Question

# In the given fig. 9.36, the cross type figure is divided into five squares. The side of each square is 2 cm, then find the radius of the circle and also the area of portion between circle and cross type figure (Take $\mathrm{\pi }=3.14$)

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Solution

## Diagonal AC of rectangle ABCD passes through the centre of the circle. We can find the length of AC using Pythagoras theorem in right triangle ABC. BC = BF + FG + GC = (2 + 2 + 2) cm = 6 cm AB = 2 cm In right $△$ABC, we have: ${\mathrm{AC}}^{2}={\mathrm{AB}}^{2}+{\mathrm{BC}}^{2}\left(\mathrm{Pythagoras}\mathrm{theorem}\right)\phantom{\rule{0ex}{0ex}}⇒{\mathrm{AC}}^{2}={2}^{2}+{6}^{2}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{AC}}^{2}=4+36=40\phantom{\rule{0ex}{0ex}}⇒\mathrm{AC}=2\sqrt{10}\mathrm{cm}\phantom{\rule{0ex}{0ex}}$ Now, OA = $\frac{\mathrm{AC}}{2}=\frac{2\sqrt{10}}{2}\mathrm{cm}=\sqrt{10}\mathrm{cm}\left(\mathrm{As}\mathrm{diagonals}\mathrm{of}\mathrm{a}\mathrm{rectangle}\mathrm{are}\mathrm{equal}\mathrm{and}\mathrm{bisect}\mathrm{each}\mathrm{other}\right)$ i.e., radius of the circle = $\sqrt{10}$ cm Now, area of square ABFE = (side)2 = (2)2 = 4 cm2 Similarly, area of square RSFG = area of square DCGH = area of square PQHE = area of square EFGH = 4 cm2 $\therefore$ Area of 5 squares = 5 (4 cm2) = 20 cm2 and area of the circle =$\pi$r2 = 3.14 × $\sqrt{10}$ × $\sqrt{10}$ = 31.4 cm2 $\therefore$ Area of the portion between the circle and the cross-type figure = area of the circle − area of the 5 squares = (31.4 – 20) cm2 = 11.4 cm2

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