# In figure, OCDE is a rectangle inscribed in a quadrant of a circle of radius 10 cm.

Question:

In figure, OCDE is a rectangle inscribed in a quadrant of a circle of radius 10 cm. If OE = 2√5 cm, find the area of the rectangle.

Solution:

Given OD = 10 cm and OE = 2√5cm

By using Pythagoras theorem

$\therefore \mathrm{OD}^{2}=\mathrm{OE}^{2}+\mathrm{DE}^{2}$

$\Rightarrow \mathrm{DE}=\sqrt{\mathrm{OD}^{2}-\mathrm{OE}^{2}}=\sqrt{10^{2}-(2 \sqrt{5})^{2}}=4 \sqrt{5} \mathrm{~cm}$

∴ Area of rectangle OCDE = OE × DE

$=2 \sqrt{5} \times 4 \sqrt{5} \mathrm{~cm}^{2}=40 \mathrm{~cm}^{2}$