Question

the figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (Use π = 3.14) class 10 maths.,​

Answer 1

The area of the shaded region = 228 cm².

Given :

A square OABC is inscribed in a quadrant OPBQ.

OA = 20 cm (given, a square that means, all the sides are equal I.e. all the sides of the square is 20 cm).

π = 3.14.

To Find :

The area of the shaded region.

Solution :

Since,

OABC is a square.

So, by the pythagoras theorem,

$\blue{ \boxed{ \green{ \bf{H}^{2} = {B}^{2} + {P}^{2}} }}$

Where,

H = hypotenuse.

B = base.

P = perpendicular.

We have,

H = OB = ?

B = OA = 20 cm.

P = AB = 20 cm.

Substitute all the values in the Pythagoras theorem,

$\bf \implies {OB}^{2} = {OA}^{2} + {AB}^{2} \\ \\ \\ \bf \implies {OB}^{2} = {(20 \: cm)}^{2} + {(20 \: cm)}^{2} \\ \\ \\ \bf \implies {OB}^{2} = 400 \: {cm}^{2} + 400 \: {cm}^{2} \\ \\ \\ \bf \implies {OB}^{2} = 800 \: {cm}^{2} \\ \\ \\ \bf \implies OB = \sqrt{800 \: {cm}^{2} } \\ \\ \\ \bf \implies OB = \sqrt{2 \times 2 \times 2 \times 10 \times 10} \: cm \\ \\ \\ \bf \implies OB = \sqrt{ {(2)}^{2} \times 2 \times {(10)}^{2} } \: cm \\ \\ \\ \bf \implies OB = 2 \times 10 \sqrt{2} \: cm \\ \\ \\ \bf \implies OB = 20 \sqrt{2} \: cm$

But OB is a radius of the quadrant. (i.e. r = 20√2 cm).

Now we have to find the area of the shaded region.

Area of shaded region = Area of quadrant OPBQ – Area of square OABC.

$\bf \implies \dfrac{\pi {r}^{2} }{4} - {a}^{2} \\ \\ \\ \bf \implies \dfrac{3.14 \times {(20 \sqrt{2} \: cm )}^{2} }{4} - {(20 \: cm)}^{2} \\ \\ \\ \bf \implies \frac{3.14 \times \not400 \times 2}{ \not4} \: {cm}^{2} - 400 \\ \\ \\ \bf \implies3.14 \times 100 \times 2 \: {cm}^{2} - {400 \: cm}^{2} \\ \\ \\ \bf \implies 6.28 \times 100 \: {cm}^{2} - 400 \: {cm}^{2} \\ \\ \\ \bf \implies 628 \: {cm}^{2} - 400 \: {cm}^{2} \\ \\ \\ \bf \implies 228 \: {cm}^{2}$

Hence,

The area of the shaded region is 228 cm².