Question

A rangoli has been drawn on the floor of a house. ABCD and PQRS both are in the shape of a rhombus. Find the radius of semi-circle drawn on each side of rhombus ABCD.

Answer 1

In rhombus ABCD,
AO = OP + PA = 2 + 2 = 4 units and OB = OQ+QB = 2 + 1 = 3 units.
We know that, diagonals of the rhombus bisect each other at ${90}^{\circ }$.
Now,
In $\Delta OAB,(AB{)}^2=(OA{)}^2+(OB{)}^2$ [by Pythagoras theorem]
$\Rightarrow (AB{)}^2=(4{)}^2+(3{)}^2=25\Rightarrow AB=\sqrt{25}\Rightarrow AB=5units$
Since, AB is diameter of semi-circle.
$\therefore$ Radius $=\frac{Diameter}{2}=\frac{AB}{2}=\frac{5}{2}=2.5units$
Hence, radius of semi-circle is 2.5 units.