Question

ABCD is a square of side 5 cm. There are 2 Quadrants (ACD and ABD) forming inside the square. Find the area of the shaded region.

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Answer 1

Given :

ABCD is a square

Sids of square = 5cm

To Find : Area of shaded region

Construction : Draw AP and AB

Solution :

Triangle PAB ,

AB = AP = BP = r

Triangle PAB is equilateral triangle

So , sides of equilateral triangle is 60°

Let AP be region 1

1 = area of sector PAB - area of triangle PAB

$- > \frac{60}{360} \pi {r}^{2} - \frac{ \sqrt{3} }{4} {r}^{2}$

$- > {r}^{2} ( \frac{\pi}{6} - \frac{ \sqrt{3} }{4} )$

Let sector PAB be region 2

$- > \frac{tita}{360} \pi {r}^{2}$

$- > \frac{60}{360} \pi {r}^{2}$

$- > \frac{\pi {r}^{2} }{6}$

Now region 1 + 2 =

$- > \frac{\pi {r}^{2} }{6} - \frac{ \sqrt{3} }{4} {r}^{2} + \frac{\pi {r}^{2} }{6}$

$- > \frac{\pi {r}^{2} }{3} - \frac{ \sqrt{3} }{4} {r}^{2}$

Area of shaded region = Area of sector ADB - ( 1 + 2)

$- > \frac{\pi {r}^{2} }{4} - ( \frac{\pi {r}^{2} }{3} - \frac{ \sqrt{3} }{4} {r}^{2} )$

$- > \frac{\pi {r}^{2} }{4} - \frac{\pi {r}^{2} }{3} + \frac{ \sqrt{3} }{4} {r}^{2}$

$- > \frac{ \sqrt{3} }{4} {r}^{2} - \frac{\pi {r}^{2} }{12}$

Total shaded region :

$- > 2( \frac{ \sqrt{3} }{4} {r}^{2} - \frac{\pi {r}^{2} }{12} )$

$- > \frac{ \sqrt{3} }{2} {r}^{2} - \frac{\pi {r}^{2} }{6}$

$- > \frac{ {r}^{2} }{2} ( \sqrt{3} - \frac{\pi}{3} )$

Side of square = 5cm

Total shaded region =

$- > \frac{ {5}^{2} }{2} ( \sqrt{3} - \frac{\pi}{3} )$

$- > \frac{25}{2} ( \sqrt{3} - \frac{\pi}{3} )$

$- > 12.5( \sqrt{3} - \frac{\pi}{3} )$

$- > 8.525 \: {cm}^{2}$

I hope it helps u dear friend ^_^♡♡