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.

Answer 2

$\huge\bold{\underline{Question:}}$

The figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region.

$\huge\bold{\underline{Answer:}}$

GIVEN:

• A square OABC is inscribed in a quadrant OPBQ.

• OA = 20 cm

TO FIND:

Find the area of the shaded region.

SOLUTION:

$\boxed{\bf{\pink{Area\:of\:square=Side×Side}}}$

= 20 × 20 cm²

= 400 cm²

Area of quadrant:

we need to find the radius

Joining OB

Also all angles of a square are 90°

.°. Angle BOA = 90°

Hence ∆ OBA is a right triangle

In ∆OBA by pythagiras theorem

$\sf{\implies (Hypotenuse)^{2} = (perpendicular)^{2} + (base)^{2}}$

$\sf{\implies (OB)^{2}=(AB)^{2} + (OA)^{2}}$

$\sf{\implies (OB)^{2} = (20)^{2} + (20)^{2}}$

$\sf{\implies (OB)^{2} = 400 + 400}$

$\sf{\implies (OB)^{2} = 800}$

$\sf{\implies OB = 20√2 cm}$

We know that,

$\boxed{\sf{\pink{Area\:of\:quadrant=\dfrac{1}{4}πr²}}}$

$\sf{\implies \dfrac{1}{4}× 3.14 × (20√2)^{2}}$

$\sf{\implies 3.14 × 5 × 20 × 2}$

$\sf{\implies 628\: cm²}$

Now,

Area of shaded region =Area of quadrant OBPQ – Area of square OABC

$\sf{\implies Area\:of\:shaded\:region = 628 - 400}$

$\sf{\implies Area\:of\:shaded\:region= 228\:cm²}$

Therefore,the area of shaded region = 228 cm²

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