Question

In fig. from a rectangular region ABCD with AB= 20 cm, a right triangle AED with AE= 9 cm and DE= 12 cm, is cut off. On the other end, taking BC as diameter, a semicircle is added on outside the region. Find the area of the shaded region.

Answer 1

GIVEN:
Length of a rectangle (AB) = DC = 20 cm
Breadth of a rectangle( BC) = AD=15 cm
AE  = 9 cm ,  ED = 12 cm

AREA OF RECTANGLE = length X breadth= 20 × 15 = 300 cm²

Diameter of Semicircle = Breadth of a rectangle = 15 cm
Radius of Semicircle = diameter/2 = 15 /2 cm
AREA OF SEMICIRCLE= ½ πr² = ½ ×(3.14) (15/2)² = (1.57 × 225) /4 = 353.25/ 4 = 88.31 cm²

Area of right angled ∆ = ½ × Base × height
AREA OF RIGHT ANGLED ∆AED = ½ × AE × ED
= ½ × 9 × 12  = 9 × 6 = 54 cm²

Area of shaded region = Area of rectangle - Area of right angled ∆AED  + Area of semicircle
AREA OF SHADED REGION  = 300 - 54 + 88.31 = 246 + 88.31= 334.31 cm²

Hence, the Area of shaded region is 334.31 cm²

HOPE THIS WILL HELP YOU...

Answer 2

Answer:

Step-by-step explanation:

Length of a rectangle (AB) = DC = 20 cm

Breadth of a rectangle( BC) = AD=15 cm

AE  = 9 cm ,  ED = 12 cm

AREA OF RECTANGLE = length X breadth= 20 × 15 = 300 cm²

Diameter of Semicircle = Breadth of a rectangle = 15 cm

Radius of Semicircle = diameter/2 = 15 /2 cm

AREA OF SEMICIRCLE= ½ πr² = ½ ×(3.14) (15/2)² = (1.57 × 225) /4 = 353.25/ 4 = 88.31 cm²

Area of right angled ∆ = ½ × Base × height

AREA OF RIGHT ANGLED ∆AED = ½ × AE × ED

= ½ × 9 × 12  = 9 × 6 = 54 cm²

Area of shaded region = Area of rectangle - Area of right angled ∆AED  + Area of semicircle

AREA OF SHADED REGION  = 300 - 54 + 88.31 = 246 + 88.31= 334.31 cm²

Hence, the Area of shaded region is 334.31 cm²