – 1.25 – 0.25 = – 1.50
a₃ = a₂ + d = – 1.50 – 0.25 = – 1.75
a₄ = a₃ + d = – 1.75 – 0.25 = – 2.00
Clearly, the series will be 1.25, – 1.50, – 1.75, – 2.00 ……..
First four terms of this A.P. will be – 1.25, – 1.50, – 1.75 and – 2.00.

3. For the following A.P.s, write the first term and the common difference.
(i) 3, 1, – 1, – 3 …
(ii) -5, – 1, 3, 7 …
(iii) 1/3, 5/3, 9/3, 13/3 ….
(iv) 0.6, 1.7, 2.8, 3.9 …

Answer

(i) 3, 1, – 1, – 3 …
Here, first term, a = 3
Common difference, d = Second term – First term
= 1 – 3 = – 2

(ii) – 5, – 1, 3, 7 …
Here, first term, a = – 5
Common difference, d = Second term – First term
= ( – 1) – ( – 5) = – 1 + 5 = 4
(iii) 1/3, 5/3, 9/3, 13/3 ….
Here, first term, a = 1/3

(iv) 0.6, 1.7, 2.8, 3.9 …
Here, first term, a = 0.6
Common difference, d = Second term – First term
= 1.7 – 0.6
= 1.1

4. Which of the following are APs? If they form an A.P. find the common difference d and write three more terms.
(i) 2, 4, 8, 16 …
(ii) 2, 5/2, 3, 7/2 ….
(iii) -1.2, -3.2, -5.2, -7.2 …
(iv) -10, – 6, – 2, 2 …
(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2
(vi) 0.2, 0.22, 0.222, 0.2222 ….
(vii) 0, – 4, – 8, – 12 …
(viii) -1/2, -1/2, -1/2, -1/2 ….
(ix) 1, 3, 9, 27 …
(x) a, 2a, 3a, 4a …
(xi) a, a², a³, a⁴ …
(xii) √2, √8, √18, √32 …
(xiii) √3, √6, √9, √12 …
(xiv) 1², 3², 5², 7² …
(xv) 1², 5², 7², 7³ …

Answer

(i) 2, 4, 8, 16 …
Here,
a₂ – a₁ = 4 – 2 = 2
a₃ – a₂ = 8 – 4 = 4
a₄ – a₃ = 16 – 8 = 8
⇒ a_n+1 – a_n is not the same every time.

a₂ – a₁ = 5/2 – 2 = 1/2
a₃ – a₂ = 3 – 5/2 = 1/2
a₄ – a₃ = 7/2 – 3 = 1/2
⇒ a_n+1 – a_n is same every time.
Therefore, d = 1/2 and the given numbers are in A.P.
Three more terms are
a₅ = 7/2 + 1/2 = 4
a₆ = 4 + 1/2 = 9/2
a₇ = 9/2 + 1/2 = 5

(iii) -1.2, – 3.2, -5.2, -7.2 …
Here,
a₂ – a₁ = ( -3.2) – ( -1.2) = -2
a₃ – a₂ = ( -5.2) – ( -3.2) = -2
a₄ – a₃ = ( -7.2) – ( -5.2) = -2
⇒ a_n+1 – a_n is same every time.
Therefore, d = -2 and the given numbers are in A.P.
Three more terms are
a₅ = – 7.2 – 2 = – 9.2
a₆ = – 9.2 – 2 = – 11.2
a₇ = – 11.2 – 2 = – 13.2

(iv) -10, – 6, – 2, 2 …
Here,
a₂ – a₁ = (-6) – (-10) = 4
a₃ – a₂ = (-2) – (-6) = 4
a₄ – a₃ = (2) – (-2) = 4
⇒ a_n+1 – a_n is same every time.
Therefore, d = 4 and the given numbers are in A.P.
Three more terms are
a₅ = 2 + 4 = 6
a₆ = 6 + 4 = 10
a₇ = 10 + 4 = 14

(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2
Here,
a₂ – a₁ = 3 + √2 – 3 = √2
a₃ – a₂ = (3 + 2√2) – (3 + √2) = √2
a₄ – a₃ = (3 + 3√2) – (3 + 2√2) = √2
⇒ a_n+1 – a_n is same every time.
Therefore, d = √2 and the given numbers are in A.P.
Three more terms are
a₅ = (3 + √2) + √2 = 3 + 4√2
a₆ = (3 + 4√2) + √2 = 3 + 5√2
a₇ = (3 + 5√2) + √2 = 3 + 6√2

(vi) 0.2, 0.22, 0.222, 0.2222 ….
Here,
a₂ – a₁ = 0.22 – 0.2 = 0.02
a₃ – a₂ = 0.222 – 0.22 = 0.002
a₄ – a₃ = 0.2222 – 0.222 = 0.0002
⇒ a_n+1 – a_n is not the same every time.

(xiii) √3, √6, √9, √12 …
Here,
a₂ – a₁ = √6 – √3 = √3 × 2 -√3 = √3(√2 – 1)
a₃ – a₂ = √9 – √6 = 3 – √6 = √3(√3 – √2)
a₄ – a₃ = √12 – √9 = 2√3 – √3 × 3 = √3(2 – √3)
⇒ a_n+1 – a_n is not the same every time.

Or, 1, 9, 25, 49 …..
Here,
a₂ − a₁ = 9 − 1 = 8
a₃ − a₂ = 25 − 9 = 16
a₄ − a₃ = 49 − 25 = 24
⇒ a_n+1 – a_n is not the same every time.

(xv) 1², 5², 7², 73 …
Or 1, 25, 49, 73 …
Here,
a₂ − a₁ = 25 − 1 = 24
a₃ − a₂ = 49 − 25 = 24
a₄ − a₃ = 73 − 49 = 24
i.e., a_k+1 − a_k is same every time.
⇒ a_n+1 – a_n is same every time.
Therefore, d = 24 and the given numbers are in A.P.
Three more terms are
a₅ = 73+ 24 = 97
a₆ = 97 + 24 = 121
a₇ = 121 + 24 = 145

(iv) 1/15, 1/12, 1/10, …… , to 11 terms
For this A.P.,


2. Find the sums given below
(i) 7 + + 14 + ……………… +84
(ii)+ 14 + ………… + 84
(ii) 34 + 32 + 30 + ……….. + 10
(iii) − 5 + (− 8) + (− 11) + ………… + (− 230)

Answer

(i) For this A.P.,
a = 7
l = 84
d = a₂ − a₁ = – 7 = 21/2 – 7 = 7/2
Let 84 be the n^th term of this A.P.
l = a (n – 1)d
84 = 7 + (n – 1) × 7/2
77 = (n – 1) × 7/2
22 = n − 1
n = 23
We know that,
S_n = n/2 (a + l)
S_n = 23/2 (7 + 84)
= (23×91/2) = 2093/2
= 


(ii) 34 + 32 + 30 + ……….. + 10
For this A.P.,
a = 34
d = a₂ − a₁ = 32 − 34 = −2
l = 10
Let 10 be the n^th term of this A.P.
l = a + (n − 1) d
10 = 34 + (n − 1) (−2)
−24 = (n − 1) (−2)
12 = n − 1
n = 13
S_n = n/2 (a + l)
= 13/2 (34 + 10)
= (13×44/2) = 13 × 22
= 286

(iii) (−5) + (−8) + (−11) + ………… + (−230) For this A.P.,
a = −5
l = −230
d = a₂ − a₁ = (−8) − (−5)
= − 8 + 5 = −3
Let −230 be the n^th term of this A.P.
l = a + (n − 1)d
−230 = − 5 + (n − 1) (−3)
−225 = (n − 1) (−3)
(n − 1) = 75
n = 76
And,
S_n = n/2 (a + l)
= 76/2 [(-5) + (-230)]
= 38(-235)
= -8930

3. In an AP
(i) Given a = 5, d = 3, a_n = 50, find n and S_n.
(ii) Given a = 7, a₁₃ = 35, find d and S₁₃.
(iii) Given a₁₂ = 37, d = 3, find a and S₁₂.
(iv) Given a₃ = 15, S₁₀ = 125, find d and a₁₀.
(v) Given d = 5, S₉ = 75, find a and a₉.
(vi) Given a = 2, d = 8, S_n = 90, find n and a_n.
(vii) Given a = 8, a_n = 62, S_n = 210, find n and d.
(viii) Given a_n = 4, d = 2, S_n = − 14, find n and a.
(ix) Given a = 3, n = 8, S = 192, find d.
(x) Given l = 28, S = 144 and there are total 9 terms. Find a.

Answer

(i) Given that, a = 5, d = 3, a_n = 50
As a_n = a + (n − 1)d,
⇒ 50 = 5 + (n – 1) × 3
⇒ 3(n – 1) = 45
⇒ n – 1 = 15
⇒ n = 16
Now, S_n = n/2 (a + a_n)
S_n = 16/2 (5 + 50) = 440


(ii) Given that, a = 7, a₁₃ = 35
As a_n = a + (n − 1)d, ⇒ 35 = 7 + (13 – 1)d
⇒ 12d = 28
⇒ d = 28/12 = 2.33
Now, S_n = n/2 (a + a_n)
S₁₃ = 13/2 (7 + 35) = 273

(iii)Given that, a₁₂ = 37, d = 3 As a_n = a + (n − 1)d,
⇒ a₁₂ = a + (12 − 1)3
⇒ 37 = a + 33
⇒ a = 4
S_n = n/2 (a + a_n)
S_n = 12/2 (4 + 37)
= 246

(iv) Given that, a₃ = 15, S₁₀ = 125
As a_n = a + (n − 1)d,
a₃ = a + (3 − 1)d
15 = a + 2d … (i)
S_n = n/2 [2a + (n – 1)d]
S₁₀ = 10/2 [2a + (10 – 1)d]
125 = 5(2a + 9d)
25 = 2a + 9d … (ii)
On multiplying equation (i) by (ii), we get
30 = 2a + 4d … (iii)
On subtracting equation (iii) from (ii), we get
−5 = 5d
d = −1
From equation (i),
15 = a + 2(−1)
15 = a − 2
a = 17
a₁₀ = a + (10 − 1)d
a₁₀ = 17 + (9) (−1)
a₁₀ = 17 − 9 = 8

(v) Given that, d = 5, S₉ = 75
As S_n = n/2 [2a + (n – 1)d]
S₉ = 9/2 [2a + (9 – 1)5]
25 = 3(a + 20)
25 = 3a + 60
3a = 25 − 60
a = -35/3
a_n = a + (n − 1)d
a₉ = a + (9 − 1) (5)
= -35/3 + 8(5)
= -35/3 + 40
= (35+120/3) = 85/3

(vi) Given that, a = 2, d = 8, S_n = 90
As S_n = n/2 [2a + (n – 1)d]
90 = n/2 [2a + (n – 1)d]
⇒ 180 = n(4 + 8n – 8) = n(8n – 4) = 8n² – 4n
⇒ 8n² – 4n – 180 = 0
⇒ 2n² – n – 45 = 0
⇒ 2n² – 10n + 9n – 45 = 0
⇒ 2n(n -5) + 9(n – 5) = 0
⇒ (2n – 9)(2n + 9) = 0
So, n = 5 (as it is positive integer)
∴ a₅ = 8 + 5 × 4 = 34

(vii) Given that, a = 8, a_n = 62, S_n = 210
As S_n = n/2 (a + a_n)
210 = n/2 (8 + 62)
⇒ 35n = 210
⇒ n = 210/35 = 6
Now, 62 = 8 + 5d
⇒ 5d = 62 – 8 = 54
⇒ d = 54/5 = 10.8

(viii) Given that, a_n = 4, d = 2, S_n = −14
a_n = a + (n − 1)d
4 = a + (n − 1)2
4 = a + 2n − 2
a + 2n = 6
a = 6 − 2n … (i)
S_n = n/2 (a + a_n)
-14 = n/2 (a + 4)
−28 = n (a + 4)
−28 = n (6 − 2n + 4) {From equation (i)}
−28 = n (− 2n + 10)
−28 = − 2n² + 10n
2n² − 10n − 28 = 0
n² − 5n −14 = 0
n² − 7n + 2n − 14 = 0
n (n − 7) + 2(n − 7) = 0
(n − 7) (n + 2) = 0
Either n − 7 = 0 or n + 2 = 0
n = 7 or n = −2
However, n can neither be negative nor fractional.
Therefore, n = 7
From equation (i), we get
a = 6 − 2n
a = 6 − 2(7)
= 6 − 14
= −8

(ix) Given that, a = 3, n = 8, S = 192
As S_n = n/2 [2a + (n – 1)d]
192 = 8/2 [2 × 3 + (8 – 1)d]
192 = 4 [6 + 7d]
48 = 6 + 7d
42 = 7d
d = 6

(x) Given that, l = 28, S = 144 and there are total of 9 terms.
S_n = n/2 (a + l)
144 = 9/2 (a + 28)
(16) × (2) = a + 28
32 = a + 28
a = 4

Page No: 113

4. How many terms of the AP. 9, 17, 25 … must be taken to give a sum of 636?

Answer

Let there be n terms of this A.P.
For this A.P., a = 9
d = a₂ − a₁ = 17 − 9 = 8
As S_n = n/2 [2a + (n – 1)d]
636 = n/2 [2 × a + (8 – 1) × 8]
636 = n/2 [18 + (n– 1) × 8]
636 = n [9 + 4n − 4]
636 = n (4n + 5)
4n² + 5n − 636 = 0
4n² + 53n − 48n − 636 = 0
n (4n + 53) − 12 (4n + 53) = 0
(4n + 53) (n − 12) = 0
Either 4n + 53 = 0 or n − 12 = 0
n = (-53/4) or n = 12
n cannot be (-53/4). As the number of terms can neither be negative nor fractional, therefore, n = 12 only.

5. The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

Answer

Given that,
a = 5
l = 45
S_n = 400
S_n = n/2 (a + l)
400 = n/2 (5 + 45)
400 = n/2 (50)
n = 16
l = a + (n − 1) d
45 = 5 + (16 − 1) d
40 = 15d
d = 40/15 = 8/3

6. The first and the last term of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

Answer

Given that,

S₇ = 49
S₁₇ = 289
S₇  = 7/2 [2a + (n – 1)d]
S₇ = 7/2 [2a + (7 – 1)d]
49 = 7/2 [2a + 16d]
7 = (a + 3d)
a + 3d = 7 … (i)
Similarly,
S₁₇ = 17/2 [2a + (17 – 1)d]
289 = 17/2 (2a + 16d)
17 = (a + 8d)
a + 8d = 17 … (ii)
Subtracting equation (i) from equation (ii),
5d = 10
d = 2
From equation (i),
a + 3(2) = 7
a + 6 = 7
a = 1
S_n = n/2 [2a + (n – 1)d]
= n/2 [2(1) + (n – 1) × 2]
= n/2 (2 + 2n – 2)
= n/2 (2n)
= n²

10. Show that a₁, a₂ … , a_n , … form an AP where a_n is defined as below
(i) a_n = 3 + 4n
(ii) a_n = 9 − 5n
Also find the sum of the first 15 terms in each case.

Answer

(i) a_n = 3 + 4n
a₁ = 3 + 4(1) = 7
a₂ = 3 + 4(2) = 3 + 8 = 11
a₃ = 3 + 4(3) = 3 + 12 = 15
a₄ = 3 + 4(4) = 3 + 16 = 19
It can be observed that
a₂ − a₁ = 11 − 7 = 4
a₃ − a₂ = 15 − 11 = 4
a₄ − a₃ = 19 − 15 = 4
i.e., a_{k + 1} − a_k is same every time. Therefore, this is an AP with common difference as 4 and first term as 7.
S_n = n/2 [2a + (n – 1)d]
S₁₅ = 15/2 [2(7) + (15 – 1) × 4]
= 15/2 [(14) + 56]
= 15/2 (70)
= 15 × 35
= 525

(ii) a_n = 9 − 5n
a₁ = 9 − 5 × 1 = 9 − 5 = 4
a₂ = 9 − 5 × 2 = 9 − 10 = −1
a₃ = 9 − 5 × 3 = 9 − 15 = −6
a₄ = 9 − 5 × 4 = 9 − 20 = −11
It can be observed that
a₂ − a₁ = − 1 − 4 = −5
a₃ − a₂ = − 6 − (−1) = −5
a₄ − a₃ = − 11 − (−6) = −5
i.e., a_{k + 1} − a_k is same every time. Therefore, this is an A.P. with common difference as −5 and first term as 4.
S_n = n/2 [2a + (n – 1)d]
S₁₅ = 15/2 [2(4) + (15 – 1) (-5)]
= 15/2 [8 + 14(-5)]
= 15/2 (8 – 70)
= 15/2 (-62)
= 15(-31)
= -465

11. If the sum of the first n terms of an AP is 4n − n², what is the first term (that is S₁)? What is the sum of first two terms? What is the second term? Similarly find the 3^rd, the10^th and the n^th terms.

Answer

Given that,
S_n = 4n − n²
First term, a = S₁ = 4(1) − (1)² = 4 − 1 = 3
Sum of first two terms = S₂
= 4(2) − (2)² = 8 − 4 = 4
Second term, a₂ = S₂ − S₁ = 4 − 3 = 1
d = a₂ − a = 1 − 3 = −2
a_n = a + (n − 1)d
= 3 + (n − 1) (−2)
= 3 − 2n + 2
= 5 − 2n
Therefore, a₃ = 5 − 2(3) = 5 − 6 = −1
a₁₀ = 5 − 2(10) = 5 − 20 = −15
Hence, the sum of first two terms is 4. The second term is 1. 3^rd, 10^th, and n^th terms are −1, −15, and 5 − 2n respectively.

12. Find the sum of first 40 positive integers divisible by 6.

Answer

The positive integers that are divisible by 6 are
6, 12, 18, 24 …
It can be observed that these are making an A.P. whose first term is 6 and common difference is 6.
a = 6
d = 6
S₄₀ = ?
S_n = n/2 [2a + (n – 1)d]
S₄₀ = 40/2 [2(6) + (40 – 1) 6]
= 20[12 + (39) (6)]
= 20(12 + 234)
= 20 × 246
= 4920

13. Find the sum of first 15 multiples of 8.

Answer

The multiples of 8 are
8, 16, 24, 32…
These are in an A.P., having first term as 8 and common difference as 8.
Therefore, a = 8
d = 8
S₁₅ = ?
S_n = n/2 [2a + (n – 1)d]
S₁₅ = 15/2 [2(8) + (15 – 1)8]
= 15/2[6 + (14) (8)]
= 15/2[16 + 112]
= 15(128)/2
= 15 × 64
= 960

14. Find the sum of the odd numbers between 0 and 50.

Answer

The odd numbers between 0 and 50 are
1, 3, 5, 7, 9 … 49
Therefore, it can be observed that these odd numbers are in an A.P.
a = 1
d = 2
l = 49
l = a + (n − 1) d
49 = 1 + (n − 1)2
48 = 2(n − 1)
n − 1 = 24
n = 25
S_n = n/2 (a + l)
S₂₅ = 25/2 (1 + 49)
= 25(50)/2
=(25)(25)
= 625

15. A contract on construction job specifies a penalty for delay of completion beyond a certain dateas follows: Rs. 200 for the first day, Rs. 250 for the second day, Rs. 300 for the third day, etc., the penalty for each succeeding day being Rs. 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days.

Answer

It can be observed that these penalties are in an A.P. having first term as 200 and common difference as 50.
a = 200
d = 50
Penalty that has to be paid if he has delayed the work by 30 days = S₃₀
= 30/2 [2(200) + (30 – 1) 50]

= 15 [400 + 1450]
= 15 (1850)
= 27750
Therefore, the contractor has to pay Rs 27750 as penalty.

16. A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each of the prizes.

Answer

Let the cost of 1^st prize be P.
Cost of 2^nd prize = P − 20
And cost of 3^rd prize = P − 40
It can be observed that the cost of these prizes are in an A.P. having common difference as −20 and first term as P.
a = P
d = −20
Given that, S₇ = 700
7/2 [2a + (7 – 1)d] = 700

a + 3(−20) = 100
a − 60 = 100
a = 160
Therefore, the value of each of the prizes was Rs 160, Rs 140, Rs 120, Rs 100, Rs 80, Rs 60, and Rs 40.

17. In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of class I will plant 1 tree, a section of class II will plant 2 trees and so on till class XII. There are three sections of each class. How many trees will be planted by the students?

Answer

It can be observed that the number of trees planted by the students is in an AP.
1, 2, 3, 4, 5………………..12
First term, a = 1
Common difference, d = 2 − 1 = 1
S_n = n/2 [2a + (n – 1)d]
S₁₂ = 12/2 [2(1) + (12 – 1)(1)]
= 6 (2 + 11)
= 6 (13)
= 78
Therefore, number of trees planted by 1 section of the classes = 78
Number of trees planted by 3 sections of the classes = 3 × 78 = 234