8th Maths : Chapter 5 : Geometry : Congruent Triangles and Similar Triangles : Exercise 5.1 : Text Book Back Numerical problems, Exercises Questions with Answers, Solution

**Exercise 5.1**

**1. Fill in the blanks with the correct
term from the given list.**

(in proportion, similar,
corresponding, congruent, shape, area, equal)

(i) Corresponding
sides of similar triangles are _______. **[Answer: in proportion]**

(ii) Similar
triangles have the same _________ but not necessarily the same size. **[Answer: shape]**

(iii) In
any triangle ______ sides are opposite to equal angles. **[Answer: equal]**

(iv) The
symbol ≡ is used to represent _______ triangles. **[Answer: congruent]**

(v) The symbol
~ is used to represent ________ triangles. **[Answer: similar]**

**2. In the given figure, ****∠****CIP ≡ ****∠****COP and ****∠****HIP ≡ ****∠****HOP . Prove that IP ≡ OP.**

**Solution:**

__Statements : ____Reasons__

**1. CI = CO :** ∵ CIP ≡ COP, by CPCTC

**2. IP = OP :** By CPCTC

**3. CP = CP : **By CPCTC

**4. Also HI = HO : **CPCTC ΔHIP ≡ HOP given

**5. IP = OP : **By CPCTC and (4)

**6. ****∴ ****IP ****≡**** OP : **By (2) and (4)

** **

**3. In the given figure, AC ≡ AD and ****∠****CBD ≡ ****∠****DEC . Prove that ∆ BCF ≡ ∆EDF .**

**Solution:**

__Statements : ____Reasons__

**1. ****∠****BFC ****=**** ****∠****EFD : **vertically opposite angles

**2. ****∠****CBD ****=**** ****∠****DEC : **Angles on the same base given

**3. ****∠****BCF ****=**** ****∠****EDF : **Remaining angles of ΔBCF and ΔEDF

**4. ****Δ****BCF ****≡**** ****Δ****EDF : **By (1) and (2) AAA criteria

** **

**4. In the given figure, ∆ BCD is isosceles with base BD and **

**Solution:**

__Statements : ____Reasons__

**1. ****∠****BAE ****≡**** ****∠****DEA : **Given

**2. AC = EC : **By (1) sides opposite to equal angles are equal

**3. BC = DC : **Given BCD is isosceles with base BD

**4. AC − BC = EC −
DC : **2 − 3

**5. AB ****≡**** ED : **By 4

** **

**5. In the given figure, D is the midpoint of OE and **

**Solution:**

__Statements :____ Reasons__

**1. OD = ED : **D is the midpoint of OE (given)

**2. DC = DC : **Common side

**3. ****∠****CDE = ****∠****CDO = 90° : **Linear pair and given ∠CDE = 90°

**4. ΔODC ****≡**** ΔEDC : **By RHS criteria

** **

**6. Is PRQ **

**Solution:**

In ΔPRQ and ΔPSQ

∠PRQ = ∠PSQ = 90° given

PR = QS = 3 cm given

PQ = PQ = 5 cm common

It satisfies RHS criteria

∴ ΔPRQ congruent to ΔQSP.

** **

**7. From the given figure, prove that
****∆ABC
~ ∆***EDF*

**Solution:**

From the ΔABC, AB = AC

It is an isosceles triangle

Angles opposite to equal sides are equal

∴ ∠B = ∠C = 65°

∴ ∠B + ∠C = 65° + 65°

= 130°

We know that sum of three angles is a triangle = 180°

∠A + ∠B + ∠C = 180°

∠A + 130° = 180°

∠A = 180° − 130°

∠A = 50°

From ΔEDF, ∠E = 50°

∴ Sum of Remaining angles
= 180° − 50° = 130°

DE = FD

∴ ∠D = ∠F

From ΔABC and ΔEDF ∴

∠D = 130/2 = 65°

∠A = ∠E = 50°

∠B = ∠D = 65°

∠C = ∠F = 65°

∴ By AAA criteria ΔEDF ≈ ΔABC

** **

**8. In the given figure YH || TE . Prove that ∆WHY
~ ∆WET and also find HE and TE.**

Solution:

__Statements : ____Reasons__

**1. ****∠E****WT = ****∠****HWY :** Common Angle

**2. ****∠E****TW = ****∠****HYW :** Since YH ‖‖ TE, corresponding angles

**3. ****∠****WET = ****∠****WHY :** Since YH ‖‖ TE corresponding angles

**4. ΔWHY****∼**** ΔWET : **By AAA criteria

Also ΔWHY ~ ΔWET

∴ Corresponding sides are
proportionated

WH / WE = HY / TE = WY / WT

6 / [ 6 + HE ] = 4/TE = 4/16

6 / [6 + HE] = 4/16

6 + HE = [6/4] × 16

6 + HE = 24

∴ HE = 24 − 6

HE = 18

Again 4/TE = 4/16

ET = 4/4

TE = 16

** **

**9. In the given figure, if ∆ EAT ~ ∆BUN , find the measure of all angles.**

**Solution:**

Given ΔEAT ≡ ΔBUN

∴ Corresponding angles are
equal

∴ ∠E = ∠B ...(1)

∠A = ∠U

∠T = ∠N

∠E = *x*°

∠A = 2*x*°

Sum of three angles of a triangle = 180°

In ΔEAT, *x *+ 2*x* +
∠T = 180°

∠T = 180° − (*x*° + 2*x*°)

∠T = 180° − 3*x*° ……..(4)

Also in ΔBUN

(*x *+ 40)° +* x*° + ∠U = 180°

*x* + 40° + *x* + ∠U = 180°

2*x*° + 40° + ∠U = 180°

∠U = 180° − 2*x* − 40° = 140°− 2*x*°

Now by (2)

∠A = ∠U

2*x* = 140° − 2*x*°

2*x* + 2*x* = 140°

4*x* = 140°

*x* = 140°/4 = 35°

∠A = 2*x*° = 2 × 35° = 70°

∠N = *x* + 40° = 35° + 40° = 75°

∴ ∠T = ∠N = 75°

∠E = ∠B = 35°

∠A = ∠U = 70°

** **

**10. In the given figure, UB ||AT
and CU **

**Solution:**

__Statements:____ Reasons__

**1. ****∠****CUB = ****∠****CBU : **∵ In ΔCUB, CU = CB

**2. ****∠****CUB = ****∠****CAB : **∵ UB **|| **AT,
corresponding angle if CA is the transversal.

**3. ****∠****CBU = ****∠****CTA : **CT is transversal UB **|| **AT, corresponding angle common
angle.

**4. ****∠****UCB****=**** ****∠****ACT : **common angle

**5. ΔCUB ****∼**** ΔCAT : **By AAA criteria

**6. CA =
CT :** ∵ ∠CAT = ∠CTA

** 7. Also ****Δ****CAT is isosceles : **By 1, 2 and 3 and sides opposite to equal angles are equal.

** **

**Objective
Type Questions**

**11. Two similar triangles will always
have ________angles **

(A) acute

(B) obtuse

(C) right

(D) matching

**[Answer: (D) matching]**

**12. If in triangles PQR and XYZ, PQ/
XY = QR/YZ then they will be similar if**

(A) ∠Q = ∠Y

(B) ∠P = ∠Y

(C) ∠Q = ∠X

(D) ∠P = ∠Z

**[Answer: (a) ****∠****Q = ****∠****Y]**

**13. A flag pole 15 m high casts a shadow of 3 m at 10 a.m. The shadow cast by a building
at the same time is 18.6 m. The height
of the building is**

(A) 90 *m*

(B) 91 *m*

(C) 92 *m*

(D) 93 *m*

**[Answer: (D) 93 m]**

**14. If ****∆ABC**** ~ ****∆PQR**** in
which ****∠***A*** = 53º and ****∠***Q*** = 77º
, then ****∠***R*** is**

(A) 50°

(B) 60°

(C) 70°

(D) 80°

**[Answer: (A) 50**°**]**

**15. In the figure, which of the following
statements is true?**

(A) *AB* = *BD*

(B) *BD* < *CD*

(C) *AC* = *CD*

(D) *BC* = *CD*

**[Answer: (C) AC = CD]**

**Answer:**

**Exercise 5.1 **

**1. (i) in proportion
(ii) shape (iii) equal (iv) congruent (v) similar**

**6. Yes, RHS Congruence
**

**8. HE =18,TE =16 **

**9. ****∠****T = ****∠****N = 75º,****∠****E = ****∠****B = 35º,****∠****A = ****∠****U = 70º**

**11. (D) matching **

**12.(A) ****∠****Q = ****∠****Y **

**13. (D) 93 m
**

**14. (A) 50º **

**15. (C) AC = CD**

Tags : Questions with Answers, Solution | Geometry | Chapter 5 | 8th Maths , 8th Maths : Chapter 5 : Geometry

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8th Maths : Chapter 5 : Geometry : Exercise 5.1 (Congruent and Similar Triangles) | Questions with Answers, Solution | Geometry | Chapter 5 | 8th Maths

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