Question

In the given figure, O is the centre of circle and AB is the diameter. Find the area of shaded region.

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

6(π - 2) cm²

or

6.85 cm² (approx)

Step-by-step explanation:

Given

AC = BC = 4 cm

$\measuredangle C = 90^\circ$

So, apply Pythagoras theorem to find the diameter AB

By Pythagoras Theorem,

AC² + BC² = AB²

→ 4² + 4² = AB²

→ 16 + 16 = AB²

→ 32 = AB²

→ AB = √32

→ AB = 4√2 cm

Since, AB is the diameter, radius =

$\frac{4\sqrt2}{2} \\\\\implies 2\sqrt2 \ cm$

Now, we will find area of semicircle ACB

Area of semicircle =

$\frac{\pi r^2}{2} \\\\\implies \frac{\pi \times (2\sqrt 2)^2}{2} \\\\\implies \frac{\pi \times 8}{2}\\\\\implies 4\pi$

And, area of Triangle ACB

=$\frac{1}{2} \times b \times h\\\\\implies \frac{1}{2} \times 4 \times 4\\\\\implies 8 cm^2$

So, area of shaded region =

$4 \pi - 8 \\\\\implies 4(\pi - 2) cm^2$

Now, OH and OE are the radius of circle

→ OH = OE = 2√2 cm

In right angled triangle HOE, HE is the hypotenuse. By Pythagoras theorem,

OH² + OE² = HE²

→ (2√2)² + (2√2)² = HE²

→ 8 + 8 = HE²

→ 16 = HE²

→ HE = √16

→ HE = 4 cm

Now, area of sector OHE =

$\frac{\theta}{360^\circ} \times \pi r^2 \\\\\\\implies \frac{90^\circ}{360^\circ}\times \pi \times (2\sqrt2)^2\\\\\implies \frac{1}{4} \times \pi \times 8\\\\\implies 2 \pi$

And, area of triangle HOE =

$\frac{1}{2} \times 2\sqrt2 \times 2\sqrt2\\\\\implies \frac{8}{2}\\\\\implies 4 \: cm^2$

So, area of that shaded region = 2π - 4 cm²

= 2(π - 2)

So, total area of shaded region =

4(π - 2) + 2(π - 2)

→ (π - 2)(4 + 2)

→ 6(π - 2) cm²

Further solving,

6π - 12

⇒ 18.85 - 12 (approx)

⇒ 6.85 cm² (approx)