Question

A circle with diameter 8 cm passes through two vertices (corners) of a square with sides of 5 cm. 5 cm 8 cm 026 2.9 © If the two vertices of the square are adjacent (shown in the diagram), calculate the area of the region where the circle and square overlap. by Calculate the area of the region where the circle and square overlap if the circle passes through opposite vertices of the square.

Answer 1

Answer A sem fcm ko By applying cosine rule .2 2 coso = 4+4 2*4X4 32-25 16+16-25 32 » Caso = 1 32 0 cas" (1/2) i. O = 77.4' Area of triangle triangle OAB OAB = Źr? sino = ²x4² sin (77.4) - 7.80ccm Area of sector OAB = 782 360 = 77.4 x 24 x 4² 360 = 10.811 cm? Hence area of overlap = Area of - Are of A0A8 Sector - 10.819-7806

= 3.005 m 2 Therefore area of overlap part = 3.005 cm? 5 Answer diagonal of square 5 -Sza s 4² +4²_155272 2xyxy -18 -9 9 / "O = cos" (-8) = √2+5=582cm According to cosine rule, coso = q*+-(582) 2*4*4 16+16 - 50 32-50 lo ha 32 32 = 124.23° Źr² sing = 2 xq?nsin (124:23) = 6.614 cm² Area of sector OAC = 0 xar2 360 124:23,2{x4? 360 = 17.353 cm? in area of shaded - Area of sector - Area of QORG segment OAC Area of AOAC =

- 17.353-6.614 10.739 cm Now area of AACD = area of squass 2 5- = 2/2 = 12.5 cm? "" area of overlap = area of JACD + Area of segment = 12.5 can'+ 10.739 cm? = 23.239 cm2 Ang 5