Question

22.Find the area of the segment shaded in the figure in which PQ = 12cm.,PR = 5cm and QR is the diameter of the circle with center 'O'

Answer 1

$\large{\underline{\bf{\pink{Answer:-}}}}$

✰ Area of shaded region = 2038/56 cm²

$\large{\underline{\bf{\blue{Explanation:-}}}}$

✰✰ Area of shaded region

= area of semicircle - area of ∆PQR.

$\large{\underline{\bf{\green{Given:-}}}}$

✰ PQ = 12cm

✰ PR = 5cm

✰ QR is the diameter of the circle.

$\large{\underline{\bf{\green{To\:Find:-}}}}$

✰ we need to find the area of shaded region.

$\huge{\underline{\bf{\red{Solution:-}}}}$

It is given that QR is a diameter,

So, it forms a semicircle.

we know that that the angle in a semicircle is a right angle.

So ,$\angle RPQ = 90\degree$

So, ,$\angle RPQ$ is right angled triangle.

By using Pythagoras theorem:-

(Hypotenuse)² = (height)² + (base)²

$:\implies\:(QR)^2=(PQ)^2+(PR)^2$

$:\implies\:(QR)^2=(12)^2+(5)^2$

$:\implies\:(QR)^2=144+25$

$:\implies\:QR=\sqrt{169}$

$:\implies\:\bf\:QR=13cm$

So, QR = 13cm

Then , Radius = 13/2 cm

${\underline {\boxed{\bf{\:Area\:of\: circle=\pi\: r^2}}}}$

${\underline{\boxed {\bf{\:Area\:of\: semicircle=\frac{1}{2}\times\: \pi\:r^2}}}}$

$:\implies\:\frac{1}{2}\times\frac{22}{7}\times(\frac{13}{2})^2$

$:\implies\:\frac{1}{2}\times\frac{22}{7}\times\frac{169}{4}$

$:\implies\:\frac{3718}{56}cm^2$

Area of ∆ PQR :-

$\bf\:\frac{1}{2}\times\:Base\times\: Height$

$:\implies\:\frac{1}{2}\times\:PQ\times\:PR$

$:\implies\frac{1}{\cancel{2}}\times{\cancel{12}}\times\:5$

$:\implies\:\bf 30cm^2$

Now,

Area of shaded region

= area of semicircle - area of ∆PQR

$:\implies\:\frac{3718}{56}-30$

$:\implies\:\frac{3718-1680}{56}$

$:\implies\:\bf\frac{2038}{56}cm^2$

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Answer 2

Answer:

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