Selina Concise Class 9 Maths Chapter 20 Area and Perimeter Of A Plane Figures Exercise 20A Solutions
EXERCISE – 20A
(1) Find the area of a triangle whose Sides are 18 cm, 24 cm and 30 cm. Also, Find the length of attitude corresponding to the largest side of the triangle.
Solution:
Given:
a = 18 cm
b = 24 cm
c = 30 cm
S = a + b + c / 2
= 18 + 24 + 30 / 2
S = 72 / 2
S = 36
By using formula,
Area (△ABC) = √S ( s – a) (s – b) (s – c)
= √36 (36 – 18) (36 – 24) (36 – 30)
Area (△ABC) = √36 (18) (12) (6)
= √36 × 18 × 12 × 6
= √6 × 6 × 2 × 3 × 3 × 2 × 6 × 6
= 6 × 6 × 3 × 2
= 36 × 6
Area (△ ABC) = 216 cm2 sq. cm.
Area of (△ ABC) = 1/2 × Base × height
= 1/2 × 30 × H
216 = 1/2 × 30 × H
216 × 2/ 30 = H
144/10 = H
14.4 cm = H
∴ Height = 14. 4 cm.
(2) The length of the sides of a triangle are in the ratio 3: 4: 5. Find the area of the triangle if its perimeter is 144 cm.
Solution:
Given:
a = 3x
b = 4x
c = 5x
Also, given the Perimeter is 144 cm.
Perimeter = a + b+ c
144 = 3x + 4x + 5x
144 = 12x
144/ 12 = x
12 = x
∴ x = 12
a = 3 × (12) = 36
b = 4 × (12) = 48
c = 5 × (12) = 60
S = a + b + c/ 2 = 144/2 = 72.
Area of triangle =
√S( s – a) (s – h) (s – c)
= √72 ( 72 – 36) ( 72 – 48) ( 72 – 60)
= √72 × 36 × 24 × 12
= √2 × 36 × 36 × 2 × 12 × 12
= 36 × 12 × 2
= 36 × 24
= 864 cm2
∴ Area of triangle = 864 cm2
(3) ABC is a triangle in which AB = AC = 4cm and < A =900 calculate:
(i) The area of △ ABC.
(ii) The length of perpendicular from A to BC.
Solution:
(i) Area of triangle
ABC = 1/2 × Base × height
= 1/2 × Ac × AB
= 1/2 × 4 × 4
= 2 × 4
= 8 cm2
(ii) Area of (△ ABC) = 1/2 × Base × height
8 = 1/2 × BC × AD
8 = 1/2 × BC × AD —– (i)
Here, we have to find BC – by Pythagoras Theorem,
(hypotenuse)2 = (Base)2 + (height)2
BC2 = (4)2 + (4)2
BC2 = 16 + 16
Taking Square root on both Sides,
BC = √16 + 16
BC = √32
BC = 4√2
Equation (i)
8 = 1/2 × 4√2 × AD
8 = 2 √2 × AD
8/2√2 = AD
4√2 = AD
2 × √2 × √2/√2 = AD
2√2 = AD
2 × 1. 4/4 = AD
2.828 = AD
2.83 cm = AD
Hence, the length of perpendicular from A to BC is 2. 83 cm.
(4)The area of a equilateral triangle is 36 √3 sq . cm. Find it’ s Perimeter.
Solution:
Given: The area of a equilateral triangle is 36 √3 Sq. cm.
√3/4 a2 = 36 √3
a2 = 36 × 4
a2 = 36 × 4
Taking Square root on both sides.
√a2 = √36 × 4
a = 6 × 2
a = 12
Now, we have to find perimeter
Perimeter = 3a
= 3 × 12
= 36 cm.
(5) Find the area of an isosceles triangle whose perimeter is 36 cm and base is 16 cm.
Solution:
Given: the perimeter of a triangle is 36 cm and base = 16 cm.
∴ The given triangle is isosceles triangle.
∴ Perimeter = 36 cm.
a + a + 16 = 36
2a + 16 = 36
2a = 36 – 16
2a = 20
a = 20/2
a = 10
Now, we have to find area of an isosceles triangle –
Area of isosceles = b/4 √4a2 – b2
= 16/4 √4 (10)2 – (16)2
= 4 √4 × 100 – 256
= 4 √400 – 256
= 4 √144
= 4 × 12
= 48 cm2
∴ Area of isosceles triangle is 48 cm2
Here is your solution of Selina Concise Class 9 Maths Chapter 20 Area and Perimeter Of A Plane Figures Exercise 20A
Dear Student, I appreciate your efforts and hard work that you all had put in. Thank you for being concerned with us and I wish you for your continued success.
For more solutions, See below ⇓