Question

PQRS is a diameter of a circle of radius 6 cm. the length PQ, QR and RS are equal. Semicircles are drawn on PQ and Qs and diameter as shown in the figure. find the perimeter and ar. of shaded portion.


please,
solve it urgently​

Answer 1

Step-by-step explanation:

Required area= Area of semicircle with PS as diameter + Area of semicircle with PQ as diameter Area of semicircle with QS as diameter

Answer 2

Given Data :

It is stated that a circle is having radius 6 cm. On the diameter of the circle PS there are some points R, S respectively.

It is also said that these points are equidistant from each other i.e, PQ = QR = RS. Then, semicircle is drawn from point P to Q and from point Q to S respectively. So, we have to find out the area of the shaded part in the figure.

★ What we will do?

First we will find out the diameter which is nothing but the double of given radius. Then, as according to the given information all points are plotted at the equal distance. From this, we will find the measurement of the line segments PQ, QR and RS.

There after we calculate the area of semicircle with arc PBQ and diameter PQ and then the area of semicircle with arc PDS and diameter PS and area of semicircle with arc QCS and diameter QS respectively. For area of the portion, we will subtract the Area of Semicircle with diameter QS from Area of semicircle PS and will add the area of semicircle with diameter PQ. There we get our answer.

Let's do it now :-

We know,

• Radius → 6 cm

Then,

• Diameter → 2 × R

• Diameter → 12 cm

We also know,

Diameter = PQ + QR + RS = 12 cm

∴ PQ = QR = RS { Equidistant }

→ PQ + PQ + PQ = 12 cm

→ 3 PQ = 12

$\dashrightarrow\:\:\sf PQ = \dfrac{12}{3}$

$\dashrightarrow\:\: \underline {\bf {PQ = 4 cm = QR = RS}}$

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Now, for area of semicircle with arc PBQ and diameter PQ.

Diameter is 4 cm then radius will be 2 cm. ( $\sf \therefore Radius = \dfrac{Diameter}{2}$ )

${\bullet \: \: {\boxed{\sf{Area \: of \: circle = \pi r^{2} }}}}$

Semicircle is half of circle so, it's area will be :-

$\dashrightarrow\:\:\sf Area \: of \: semicircle = \dfrac{1}{2} \times \pi r^{2}$

Radius → 2 cm

$\dashrightarrow\:\:\bf Area \: of \: semicircle = \dfrac{1}{2} \times \pi 2^{2}$

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Now, for area of semicircle with arc PDS and diameter PS.

Radius = 6 cm

$\dashrightarrow\:\:\sf Area \: of \: semicircle = \dfrac{1}{2} \times \pi r^{2}$

$\dashrightarrow\:\:\bf Area \: of \: semicircle = \dfrac{1}{2} \times \pi 6^{2}$

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Then,

Area of semicircle with arc QCS and diameter QS.

Radius = 4 cm ( RS)

$\dashrightarrow\:\:\sf Area \: of \: semicircle = \dfrac{1}{2} \times \pi r^{2}$

$\dashrightarrow\:\:\bf Area \: of \: semicircle = \dfrac{1}{2} \times \pi 4^{2}$

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FOR SHADED PORTION AREA :-

(Area of semicircle with arc PDS - Area of semicircle with arc with QCS) + Area of semicircle with arc PBQ.

Substituting the respective values, we get :-

$:\implies \sf \dfrac{1}{2} \pi 6^{2} - \dfrac{1}{2} \times \pi 4^{2} + \dfrac{1}{2} \times \pi 2^{2}$

Taking $\bf \dfrac{1}{2} \pi$ common,

$:\implies \sf \dfrac{1}{2} \pi ( 6^{2} - 4^{2} + 2^{2} )$

$:\implies \sf \dfrac{1}{2} \pi ( 36 - 16 + 4 )$

$:\implies \sf \dfrac{1}{2} \pi ( 40 - 16 )$

$:\implies \sf \dfrac{1}{2} \pi ( 24 )$

Putting the value of $\sf \pi = \dfrac{22}{7}$

$:\implies \sf \dfrac{1}{2} \times \dfrac{22}{7} \times 24$

$:\implies \sf \dfrac{22}{7} \times 12$

$:\implies \sf \dfrac{264}{7}$

Hence,

$:\implies \underline{ \boxed{\sf Area \: of \: shaded \: portion = 37. 71 \: cm ^{2} }}$

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Note~ → Perimeter and circumference are both the same things.

For perimeter of Shaded portion :

${\bullet \: \:{\boxed{\sf{Perimeter \: (Circumference) \: of \: circle = 2 \pi r }}}}$

Perimeter of semicircle will be :-

$\dashrightarrow\:\:\sf Perimeter \: of \: semicircle = \dfrac{1}{2} \times 2 \pi r$

$\dashrightarrow\:\:\sf Perimeter \: of \: semicircle = \pi r$

Now, for perimeter of semicircle with arc PBQ.

• Radius → 2 cm

$\dashrightarrow\:\:\sf Perimeter \: of \: semicircle = 2 \pi$

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Now, for perimeter of semicircle with arc PDS.

• Radius = 6 cm

$\dashrightarrow\:\:\sf Perimeter \: of \: semicircle = 6 \pi$

________________

Then,

Perimeter semicircle with arc QCS.

• Radius = 4 cm ( RS)

$\dashrightarrow\:\:\sf Perimeter \: of \: semicircle = 4 \pi$

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FOR SHADED PORTION PERIMETER :-

(Perimeter of semicircle with arc PDS - Perimeter of semicircle with arc with QCS) + Perimeter of semicircle with arc PBQ.

By Plugging up the respective values, we get :-

$:\implies \sf ( 6 \pi - 4 \pi ) + 2 \pi$

Taking $\bf \pi$ common,

$:\implies \sf \pi ( 6 - 4 ) + 2$

$:\implies \sf \pi 2 + 2$

$:\implies \sf 4 \pi$

Putting the value of $\sf \pi = \dfrac{22}{7}$

$:\implies \sf 4 \times \dfrac{22}{7}$

$:\implies \sf \dfrac{88}{7}$

$:\implies \underline{ \boxed{\sf Circumference \: of \: shaded \: portion = 12. 57 \: cm }}$

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