Q. 94.5( 171 Votes )

Answer :

Radius (*r _{1}*) of larger circle = 7 cm

As OD =

Radius (*r _{2}*) of smaller circle = cm

Area of smaller circle = π (r_{2})^{2}

= x x

= cm^{2}

Area of semi-circle ACB = 1/2 Πr_{1}^{2}

= x x (7)^{2}

= 77 cm^{2}

Area of triangle ABC = x AB x OC

As AB = OB + OA

⇒ AB = 7 + 7

⇒ AB = 14 cm

Area of triangle ABC = x 14 x 7

= 49 cm^{2}

Area of the shaded region

= Area of smaller circle + Area of semi-circle ACB - Area of ΔABC

= + 77 – 49

= 28 +

= 28 + 38.5

= 66.5 cm^{2}

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PREVIOUSFig. 12.26 depicts a racing track whose left and right ends are semicircular.The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide, find:(i) The distance around the track along its inner edge(ii) The area of the track.NEXTThe area of an equilateral triangle ABC is 17320.5cm2. With each vertex of the triangle as centre, acircle is drawn with radius equal to half the length of the side of the triangle (see Fig. 12.28). Find the area of the shaded region. (Use π = 3.14 and= 1.73205)