Question

In the following figure, shows a kite in which BCD is the shape of a quadrant of a circle of radius 42 cm. ABCD is a square and Δ CEF is an isosceles right angled triangle whose equal sides are 6 cm long. Find the area of the shaded region.​

Answer 1

Answer:

The area of shaded region is  1404 cm².

Step-by-step explanation:

Given :

Radius of a quadrant of a circle , r = 42 cm.  

Equal sides of  an isosceles right angled ∆ =  6 cm  

Area of shaded region ,A = Area of quadrant +  Area of isosceles ∆

A = ¼ πr² + ½ × base ×  height

A = ¼ × 22/7 × 42²  + ½ × 6 × 6

A = ½ × 11 × 6 × 42 + 18

A = 11 × 3 × 42 + 18

A = 33 × 42 + 18

A = 1386 + 18  

A = 1404 cm²

Area of shaded region = 1404 cm²

Hence, the area of shaded region is  1404 cm².

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Answer 2

$\huge\mathcal\purple{Solution}$

It is given that ...

The radius of the quadrant is 42 cm .

The equal sides of the isoceles triangle is 6 cm .

The area of shaded region = Area of quadrant + area of isoceles triangle

$= \frac{1}{4} \pi {r}^{2} + \frac{1}{2} \times base \times height$

$= \frac{1}{4} \times \frac{22}{7} \times 42 \times 42 + \frac{1}{2} \times 6 \times 6$

$= 11 \times 3 \times 42 + 3 \times 6$

$= 1386 + 18$

$= 1404 {cm}^{2}$

Hence , the area of the shaded region is 1404 cm^2 .