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

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

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, PR² = PQ² + QR², 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, l² = ……….
Answer:
n² – m²
Hint:

Question (iii)
If the sides of a triangle are in the ratio 5:12:13 then, it is ………
Answer:
a right angled triangle.
13² = 169
5² = 25
12² = 144
169 = 25 + 144
132 = 5² + 12²
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 BE² = TE x ………

Answer:
EN

Question 2.
Say True or False.

Question (i)
8, 15, 17 is a Pythagorean triplet.
Answer:
True
Hint:
17² = 289
15² = 225
8² = 64
64 + 225 = 289 ⇒ 17² = 15² + 8²

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:

49² = 2401
9² = 81
40² = 1600
40² + 9² = 1600 + 81 + 1681
49² = 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,

1. 8, 15, 17
2. 12, 13, 15
3. 30, 40, 50
4. 9, 40, 41
5. 24, 45, 51

Solution:
1. 8, 15, 17
Take a = 8,
b = 15 and
c = 17
Now a² + b² = 8² + 15² = 64 + 225 = 289
17² = 289 = c²
∴ a² + b² = c²
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 a² + b² = 12² + 13² = 144 + 169 = 313
15² = 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 a² + b² = 30² + 40² = 900 + 1600 = 2500
c² = 50² = 2500
∴ a² + b² = c²
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 a² + b² = 9² + 40² = 81 + 1600= 1681
c² = 41² = 1681
∴ a² + b² = c²
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
a² + b² = 24² + 45² = 576 + 2025 = 2601
c² = 51² = 2601
a² + b² = c²
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

BC² = AB² + AC²
Take AB² + AC² = 9² + 40² = 81 + 1600 = 1681
BC² = AB² + AC² = 1681 = 41²
BC² = 41² ⇒ BC = 41
∴ x = 41

(ii) From ∆ PQR, by Pythagoras theorem,

PR² = PQ² + QR²
34² = y² + 30²
⇒ y² = 34² – 30²
= 1156 – 900
= 256 = 16²
y² = 16² ⇒ y = 16

(iii) From ∆ XYZ, by Pythagoras theorem,

YZ² = XY² + XZ²
⇒ XY² = YZ² – XZ²
Z² = 39² – 36²
= 1521 – 1296 = 225 = 15²
Z² = 15²
⇒ 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,
AB² = AD² + BD²
⇒ AD² = AB² – BD²
= 13² – 12² = 169 – 144 = 25
AD² = 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,
PR² = PQ² + QR²
(x + 1)² = 7² + x²
x² +2 × x × 1 + 1² = 49+ x²
2x + 1 = 49
2x = 49 – 1 = 48
x = 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,
AC² = AB² + BC²
= 24² + 18² = 576 + 324 = 900 = 30²
∴ 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,
AS² = AP² + PS²
= 80² + 150²
= 6400 + 22500 = 28900 = 170²
∴ 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,
OA² = OT² – TA² = 6² – 3²
i.e. h² = 36 – 9 = 27 cm.
Now, by altitude – on – hypotenuse theorem
h² = xy
27 = x × 3
x = 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
PQ² = PR² + RQ² = 20² + 15²
= 400 + 225 = 625 = 25²
∴ PQ = 25 cm
Now by altitude – on – hypotenuse theorem,
RQ² = q x r
15² = q x 25
q = 225/25 = 9 cm ⇒ SQ = 9 cm
PR² = P x r
20² = P x 25
P = 400/25 = 16 cm ⇒ PS = 16 cm
Answer:

1. PQ = 25 cm
2. PS = 16 cm
3. 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 c² = a² + b² = 12² + 16^2
2 = 400 ⇒ c = 20 cm

Question 13.
The area of a rectangle of length 21 cm and diagonal 29 cm is ………. cm²
(a) 609
(b) 580
(c) 420
(d) 210
Answer:
(c) 420
Hint:

Length = 21 cm
Diagonal = 29 cm
By the converse of Pythagoras theorem,
AB² + BC² = AC²
21² + x² = 29²
x² = 841 – 441 = 400 = 20²
x = 20 cm
Now area of the rectangle = length x breadth.
i.e. AB x BC = 21 cm x 20 cm = 420 cm²

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
c² = a² + b²
In an isosceles triangle, a = b
c² = a² + a² = 2a²
⇒ 2a² = 50
a² = 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

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