A design has been drawn on a tile. $ABCD$ and $PQRS$ both are in the shape of a rhombus. Find the total area of semi circles drawn on each side of the rhombus $ABCD$."

Given: A design has been drawn on a tile. $ABCD$ and $PQRS$ both are in the shape of a rhombus.

To do: To find the total area of semi circles drawn on each side of the rhombus $ABCD$.

Solution:

Let $O$ is the point where diagonals intersect each other.

Therefore, $OA=OP+AP=3\ cm+3\ cm=6\ cm$

$OB=OQ+QB=3\ cm+5\ cm=8\ cm$

As known, diagonals of a rhombus intersect each other at $90^{\circ}$.

$AB^2=OA^2+OB^2=6^2+8^2=36+64=100$

$\Rightarrow AB=\sqrt{100}$

$\Rightarrow AB=10\ cm$

Here, $AB$ is the diameter of the semi-circle.

Therefore, radius of the semi-circle $r=\frac{10}{2}=5\ cm$

$\therefore$ Area of the semi-circle $=\frac{1}{2}\pi r^2$

$=\frac{1}{2}\times(5)^2\pi$

$=\frac{25}{2}\pi$

$\therefore$ Total area of semi circles on each side $ABCD=4\times\frac{25}{2}\pi=50\pi\ cm^2$

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Updated on: 10-Oct-2022

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