Samacheer Kalvi 8th Maths Guide Chapter 5 Geometry Ex 5.1

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 5 Geometry Ex 5.1 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 5 Geometry Ex 5.1

Question 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

Question 2.
In the figure, ∠CIP ≡ ∠COP and ∠HIP ≡ ∠HOP. Prove that IP ≡ OP.
Samacheer Kalvi 8th Maths Guide Answers Chapter 5 Geometry Ex 5.1 1
Answer:

Statements	Reasons
1. CI = CO	∵ CIP ≡ COP, by CPCTC
2. IP = OP	By CPCTC
3. CP = CP	By CPCTC
4. Also HI = HO	CPCTC ∆HIP ≡ HOP given
5. IP = OP	By CPCTC and (4)
6. ∴ IP ≡ OP	By (2) and (4)

Question 3.
In the given figure, AC ≡ AD and ∠CBD ≡ ∠DEC. Prove that ∆ BCF ≡ ∆ EDF.
Samacheer Kalvi 8th Maths Guide Answers Chapter 5 Geometry Ex 5.1 2
Answer:

Statements	Reasons
1. ∠BCF = ∠EFD	Vertically opposite angles
2. ∠CBD = ∠DEC	Angles on the same base given
3. ∠BCF = ∠EDF	Remaining angles of ∆BCF and ∆EDF
4. ∆BCF ≡ ∆EDF	By (1) and (2) AAA criteria

Question 4.
In the given figure, △ BCD is isosceles with base BD and ∠BAE ≡∠DEA. Prove that AB ≡ ED.
Samacheer Kalvi 8th Maths Guide Answers Chapter 5 Geometry Ex 5.1 3
Answer:

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

Question 5.
In the given figure, D is the midpoint of OE and ∠CDE = 90°. Prove that △ODC ≡ △EDC
Samacheer Kalvi 8th Maths Guide Answers Chapter 5 Geometry Ex 5.1 4
Answer:

Statements	Reasons
1. OD = ED	D is the midpoint OE (given)
2. DC = DC	Common side
3. ∠CDE = ∠CDO = 90°	Linear pair and given ∠CDE = 90°
4. △ODC ≡ △EDC	By RHS criteria

Question 6.
Is △PRQ ≡ △QSP? Why?
Samacheer Kalvi 8th Maths Guide Answers Chapter 5 Geometry Ex 5.1 5
Answer:
In △PRQ and △PSQ
∠PRQ = ∠PSQ = 90° given
PR = QS = 3 cm given
PQ = PQ = 5 cm common
It satisfies RHS criteria
∴ △PRQ congruent to △QSP.

Question 7.
From the given figure, prove that △ABC ~ △EDF
Samacheer Kalvi 8th Maths Guide Answers Chapter 5 Geometry Ex 5.1 6
Answer:
From the △ABC, 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 △EDF, ∠E = 50°
∴ Sum of Remaining angles = 180° – 50° = 130°
DE = FD
∴ ∠D = ∠F
From △ABC and △EDF ∴ △D = \(\frac{130}{2}\) = 65°
∠A = ∠E = 50°
∠B = ∠D = 65°
∠C = ∠F = 65°
∴ By AAA criteria △EDF ~ △ABC

Question 8.
In the given figure YH || TE. Prove that △WHY ~ △WET and also find HE and TE.
Samacheer Kalvi 8th Maths Guide Answers Chapter 5 Geometry Ex 5.1 7
Answer:

Statements	Reasons
1. ∠EWT = ∠HWY	Common angle
2. ∠ETW = ∠HYW	Since YH \|\| TE, corresponding angles
3. ∠WET = ∠WHY	Since YH \|\| TE corresponding angles
4. △WHY ~ △WET	By AAA criteria

Also △ WHY ~ △WET
∴ Corresponding sides are proportionated
Samacheer Kalvi 8th Maths Guide Answers Chapter 5 Geometry Ex 5.1 8
⇒ 6+HE = \(\frac{6}{4}\) × 16
⇒ 6 + HE = 24
∴ HE = 24 – 6
HE = 18
Again \(\frac{4}{\mathrm{ET}}=\frac{4}{16}\)
ET = \(\frac{4}{4}\)
ET = 16

Question 9.
In the given figure, if △EAT ~ △BUN, find the measure of all angles.
Samacheer Kalvi 8th Maths Guide Answers Chapter 5 Geometry Ex 5.1 9
Answer:
Given △EAT ≡ △BUN
∴ Corresponding angles are equal
∴ ∠E = ∠B …… (1)
∠A = ∠U …… (2)
∠T = ∠N ……. (3)
∠E = x°
∠A = 2x°
Sum of three angles of a triangle = 180°
In △ EAT, x + 2x + ∠T = 180°
∠T = 180° – (x° + 2x°)
∠T = 180° – 3x°
Also in △BUN
(x + 40)° + x° + ∠U = 180°
x + 40° + x + ∠U = 180°
2x° + 40° + ∠U = 180°
∠U = 180° – 2x – 40° = 140° – 2x°
Now by (2)
∠A = ∠U
2x = 140° – 2x°
2x + 2x = 140°
4x = 140°
x = \(\frac{140}{4}\) = 35°
∠A = 2x° = 2 × 35° = 70°
∠N = x + 40° = 35° + 40° = 75°
∴ ∠T = ∠N = 75°
∠E = ∠8 = 35°
∠A = ∠U = 70°

Question 10.
In the given figure, UB || AT and CU ≡ CB Prove that △CUB ~ △CAT and hence △CAT is isosceles.
Samacheer Kalvi 8th Maths Guide Answers Chapter 5 Geometry Ex 5.1 10
Answer:

Statements	Reasons
1. ∠CUB = ∠CBU	∵ In △CUB, CU = CB
2. ∠CUB = ∠CAB	∵ UB \|\| AT, Corresponding angle if CA is the transversal.
3. ∠CBU = ∠CTA	CT is transversal UB \|\| AT, Corresponding angle commom angle.
4. ∠UCB = ∠ACT	Common angle
5. △CUB ~ △CAT	By AAA criteria
6. CA = CT	∵ ∠CAT = ∠CTA
7. Also △CAT is isoceles	By 1, 2 and 3 and sides opposite to equal angles are equal.

Objective Type Questions

Question 11.
Two similar triangles will always have _______ angles
(A) acute
(B) obtuse
(C) right
(D) matching
Answer:
(D) matching

Question 12.
If in triangles PQR and XYZ, \(\frac{\mathrm{PQ}}{\mathrm{XY}}=\frac{\mathrm{QR}}{\mathrm{ZX}}\) then they will be similar if
(A) ∠Q = ∠Y
(B) ∠P = ∠X
(C) ∠Q = ∠X
(D) ∠P ≡∠Z
Ans:
(C) ∠Q = ∠X

Question 13.
A flag pole 15 m high casts a shadow of 3m at 10 a.m. The shadow cast by a building at the same time is 18.6m. The height of the building is
(A) 90 m
(B) 91 m
(C) 92 m
(D) 93 m
Answer:
(D) 93 m

Question 14.
If ∆ABC – ∆PQR in which ∠A = 53° and ∠Q = 77°, then ∠R is
(A) 50°
(B) 60°
(C) 70°
(D) 80°
Answer:
(A) 50°

Question 15.
In the figure, which of the following statements is true?
Samacheer Kalvi 8th Maths Guide Answers Chapter 5 Geometry Ex 5.1 11
(A) AB = BD
(B) BD < CD
(C) AC = CD
(D) BC = CD
Answer:
(C) AC = CD