Exercise 2.2 (Perimeter and Area of Combined shapes)

Exercise 2.2

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

Solution:

(i)

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

= (10 × 7) – (1/2)  × (22/7) × (7/2) × (7/2)

= 70 – [(11 × 7) / (2 × 2)] =  [280 – 77] / 4 = 203/4

= 50.75 m² (approx)

Area of the figure = 50.75 m² approx.

(ii) Perimeter = sum of outside lengths

Length of the arc of quadrant circle = (1/4) × 2πr units

= (1/2)  × (22/7)  × 3.5 cm

= 11 × 0.5 cm = 5.5 cm

∴  Length of arc of 2 sectors = 2 × 5.5 cm

= 11 cm

∴ Perimeter P = 11+ 6 + 3.5 + 6 + 3.5 cm

P = 30 cm

Area = Area of 2 quadrant circle + Area of a rectangle.

= 2 × (1/4)πr² + lb sq. units

= (1/2 × 22/7 × 3.5 × 3.5) + (6 × 3.5) cm²

= (11 × 3.5 × 0.5) + 21 cm²

= (19.25 + 21) cm² = 40.25 cm²

Area = 40.25 cm² approx

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

Solution:

(i)

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 − 78.5 cm² = 21.5 cm²

Area of the unshaded part = 21.5 cm² (approximately)

(ii)

Area of the shaded part = Area of semicircle − Area of the triangle

= ( 1/2 πr²) – (1/2 bh) cm²

= (½) × 3.14 × 7 × 7 – (½) × 14 × 7 cm²

= [ 153.86 / 2 ] – 49 cm² = 76.93 – 49 cm²

= 27.93 cm²

∴  Area of the shaded part =  27.93 cm² (approximately)

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

Solution:

Area of the figure = Area of 2 parallelograms with base 8 cm and height 3 cm

= 2 × (bh) sq. units

= 2 × 8 × 3 cm² = 48 cm²

∴  Area of the given figure = 48 cm²

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 )

Solution:

Area of the figure = Area of the semicircle of radius 3 cm + 2 (Area of triangle with b = 9 cm and h = 3 cm)

= ( (1/2) πr² ) + (2 × [ 1/2 bh ] ) sq. units

= 1/2 × 3.14 × 3 × 3 + (2 × 1/2 × 9 × 3) cm²

= [28.26 / 2] + 27 cm² = 14.13 + 27 cm² = 41.13 cm²

∴ Area of the figure = 41.13 cm² (approximately)

5. The door mat which is hexagonal in shape has the following measures as given in the figure. Find its area.

Solution:

Area of the doormat = Area of 2 trapezium

Height of the trapezium h = 70/2 cm; a = 90 cm; b = 70 cm

∴ Area of the trapezium = 1/2 [ h (a + b) ] sq. units

 Area of the doormat = 2 × (1/2) × 35 (90 + 70 ) cm²

= 35 × 160 cm² = 5600 cm²

∴ Area of the door mat = 5600 cm²

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

Solution:

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 = 30 cm

∴ 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

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

Solution:

Area of the field = Area of trapezium FBCH + Area of ΔDHC + Area of ΔEGD + Area of ΔEGA + Area of ΔBFA

Area of the triangle = 1/2 bh sq. 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 ΔDHC = 1/2 × 8 × 5 m² = 20 m²    ...(2)

Area of ΔEGD = 1/2 × 8 × 15 m² = 60 m² ...(3)

Area of ΔEGA = 1/2 × 8 × (8 + 6) m² = 4 × 14 m²

= 56 m²     ...(4)

Area of ΔBFA = 1/2 × 3 × 6 m² = 9 m²         ...(4)

∴ Area of the field = 99 + 20 + 60 + 56 + 9 m²

 = 244 m²

Area of the field = 244 m²

Answer:

Exercise 2.2

1. 38 m, 50.75 m² (approximately) (ii) 30 cm,40.25 cm² (approximately)

2 (i) 21.5 cm² (approximately)  (ii) 27.93 cm² (approximately)

3. 48 cm²

4. 41.13 cm² (approximately)

5. 5600 cm²

6. 3500 cm²

7. 244 m²