Prove that the area of the semicircle drawn on the hypotenuse of a right angled triangle is equal to the sum of the areas of the semicircles drawn on the other two sides of triangle.

Q: Prove that the area of the semicircle drawn on the hypotenuse of a right angled triangle is equal to the sum of the areas of the semicircles drawn on the other two sides of triangle.

Given: $\triangle\text{ABC}$ is right angled at B. Three semi-circles taking as the sides BC, AB and AC of triangle ABC as diameter $C_1, C_2$ and $C_3$ are drawn. To prove: Area of semicircles $(C_1 + C_2)$ = Area of semi-circle $C_3$ Proof: In $\triangle\text{ABC},$ $\angle\text{B}=90^\circ$ $\therefore$ $\text{BC}^2+\text{AB}^2+\text{} =\text{AC}^{2}$ $\Rightarrow (2\text{r}_{1})^{2} + (2\text{r}_{2})^{2} = (2\text{r}_{3})^{2}$ [From figure as BC, AB and AC are diameters] $\Rightarrow 4(\text{r}_{1}^{2} + \text{r}^{2}_{2}) = 4\text{r}^{2}_{3} $ $\Rightarrow \text{r}^{2}_{1} + \text{r}^{2}_{2}+\text{r}^{2}_{3}$' $\Rightarrow \frac{1}{2} \pi \text{r}_{1}^{2} + \frac{1}{2} \pi \text{r}_{2}^{2} = \frac{1}{2} \pi \text{r}_{3}^{2}$ $\operatorname{ar}\left(\right.$ semi-circle $\left.C_1\right)+\operatorname{ar}\left(\right.$ semi-circle $\left.C_2\right)=\operatorname{ar}\left(\right.$ semi-circle $\left.C_3\right)$ Hence, proved.

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Solution

Given: $\triangle\text{ABC}$ is right angled at B. Three semi-circles taking as the sides BC, AB and AC of triangle ABC as diameter $C_1, C_2$ and $C_3$ are drawn.
To prove: Area of semicircles $(C_1 + C_2)$ = Area of semi-circle $C_3$
Proof: In $\triangle\text{ABC},$
$\angle\text{B}=90^\circ$
$\therefore$ $\text{BC}^2+\text{AB}^2+\text{} =\text{AC}^{2}$
$\Rightarrow (2\text{r}_{1})^{2} + (2\text{r}_{2})^{2} = (2\text{r}_{3})^{2}$
[From figure as BC, AB and AC are diameters]
$\Rightarrow 4(\text{r}_{1}^{2} + \text{r}^{2}_{2}) = 4\text{r}^{2}_{3} $
$\Rightarrow \text{r}^{2}_{1} + \text{r}^{2}_{2}+\text{r}^{2}_{3}$'
$\Rightarrow \frac{1}{2} \pi \text{r}_{1}^{2} + \frac{1}{2} \pi \text{r}_{2}^{2} = \frac{1}{2} \pi \text{r}_{3}^{2}$
$\operatorname{ar}\left(\right.$ semi-circle $\left.C_1\right)+\operatorname{ar}\left(\right.$ semi-circle $\left.C_2\right)=\operatorname{ar}\left(\right.$ semi-circle $\left.C_3\right)$
Hence, proved.

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