Samacheer Kalvi 8th Maths Guide Chapter 2 Measurements Ex 2.2

Samacheer Kalvi 8th Maths Guide Chapter 2 Measurements Ex 2.2

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 2 Measurements Ex 2.2

Question 1.
Find the perimeter and area of the figures given below. (π = 22/7)
(i)

Answer:

Length of the arc of the semicircle = 1/2 × 2πr units
= 22/7×7/2 m =11 m
∴ Perimeter = Sum of all lengths of sides that form the closed boundary
P = 11 + 10 + 7 + 10 m
Perimeter = 38 m
Area = Area of the rectangle – Area of semicircle
= (l × b) – 1/2 πr² sq. units

= 50.75 m² (approx)
Area of the figure = 50.75 m² approx.

 

(ii)

Answer:

Question 2.
Find the area of the shaded part in the following figures. (π = 3.14)
(i)

Answer:
Area of the shaded part = Area of 4 quadrant circles of radius 10/2 cm
= 4 × 1/4 × πr² = 3.14 × 10/2 × 10/2 cm²
= 314/4 cm² = 78.5 cm²
Area of the shaded part = 78.5 cm²
Area of the unshaded part = Area of the square – Area of shaded part
= a² – 78.5 cm² = (10 × 10) – 78.5 cm²
= 100 – 78.5cm² = 21.5cm²
Area of the unshaded part = 21.5 cm² (approximately)

(ii)

Answer:
Area of the shaded part = Area of semicircle – Area of the triangle

= 1/2 × 3.14 × 7 × 7 – 1/2 × 14 × 7cm²
= 153.86/2 – 49 cm² = 76.93 – 49 cm²
= 27.3 cm²
∴ Area of the shaded part = 27.93 cm² (approximately)

 

Question 3.
Find the area of the combined figure given which is got by joining of two parallelograms

Answer:
Area of the figure = Area of 2 parallelograms with base 8 cm and height 3 cm
= 2 × (bh) sq. units
= 2 × 8 × 3cm² = 48 cm²
∴ Area of the given figure = 48 cm²

Question 4.
Find the area of the combined figure given, formed by joining a semicircle of diameter 6 cm with a triangle of base 6 cm and height 9 cm. (π = 3.14)

Question 6.
A rocket drawing has the measures as given in the figure. Find its area.

Answer:
Area = Area of a rectangle + Area of a triangle + Area of a trapezium
For rectangle length l = 120 – 20 – 20 cm = 80 cm
Breadth b = 30 cm
For the triangle base = 30 cm
Height = 20 cm
For the trapezium height h = 20 cm
Parallel sided a = 50 cm
b = 30cm
∴ Area of the figure (l × b) + (1/2 × base × height) + 1/2 × h × (a + b)sq. units
= (80 × 30) + (1/2 × 30 × 20) + 1/2 × 20 × (50 + 30) cm²
= 2400 + 300 + 800 cm² = 3500 cm²
Area of the figure = 3500 cm²

 

Question 7.
Find the area of the irregular polygon shaped fields given below.

Answer:
Area of the field = Area of trapezium FBCH + Area of ∆DHC + Area of ∆EGD + Area of ∆EGA + Area of ∆BFA
Area of the triangle = 1/2 bhsq.units
Area of the trapezium = 1/2 × h × (a + b) sq.units
Area of the trapezium FBCH = 1/2 × (10 + 8) × (8 + 3)m² = 9 × 11 = 99 m² ….. (1)
Area of the ∆DHC = 1/2 × 8 × 5 m² = 20 m² ….. (2)
Area of ∆EGD = 1/2 × 8 × 15m² = 60 m² …….. (3)
Area of ∆EGA = 1/2 × 8 × (8 + 6)m² = 4 × 14 m²
= 56m²
Area of ∆BFA = 1/2 × 3 × 6m² = 9 m²
∴ Area of the field = 99 + 20 + 60 + 56 + 9 m²
= 244 m²
Area of the field = 244 m²

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