Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 5 Geometry Ex 5.2 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 5 Geometry Ex 5.2

Question 1.

Fill in the blanks:

(i) If in a ∆ PQR, PR^{2} = PQ^{2} + QR^{2}, then the right angle of ∆ PQR is at the vertex _______ .

Answer:

Q

Hint:

(ii) If ‘l‘ and ‘m’ are the legs and ‘n’ is the hypotenuse of a right angled triangle then, l^{2} = _______ .

Answer:

n^{2} – m^{2}

Hint:

(iii) If the sides of a triangle are in the ratio 5:12:13 then, it is _______ .

Answer:

a right angled triangle

Hint:

13^{2} = 169

5^{2} = 25

12^{2} = 144

169 = 25 + 144

∴ 13^{2} = 5^{2} + 12^{2}

(iv) The medians of a triangle cross each other at _______ .

Answer:

Centroid

(v) The centroid of a triangle divides each medians in the ratio _______ .

Answer:

2 : 1

Question 2.

Say True or False.

(i) 8, 15, 17 is a Pythagorean triplet.

Answer:

True

Hint:

17^{2} = 289

15^{2} = 225

8^{2} = 64

64 + 225 = 289 ⇒ 17^{2} = 15^{2} + 8^{2}

(ii) In a right angled triangle, the hypotenuse is the greatest side.

Answer:

True

Hint:

(iii) In any triangle the centroid and the incentre are located inside the triangle.

Answer:

True

(iv) The centroid, orthocentre, and incentre of a triangle are collinear.

Answer:

True

(v) The incentre is equidistant from all the vertices of a triangle.

Answer:

False

Question 3.

Check whether given sides are the sides of right-angled triangles, using Pythagoras theorem.

(i) 8, 15, 17

Answer:

Take a = 8 b = 15 and c = 17

Now a^{2} + b^{2} = 8^{2} + 15^{2} = 64 + 225 = 289

172 = 289 = c^{2}

∴ a^{2} + b^{2} = c^{2}

By the converse of Pythagoras theorem, the triangle with given measures is a right angled triangle.

∴ Ans: yes

(ii) 12, 13, 15

Answer:

(ii) 12, 13. 15

Take a = 12,b = 13 and c = 15

Now a^{2} + b^{2} = 12^{2} + 13^{2} = 144 + 169 = 313

15^{2} = 225 ≠ 313

By the converse of Pythagoras theorem, the triangle with given measures is not a right angled triangle.

∴ Ans: No.

(iii) 30, 40, 50

Answer:

Take a = 30, b = 40 and c = 50

Now a^{2} + b^{2} = 30^{2} + 40^{2} = 900 + 1600 = 2500

C^{2} = 50^{2} = 2500

∴ a^{2} + b^{2} = c^{2}

By the converse of Pythagoras theorem, the triangle with given measures is a right

angled triangle.

∴ Ans: yes

(iv) 9, 40, 41

Answer:

Take a = 9, b = 40 and c = 41

Now a^{2} + b^{2} = 9^{2} + 40^{2} = 81 + 1600 = 1681

c^{2} = 41^{2} = 1681

∴ a^{2} + b^{2} = c^{2}

By the converse of Pythagoras theorem, the triangle with given measures is a right

angled triangle.

∴ Ans: yes

(v) 24, 45, 51

Answer:

Take a = 24,b = 45 and c = 51

Now a^{2} + b^{2} = 24^{2} + 45^{2} = 576 + 2025 = 2601

c^{2} = 51^{2} = 2601

∴ a^{2} + b^{2} = c^{2}

By the converse of Pyhtagoreas theorem, the triangle with given measure is a right angled triangle.

∴ Ans: yes

Question 4.

Find the unknown side in the following triangles.

(i)

Answer:

From ∆ ABC, by Pythagoras theorem

BC^{2} = AB^{2} + AC^{2}

Take AB^{2} + AC^{2} = 9^{2} + 40^{2} = 81 + 1600 = 1681

BC^{2} = AB^{2} + AC^{2} = 1681 = 41^{2}

BC^{2} = 41^{2} ⇒ BC = 41

∴ x = 41

(ii)

Answer:

From ∆ PQR, by Pythagoras theorem.

PR^{2} = PQ^{2} + QR^{2}

34^{2} = y^{2} + 30^{2}

⇒ y^{2} = 34^{2} – 30^{2}

= 1156 – 900

= 256 = 16^{2}

y^{2} = 16^{2} ⇒ y = 16

(iii)

Answer:

From ∆ XYZ, by Pythagoras theorem,

= YZ^{2} = XY^{2} + XZ^{2}

⇒ XY^{2} = YZ^{2} – XZ^{2}

Z^{2} = 39^{2} – 36^{2}

= 1521 – 1296

= 225 = 15^{2}

z^{2} = 15^{2}

⇒ z = 15

Question 5.

An isosceles triangle has equal sides each 13 cm and a base 24 cm in length. Find its height.

Answer:

In an isosceles triangle the altitude dives its base into two equal parts.

Now in the figure, ∆ABC is an isosceles triangle with AD as its height

In the figure, AD is the altitude and ∆ABD is a right triangle.

By Pythagoras theorem,

AB^{2} = AD^{2} + BD^{2}

⇒ AD^{2} = AB^{2} – BD^{2}

= 13^{2} – 12^{2} = 169 – 144 = 25

AD^{2} = 25 = 5^{2}

Height: AD = 5cm

Question 6.

Find the distance between the helicopter and the ship.

Answer:

From the figure AS is the distance between the helicopter and the ship.

∆ APS is a right angled triangle, by Pythagoras theorem,

AS^{2} = AP^{2} + PS^{2}

= 80^{2} + 150^{2} = 6400 + 22500 = 28900 = 170^{2}

∴ The distance between the helicopter and the ship is 170 m

Question 7.

In triangle ABC, line I, is a perpendicular bisector of BC.

If BC = 12cm, SM = 8cm, find CS.

Answer:

Given l_{1}, is the perpendicular bisector of BC.

∴ ∠SMC = 90°and BM = MC

BC = 12cm

⇒ BM + MC = 12cm

MC + MC = 12cm

2MC = 12

MC = \(\frac{12}{2}\)

MC = 6cm

Given SM = 8 cm

By Pythagoras theorem SC^{2} = SM^{2} + MC^{2}

SC^{2} = 8^{2} + 6^{2}

SC^{2} = 64 + 36

CS^{2} = 100

CS^{2} = 10^{2}

CS = 10 cm

Question 8.

Identify the centroid of ∆PQR.

Answer:

In ∆PQR, PT = TR ⇒ QT is a median from vertex Q.

QS = SR ⇒ PS is a median from vertex P.

QT and PS meet at W and therefore W is the centroid of ∆PQR.

Question 9.

Name the orthocentre of ∆PQR.

Answer:

This is a right triangle

∴ orthocentre = P [∵ In right triangle orthocentre is the vertex containing 90°]

Question 10.

In the given figure, A is the midpoint of YZ and G is the centroid of the triangle XYZ. If the length of GA is 3 cm, find XA.

Answer:

Given A is the midpoint of YZ.

∴ ZA = AY

G is the centroid of XYZ centroid divides each median in a ratio 2 : 1 ⇒ XG : GA = 2:1

\(\frac{\mathrm{XG}}{\mathrm{GA}}=\frac{2}{1}\)

\(\frac{\mathrm{XG}}{3}=\frac{2}{1}\)

XG = 2 × 3

XG = 6 cm

XA = XG + GA = 6 + 3 ⇒ XA = 9cm

Question 11.

If I is the incentre of ∆XYZ, ∠IYZ = 30° and ∠IZY = 40°, find ∠YXZ.

Answer:

Since I is the incentre of AXYZ

∠IYZ = 30° ⇒ ∠IYX = 30°

∠IZY = 40° ⇒ ∠IZX = 40°

∴ ∠XYZ = ∠XYI + ∠IYZ = 30° + 30°

∠XYZ = 60°

∠XYZ = 80°

By angle sum property of a triangle

∠XZY + ∠XYZ + ∠YXZ = 180°

80° + 60° + ∠YXZ = 180°

140° + ∠YXZ = 180°

∠YXZ = 180° – 140°

∠YXZ = 40°

Objective Type Questions

Question 12.

If ∆GUT is isosceles and right angled, then ∠TUG is ______ .

(A) 30°

(B) 40°

(C) 45°

(D) 55°

Answer:

(C) 45°

Hint:

∠U ∠T = 45° (∵ GUT is an isosceles given)

∴ ∠TUG = 45°

Question 13.

The hypotenuse of a right angled triangle of sides 12cm and 16cm is ______ .

(A) 28 cm

(B) 20 cm

(C) 24 cm

(D) 21 cm

Answer:

(B) 20 cm

Hint:

Side take a = 12 cm

b = 16cm

The hypotenuse c^{2} = a^{2} + b^{2}

= 12^{2} + 16^{2}

= 144 + 256

c^{2} = 400 ⇒ c = 20 cm

Question 14.

The area of a rectangle of length 21cm and diagonal 29 cm is ______ .

(A) 609 cm^{2}

(B) 580 cm^{2}

(C) 420 cm^{2}

(D) 210 cm^{2}

Answer:

(C) 420 cm^{2}

length = 21 cm

diagonal = 29 cm

By the converse of Pythagoras theorem,

AB^{2} + BC^{2} = AC^{2}

21^{2} + x^{2} = 29^{2}

x^{2} = 841 – 441 400 = 20^{2}

x = 20 cm

Now area of the rectangle = length × breadth.

ie AB × BC = 21 cm × 20 cm = 420 cm^{2}

Question 15.

The sides of a right angled triangle are in the ratio 5:12:13 and its perimeter is 120 units then, the sides are .

(A) 25, 36, 59

(B) 10, 24, 26

(C) 36, 39, 45

(D) 20, 48, 52

Answer:

(D) 20,48,52

Hint:

The sides of a right angled triangle are in the ratio 5 : 12 : 13

Take the three sides as 5a, 12a, 13a

Its perimeter is 5a + 12a + 13a = 30a

It is given that 30a = 120 units

a = 4 units

∴ the sides 5a = 5 × 4 = 20 units

12a = 12 × 4 = 48 units

13a = 13 × 4 = 52 units