Question

In the given figure, if $\Delta OAB\cong \Delta OFE\cong \Delta OCD$, OD = 3 cm, DC = 6 cm OC = 5 cm, then area of the shaded portion is

• A. $\sqrt{14}c{m}^2$
• B. $2\sqrt{14}c{m}^2$
• C. $6\sqrt{7}c{m}^2$
• D. $6\sqrt{14}c{m}^2$

Answer 1

The correct option is D $6\sqrt{14}c{m}^2$



Given: $\Delta OAB\cong \Delta OFE\cong \Delta OCD$
We know that the areas of congruent triangles are equal.
$\therefore Ar\left(\Delta OAB\right)=Ar\left(\Delta OFE\right)=Ar\left(\Delta OCD\right)$ …..(i)
Semiperimeter, (s) of $\Delta OCD=\frac{OD+DC+OC}{2}$
$=\frac{3+6+5}{2}$ (Given)
$=7cm$
By Heron’s formula, area of $\Delta OCD=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{7\times (7-3)\times (7-6)\times (7-5)}$
$=\sqrt{7\times 4\times 1\times 2}$
$=2\sqrt{14}c{m}^2$
$\therefore$ Area of shaded portion = $3\times$ Area of $\Delta OCD$ [From (i)]
$=3\times 2\sqrt{14}$
$=6\sqrt{14}c{m}^2$
Hence, the correct answer is option (d).