In the given figure, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.

Asked by Pragya Singh | 1 year ago |  163

##### Solution :-

Radius of the quadrant ABC of circle = 14 cm

AB = AC = 14 cm

BC is diameter of semicircle.

ABC is right angled triangle.

By Pythagoras theorem in ΔABC,

BC= AB+AC2

⇒ BC= 14+142

⇒ BC = $$14\sqrt{2}$$ cm

Radius of semicircle = $$\dfrac{14\sqrt{2}}{2}$$ cm = 7$$\sqrt{2}$$ cm

Area of ΔABC = ($$\dfrac{1}{2}$$)×14×14 = 98 cm2

Area of quadrant = $$(\dfrac{1}{4})×(\dfrac{22}{7})×(14×14)$$ = 154 cm2

Area of the semicircle = $$(\dfrac{1}{2})×(\dfrac{22}{7})×7\sqrt{2}×7\sqrt{2}$$ = 154 cm2

Area of the shaded region =Area of the semicircle + Area of ΔABC – Area of quadrant

= 154 +98-154 cm= 98cm2

Answered by Abhisek | 1 year ago

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