AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions

AP State Syllabus SSC 10th Class Maths Solutions 8th Lesson Similar Triangles InText Questions

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 8 Similar Triangles InText Questions and Answers.

10th Class Maths 8th Lesson Similar Triangles InText Questions and Answers

Do these

(Page No. 194)

Question 1.
Fill in the blanks with similar/not similar.
i) All squares are ………. (similar)
ii) All equilateral triangles are ………. (similar)
iii) All isosceles triangles are ………. (similar)
iv) Two polygons with same number of sides are ………, if their corresponding angles are equal and corresponding sides are equal. (similar)
v) Reduced and enlarged photographs of an object are ………. (similar)
vi) Rhombus and squares are ……… to each other. (not similar)

AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions

Question 2.
Write True / False for the following statements.
i) Any two similar figures are congruent.
Answer:
False
ii) Any two congruent figures are similar.
Answer:
True
iii) Two polygons are similar if their corresponding angles are equal.
Answer:
False

Question 3.
Give two different examples of pair of
i) Similar figures
ii) Non-similar figures
Answer:
i) Similar figures:
a) Any two circles
b) Any two squares
c) Any two equilateral triangles
ii) Non-similar figures:
a) A square and a rhombus
b) A square and a rectangle

Question 4.
What value(s) of x will make DE || AB, in the given figure?  (Page No. 200)
AD = 8x + 9, CD = x + 3,
BE = 3x + 4, CE = x.
AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions 1
Answer:
Given : In AABC, DE // AB AD = 8x + 9, CD = x + 3,
BE = 3x + 4, CE = x
By Basic proportionality theorem,
If DE // AB then we should have
\(\frac{\mathrm{CD}}{\mathrm{DA}}\) = \(\frac{\mathrm{CE}}{\mathrm{EB}}\)
\(\frac{x+3}{8x+9}\) = \(\frac{x}{3x+4}\)
⇒ (x + 3) (3x + 4) = x (8x + 9)
⇒ x (3x + 4) + 3 (3x + 4) – 8x2 + 9x
⇒ 3x2 + 4x + 9x + 12 = 8x2 + 9x
⇒ 8x2 + 9x – 3x2 – 4x – 9x -12 = 0
⇒ 5x2 – 4x – 12 = 0
⇒ 5x2 – 10x + 6x – 12 = 0
⇒ 5x (x – 2) + 6 (x – 2) = 0
⇒ (5x + 6) (x – 2) = 0
⇒ 5x + 6 = 0 or x – 2 = 0
⇒ x = \(\frac{-6}{5}\) or x = 2;
x cannot be negative.
∴ The value x = 2 will make DE // AB.

AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions

Question 5.
In △ABC, DE || BC. AD = x, DB = x – 2, AE = x + 2 and EC = x – 1. Find the value of x.    (Page No. 200)
AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions 2
Answer:
Given: In △ABC, DE // BC
∴ By Basic proportionality theorem, we have
\(\frac{\mathrm{AD}}{\mathrm{DB}}\) = \(\frac{\mathrm{AE}}{\mathrm{EC}}\)
\(\frac{\mathrm{x}}{\mathrm{x}-2}\) = \(\frac{x+2}{x-1}\)
⇒ x (x – 1) = (x + 2) (x – 2)
⇒ x2 – x = x2 – 4
⇒ -x = -4
∴ x = 4

Try This

Question 1.
E and F are points on the sides PQ and PR respectively of △PQR. For each of the following, state whether EF || QR or not?      (Page No. 197)
i) PE = 3.9 cm, EQ = 3 cm,
PF = 3.6 cm and FR = 2.4 cm.
Answer:
AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions 3
Here
\(\frac{\mathrm{PE}}{\mathrm{EQ}}\) = \(\frac{3.9}{3}\) = \(\frac{1.3}{1}\)
\(\frac{\mathrm{PF}}{\mathrm{FR}}\) = \(\frac{3.6}{2.4}\) = \(\frac{3}{2}\)
\(\frac{\mathrm{PE}}{\mathrm{EQ}}\) ≠ \(\frac{\mathrm{PF}}{\mathrm{FR}}\)
Hence, EF is not parallel to QR.

AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions

ii) PE = 4 cm, QE = 4.5 cm,
Answer:
Here
\(\frac{\mathrm{PE}}{\mathrm{EQ}}\) = \(\frac{4}{4.5}\) = \(\frac{8}{9}\)
\(\frac{\mathrm{PF}}{\mathrm{RF}}\) = \(\frac{8}{9}\)
\(\frac{\mathrm{PE}}{\mathrm{EQ}}\) = \(\frac{\mathrm{PF}}{\mathrm{RF}}\)
∴ EF // QR
Hence, EF is parallel to QR.

iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 1.8 cm and PF = 3.6 cm.
Answer:
AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions 4
Given: PQ = 1.28 cm, PE = 1.8 cm
⇒ EQ = PE – PQ = 1.8 – 1.28
⇒ EQ = 0.52 cm
Also, PR = 2.56 cm, PE = 3.6 cm, FR = PF – PR = 3.6 cm – 2.56 cm
FR = 1.04 cm
AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions 5
∴ EF // QR (By converse of Basic proportionality theorem)
Hence, EF is parallel to QR.

Question 2.
In the following figures DE || BC.    (Page No. 198)
i) Find EC.
AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions 6
Answer:
\(\frac{\mathrm{AD}}{\mathrm{DB}}\) = \(\frac{AE}{EC}\)
⇒ \(\frac{1.5}{3}\) = \(\frac{1}{EC}\)
∴ EC = \(\frac{3}{1.5}\) = 2 cm

ii) Find AD.
AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions 7
Answer:
AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions 8

AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions

Think & Discuss

Question 1.
Can you give some more examples from your daily life where scale factor is used?    (Page No. 192)
Answer:
Scale factor is used in drawing maps, designing machines and in sculpture, etc.

Question 2.
Can you say that a square and a rhombus are similar? Discuss with your friends.Write why the conditions are not sufficient.       (Page No. 193)
Answer:
A square □ ABCD and a rhombus ▱ PQRS are not similar.
Though the ratio of their corresponding sides are equal, the corresponding angles are not equal.
AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions 9
but ∠A ≠ ∠P; ∠B ≠ ∠Q
∠C ≠ ∠R; ∠D ≠ ∠S

Try This

(Page No. 207)

Question 1.
Are the triangles similar ? If so, name the criterion of similarity. Write the similarity relation in symbolic form.
i)
AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions 10
!! ∠G = ∠I alt. int. angles for the ∠F = ∠K parallel lines GF // KI
Answer:
∠FHG = ∠IHK (Vertically opp. angle)
∴ ∠GHF and ∠IKH are similar by AAA similarity rule.
△GHF ~ △IKH.

ii)
AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions 11
Answer:
\(\frac{PQ}{LM}\) = \(\frac{6}{3}\) = 2;
\(\frac{QR}{MN}\) = \(\frac{10}{4}\) = 2.5;
\(\frac{PQ}{LM}\) ≠ \(\frac{QR}{MN}\)
△PQR and △LMN are not similar.
△PQR ~ △LMN

AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions

iii)
AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions 12
Answer:
∠A = ∠A (Common)
\(\frac{AB}{AX}\) = \(\frac{5}{3}\); \(\frac{AC}{AY}\) = \(\frac{5}{3}\)
∴ △ABC and △AXY are similar by SAS similarity condition.
△ABC ~ △AXY.

iv)
AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions 13
Answer:
AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions 14
∴ △ABC and △APJ are not similar.

v)
AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions 15
Answer:
∠A = ∠B = 90°
∠AOQ = ∠POB (∵ Vertically opposite angles)
∠Q = ∠P (alternate interior angles)
∴ △AOQ and △BOP are similar by AAA criterion.
△AOQ ~ △BOP.

AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions

vi)
AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions 16
Answer:
△ABC and △QPR are similar by AAA similarity condition.
△ABC ~ △QPR.

vii)
AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions 17
Answer:
∠A = ∠P
\(\frac{AB}{PQ}\) = \(\frac{2}{5}\); \(\frac{AC}{PR}\) = \(\frac{3}{6}\) = \(\frac{1}{2}\)
∴ \(\frac{AB}{PQ}\) ≠ \(\frac{AC}{PR}\)
Hence not similar.

viii)
AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions 18
Answer:
\(\frac{AB}{PQ}\) = \(\frac{6}{2.5}\); \(\frac{AC}{PR}\) = \(\frac{10}{5}\)
∴ △ABC and △PQR are not similar.

Question 2.
Explain why the triangles are similar and then find the value of x.
i)
AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions 19
Answer:
Given: In △PQR and △LTS
∠Q = ∠T; ∠R = ∠S = 90°
∴ ∠P = ∠T
(by angle sum property of triangles)
Hence, △PQR ~ △LTS [∵ AAA]
AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions 20

AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions

ii)
AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions 21
Answer:
Given: In △ABC and △PQC
∠B = ∠Q
[∵ ∠PQC = 180°- 110° = 70° – linear pair of angles]
∠C = ∠C [∵ Common]
∠A = ∠P [∵ Angle Sum property of triangles]
△ABC ~ △PQC by AAA similarity condition.
Then the ratio of their corresponding sides are equal.
AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions 22

iii)
AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions 23
Answer:
Given: In △ABC and △ECD
∠A = ∠E
∠ACB = ∠ECD [∵ Vertically opposite]
∴ ∠B = ∠D [∵ angle sum property]
∴ △ABC ~ △EDC
AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions 23

iv)
AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions 25
Answer:
Given: In △RAB and △RST
∠R = ∠R (common)
∠A = ∠S and ∠B = ∠T
[∵ Pair of corresponding angles for AB // ST]
∴ △RAB ~ △RST [∵ AAA similarity]
AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions 34

AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions

v)
AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions 26
Answer:
Given: In △PQR and △PMN
∠P = ∠P [∵ Common]
∠Q = ∠M [∵ Pair of corresponding angles for MN // QR]
∠R = ∠N
△PQR ~ △PMN [∵ AAA similarity]
AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions 27
[From the figure, PR = 4 + x]
⇒ 3 × 4 = 4 + x
⇒ x = 12 – 4 = 8

vi)
AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions 28
Answer:
Given: In △XYZ and △XBA,
∠X = ∠X [∵ Common]
∠B = ∠Y [∵ Pair of corresponding ∠A = ∠Z angles for AB // ZY]
∴ △XYZ ~ △XBA [∵ AAA similarity]
AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions 29
[From the figure, XZ = 7.5 + x]
\(\frac{3}{2}\) = \(\frac{7.5+\mathrm{x}}{\mathrm{x}}\)
3x = 15 + 2x;
3x – 2x = 15
x = 15

vii)
AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions 30
Answer:
Given: With the given conditions, we can’t find the value of x.
Note: If it is given that ∠A = ∠E then
we can say that △ABC ~ △EDC by AAA rule
AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions 31

AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions

viii)
AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions 32
Answer:
In △ABC and △BEC
∠ABC = ∠CEB (given)
∠C = ∠C (Common angle)
∴ △ABC ~ △BEC
(A.A. Criterion similarity)
AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions 33

Think & Discuss

(Page No. 203)

Question 1.
Discuss with your friends that in what way similarity of triangles is dif¬ferent from similarity of other polygons?
Answer:
In two triangles if the corresponding angles are equal then they are similar, whereas in two polygons if the corre-sponding angles are equal, they may not be similar, i.e., In triangles,
(Pairs of corresponding angles are equal) ⇔ (Ratio of corresponding sides are equal).
But this is not so with respect to polygons.

Do This

AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions

Question 1.
In △ACB, ∠C = 90° and CD ⊥ AB. Prove that \(\frac{\mathrm{BC}^{2}}{\mathrm{AC}^{2}}\) = \(\frac{BD}{AD}\).    (Page No. 218)
AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions 35
Answer:
Proof: △ADC and △CDB are similar.
AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions 36
[Ratio of areas of similar triangles is equal to the ratio of squares of their corresponding sides.]
From (1) and (2),
\(\frac{BD}{AD}\) = \(\frac{\mathrm{BC}^{2}}{\mathrm{AC}^{2}}\) (Q.E.D.)

Question 2.
A ladder 15 m long reaches a window which is 9 m above the ground on one side of a street. Keeping its foot at the same point, the ladder is turned to other side of the street to reach a window 12 m high. Find the width of the street.    (Page No. 218)
Answer:
AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions 37
Let A and D be the windows on the either sides of the street.
From Pythagoras theorem,
AC2 = AB2 + BC2
152 = 92 + BC2
BC2 = 225 – 81
BC = √144 = 12 ….. (1)
Also, CD2 = DE2 + CE2
152 = 122 + CE2
CE2 = 225 – 144
CE = √81 = 9
∴ BE = BC + CE = 12 + 9 = 21
Width of the street = 21 m.

AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions

Question 3.
In the given figure if AD ⊥ BC, prove that AB2 + CD2 = BD2 + AC2.    (Page No. 219)
AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions 38
Answer:
Given: In △ABC, AD ⊥ BC.
R.T.P: AB2 + CD2 = BD2 + AC2
Proof: △ABD is a right angled triangle
AB2 – BD2 = AD2 ……. (1)
△ACD is a right angle triangle
AC2 – CD2 = AD2
From (1) and (2)
AB2 – BD2 = AC2 – CD2
AB2 + CD2 = BD2 + AC2

Think & Discuss

Question 1.
For a right angled triangle with integer sides atleast one of its measurements must be an even number. Why? Discuss this with your friends and teachers.    (Page No. 215)
Answer:
AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions 39
Let l, m, n are integer sides of a right
angled triangle.
then l2 – m2 + n2
⇒ n = l2 – m2 = (l + m) (l – m)
Now
Case I: Both l, m are even the (l + m) is even then (l + m) (l – m) is also even. So ‘n’ is even. Here all are even.
Case II: Both l, m are odd then (l + m) and (l – m) become even. Then the product of even numbers is even so ‘n’ is even.
Here only ‘n’ is even.
Case III: If we consider l is even, m is’ odd then ‘n’ will be odd. So here T is even. We observe in all above three cases at least one of l, m, n is even; Hence proved.