Ex.12.3 Q13 Areas Related to Circles Solution - NCERT Maths Class 10

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Question

In Figure, a square \({OABC} \) is inscribed in a quadrant \({OPBQ.}\) If \(OA = \text {20 cm,}\) find the area of the shaded region.

(Use \( \pi =3.14\) )

Text Solution

 

What is known?

A square \({OABC}\) is inscribed in a quadrant \(OPBQ, OA =\text{ 20 cm.}\)

What is unknown?

Area of the shaded region.

Reasoning:

 Visually from the figure it is clear that 

Area of the shaded region \(= \) Area of quadrant \({OPBQ}\, -\) Area of square \({OABC}\)

Since side of the square \(OA =\text{ 20 cm}\)

\(\rm{OB =} \) Radius \(=\) Diagonal of the square \({OABC}\)

Using formula for

Area of sector of angle \(\begin{align}\theta = \frac{\theta }{{{{360}^\circ }}} \times \pi {r^2}\end{align}\)

With \(\theta = 90^\circ \) and \(r = OB \)

We can find the area of the quadrant \({OPBQ.}\)

Steps:

Join \({OB.}\)

We know \(\Delta {OBA}\) is a right angaled triangle, a \(\angle {OAB} \,= {90^ \circ }\) (angle of a square)

\(\therefore\;\)Using Pythagoras theorem

\[\begin{align} \therefore\quad {{OB}^{2}}&={{OA}^{2}}+A{{B}^{2}} \\ &={{(20)}^{2}}+{{(20)}^{2}} \\ {DB}&=\sqrt{2{{(20)}^{2}}} \\ &=20\sqrt{2}\,\rm{cm}\end{align}\]

Therefore,radius of the quadrant,\(\begin{align}r = OB = 20\sqrt 2 cm\end{align}\)

\(\text{Area of quadrant OPBQ}\)\[\begin{align}&= \frac{{{{90}^\circ }}}{{{{360}^\circ }}} \times \pi {r^2}\\ &= \frac{1}{4} \times 3.14 \times {(20\sqrt 2 )^2}\\ &= \frac{1}{4} \times 3.14 \times {400} \times 2\\&= 628{\text{ cm}^2}\end{align}\]

\(\text{Area of square OABC}\) \[\begin{align} &= {\left( {side} \right)^2}\\&= {\left( {OA} \right)^2}\\&= {\left( {20cm} \right)^2}\\&= 400c{m^2}\end{align}\]

Area of shaded region \(= \) Area of quadrant \({OPBQ}\;-\) Area of square \({OABC}\)

\[\begin{align}&= {628 - 400{\text{ cm}^2}}\\&= {228{\text{ cm}^2}}\end{align}\]