Question

ABCD is a quadrant of a circle of radius 28 cm And a semicircle BEC is drawn with BC as diameter find the area of the shaded region

Answer 1

[FIGURE IS IN THE ATTACHMENT]
Given:
Radius (r) of the circle = AB = AC = 28 cm
Area of quadrant ABPC = 1/4×π×r²
= (1/4×22/7×28×28) cm²
=22× 28= 616 cm²
Area of ∆ABC = 1/2×AC×AB = (1/2×28×28)
= 392 cm²
Area of segment BPC = Area of quadrant ABPC − Area of ∆ABC
= (616− 392) cm²
= 224 cm² .............(1)
In a right-angled ∆ BAC
BC² = BA²+ AC² (By Pythagoras theorem)
BC² = (28²+ 28²) cm²
BC²= 784 +784 cm²
BC = √16×98 = √ 16 × 49 ×2
BC = 4×7√2= 28√2
BC= 28√2 cm
BC(Diameter) = 28√2
Radius of semicircle= 28√2/2= 14√2 cm
Area of semicircle BEC= 1/2×π×r²
= ( 1/2×22/7×14√2×14√2) cm²
= 22 × √2 × 14√2 = 22×14×2 = 44 ×14
= 616 cm²
Area of the shaded portion = Area of semicircle BEC − Area of segment BPC
= 616 − 224 cm²= 392 cm²
Hence, the area of the shaded region = 392 cm²

HOPE THIS WILL HELP YOU...

Answer 2

Answer:

first you find the area of semicircle and radius of semicircle will we 28√2 you find radius by using pytho.theorem.  find area of quadrant and area of triangle .       Last step  area of shaded portion = area of semicircle -[area of quadrant - area of triangle.       I hope you understand

Step-by-step explanation: