Students can Download Maths Chapter 3 Geometry Ex 3.1 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

## Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 3 Geometry Ex 3.1

Question 1.

Fill in the blanks:

Question (i)

If in a ∆ PQR, PR^{2} = PQ^{2} + QR^{2}, then the right angle of ∆ PQR is at the vertex ………

Answer:

Q

Hint:

Question:

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

Answer:

n^{2} – m^{2}

Hint:

Question (iii)

If the sides of a triangle are in the ratio 5:12:13 then, it is ………

Answer:

a right angled triangle.

13^{2} = 169

5^{2} = 25

12^{2} = 144

169 = 25 + 144

132 = 5^{2} + 12^{2}

By Pythagoras theorem, In a right triangle, square of the hypotenuse is equal to the sum of the squares of other two sides.

Question (iv)

If a perpendicular is drawn to the hypotenuse of a right angled triangle, then each of the three pairs of triangles formed are …………

Answer:

Similar.

Question (v)

In the figure BE^{2} = TE x ………

Answer:

EN

Question 2.

Say True or False.

Question (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}

Question (ii)

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

Answer:

False

Hint:

Question (iii)

One of the legs of a right angled triangle PQR having ∠R = 90° is PQ.

Answer:

False

Hint:

In ∆ PQR, QR and RP are legs and PQ is the hypotenuse

Question (iv)

The hypotenuse of a right angled triangle whose sides are 9 and 40 is 49.

Answer:

False

Hint:

49^{2} = 2401

9^{2} = 81

40^{2} = 1600

40^{2} + 9^{2} = 1600 + 81 + 1681

49^{2} = 2401

2401 ≠ 1681

Question (v)

Pythagoras theorem is true for all types of triangles.

Answer:

False

Hint:

Pyhtagoras theorem is true for only right angled triangles.

Question 3.

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

- 8, 15, 17
- 12, 13, 15
- 30, 40, 50
- 9, 40, 41
- 24, 45, 51

Solution:

1. 8, 15, 17

Take a = 8,

b = 15 and

c = 17

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

17^{2} = 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.

Answer:

yes.

2. 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.

Answer:

No.

3. 30, 40, 50

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.

Answer:

yes.

4. 9, 40, 41

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.

Answer:

Yes.

5. 24, 45, 51

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.

Answer:

Yes.

Question 4.

Find the unknown side in the following triangles.

Solution:

(i) 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) 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) 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.

Solution:

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 Δ 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 = ^{2}

Height: AD = 5 cm

Question 6.

In the figure, find PR and QR.

Solution:

In the figure ∆ PQR is a right triangle.

By Pythagoras theorem,

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

(x + 1)^{2} = 7^{2} + x^{2}

x^{2} +2 × x × 1 + 1^{2} = 49+ x^{2}

2x + 1 = 49

2x = 49 – 1 = 48

x = \(\frac{48}{2}\) = 24

∴ PR = x + 1 = 24 + 1 = 25

QR = x = 24

Question 7.

The length and breadth of the screen of an LED – TV are 24 inches and 18 inches. Find the length of its diagonal.

Solution:

The length and breadth of a LED TV form a right angled triangle with its diagonal.

Therefore by Pythagoras theorem,

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

= 24^{2} + 18^{2} = 576 + 324 = 900 = 30^{2}

∴ AC = 30 ⇒ The length of the diagonal is 30 inches.

Question 8.

Find the distance between the helicopter and the ship.

Solution:

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 9.

From the figure, 1. If TA = 3 cm and OT = 6 cm, find TG.

Solution:

1. From the figure, if, TA = 3 cm, OT = 6 cm

By Pythagoras theorem,

OA^{2} = OT^{2} – TA^{2} = 6^{2} – 3^{2}

i.e. h^{2} = 36 – 9 = 27 cm.

Now, by altitude – on – hypotenuse theorem

h^{2} = xy

27 = x × 3

x = \(\frac{27}{3}\) = 9 cm

TG = x + 3 = 9 + 3 = 12 cm

Question 10.

If RQ = 15 cm and RP = 20 cm, find PQ, PS and SQ.

Solution:

RQ = 15cm

RP = 20 cm and ∆ PQR is a right angled triangle

By Pythagoras theorem, p 20 cm

PQ^{2} = PR^{2} + RQ^{2} = 20^{2} + 15^{2}

= 400 + 225 = 625 = 25^{2}

∴ PQ = 25 cm

Now by altitude – on – hypotenuse theorem,

RQ^{2} = q x r

15^{2} = q x 25

q = \(\frac{225}{25}\) = 9 cm ⇒ SQ = 9 cm

PR^{2} = P x r

20^{2} = P x 25

P = \(\frac{400}{25}\) = 16 cm ⇒ PS = 16 cm

Answer:

- PQ = 25 cm
- PS = 16 cm
- SQ = 9 cm

Objective Type Questions

Question 11.

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 12.

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

(a) 28 cm

(b)20cm

(c) 24 cm

(d)21cm

Answer:

(b) 20 cm

Hint:

Side take a = 12 cm

b = 16 cm

The hypotenuse c^{2} = a^{2} + b^{2} = 12^{2} + 16^{2}

^{2} = 400 ⇒ c = 20 cm

Question 13.

The area of a rectangle of length 21 cm and diagonal 29 cm is ………. cm^{2}

(a) 609

(b) 580

(c) 420

(d) 210

Answer:

(c) 420

Hint:

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 x breadth.

i.e. AB x BC = 21 cm x 20 cm = 420 cm^{2}

Question 14.

if the square of the hypotenuse of an isosceles right triangle is 50 cm2, the length of each side is ………..

(c) 10 cm

(a) 25 cm

(b) 5 cm

(c) 10 cm

(d) 20 cm

Answer:

(b) 5 cm

Hint:

By Pythagoras theorem

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

In an isosceles triangle, a = b

c^{2} = a^{2} + a^{2} = 2a^{2}

⇒ 2a^{2} = 50

a^{2} = 25 ⇒ a = 5cm

∴ The length of each sides a = 5cm, b = 5 cm.

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 x 4 = 20 units

12a = 12 x 4 = 48 units

13a = 13 x 4 = 52 units