Question

In Fig. 3, APB and AQO are semicircles, and AO = OB. If the perimeter of the figure is 40 cm, find the area of the shaded region. Use π =227

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

$96.25 \mathrm{~cm}^2$

Step-by-step explanation:

Step 1: The complete length of a shape's border is referred to as the perimeter in geometry. A shape's perimeter is calculated by summing the lengths of all of its sides and edges. Its dimensions are expressed in linear units like centimetres, metres, inches, and feet.

Step 2: The size of a patch on a surface is determined by its area. Surface area refers to the area of an open surface or the border of a three-dimensional object, whereas the area of a plane region or plane area refers to the area of a form or planar lamina.

Step 3: Let the radius of the semi-circle APB be r.

$\Rightarrow$ The radius of the semi-circle $A Q O=\frac{r}{2}$

Now,

Perimeter of the given figure = Length of ${arc} A Q O+$ Length of ${arc}$ of A P B+O B

$\pi \times \frac{\pi}{2}+\pi \times r+r$

$r\left(\frac{3}{2} \pi+1\right)$

$r\left(\frac{3}{2} \times \frac{22}{7}+1\right)$

$=r \frac{80}{14}$

$r\left(\frac{4 g}{7}\right)$

$\Rightarrow r\left(\frac{4 g}{7}\right)=40$

$\Rightarrow r=7 \mathrm{~cm}$

$\therefore$ Area of the shaded region = Area of semi-circle $A Q O+$ Area of semi-circle APB.

$\begin{aligned} & =\frac{\pi\left(\frac{7}{2}\right)^2}{2}+\frac{\pi 7^2}{2} \\ & =49 \pi\left(\frac{1}{8}+\frac{1}{2}\right) \\ & =96.25 \mathrm{~cm}^2\end{aligned}$

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