Question

The radius of the circumcircle a right- angled triangle, is 3 cm and the altitude drawn to the hypotenuse is 2 cm long. Find the sum of the areas of the three disjoint segments of the circumcircle formed by the sides of this triangle very important.

Answer 1

Dear student
$$\begin{gathered} \mathrm{r}=\mathrm{Radius}\mathrm{of}\mathrm{circumcircle}=3\mathrm{cm} \\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{in}\mathrm{a}\mathrm{right}\mathrm{angled}\mathrm{triangle},\mathrm{the}\mathrm{circumcentre}\mathrm{is}\mathrm{the}\mathrm{mid}-\mathrm{point}\mathrm{of}\mathrm{the}\mathrm{hypotenuse}. \\ \mathrm{So},\mathrm{Hypotenuse}=2\mathrm{r}=2\times 3=6\mathrm{cm} \\ \mathrm{Now},\mathrm{Altitude}\mathrm{drawn}\mathrm{to}\mathrm{hypotenuse}=2\mathrm{cm} \\ \mathrm{Then},\mathrm{area}\mathrm{of}\mathrm{triangle}=\frac{1}{2}\times 6\times 2=6{\mathrm{cm}}^2 \\ \mathrm{Area}\mathrm{of}\mathrm{circle}=3.14\times 3\times 3=28.26{\mathrm{cm}}^2 \\ \mathrm{Sum}\mathrm{of}\mathrm{area}\mathrm{of}\mathrm{disjoint}\mathrm{segments}=\mathrm{Area}\mathrm{of}\mathrm{circle}-\mathrm{Area}\mathrm{of}\mathrm{triangle} \\ =\left(28.26-6\right) \\ =22.26{\mathrm{cm}}^2 \end{gathered}$$
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