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 $\mathrm{\pi }=3.14$)

Answer 1

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:

$$\begin{gathered} {\mathrm{AC}}^2={\mathrm{AB}}^2+{\mathrm{BC}}^2(\mathrm{Pythagoras}\mathrm{theorem}) \\ \Rightarrow {\mathrm{AC}}^2=2^2+6^2 \\ \Rightarrow {\mathrm{AC}}^2=4+36=40 \\ \Rightarrow \mathrm{AC}=2\sqrt{10}\mathrm{cm} \end{gathered}$$

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