Question

In the given figure, E, F, G, H, O and I are the mid-points of the sides DC, BC, AB, AD, GF and AC respectively. If ar $\left(\Delta OGB\right)=2c{m}^{2}andar\left(\Delta DHE\right)=4c{m}^2$ and BI, GF, HF are the line segments, then the area of quadrilateral ABCD is equal to

• A.
  $32c{m}^2$
• B.
  $64c{m}^2$
• C.
  $16c{m}^2$
• D.
  $40c{m}^2$

Answer 1

The correct option is A

$32c{m}^2$

Given that $Ar\left(\Delta OGB\right)=2c{m}^{2}andAr\left(\Delta DHE\right)=4c{m}^2\text{By construction}BJ\perp GF.\therefore Ar\left(\Delta OGB\right)=\frac{1}{2}\times BJ\times GO=\frac{1}{2}\times BJ\times \frac{GF}{2}(\because GO=\frac{GF}{2})=\frac{1}{2}\times (\frac{1}{2}\times GF\times BJ)=\frac{1}{2}\times Ar\left(\Delta BGF\right)\Rightarrow Ar\left(\Delta BGF\right)=2\times Ar\left(\Delta OGB\right)=2\times 2(\because Ar(\Delta OGB)=2c{m}^2)=4c{m}^2$

Since area of the triangle formed by joining the mid-points of the sides of a triangle is equal to one-fourth area of the given triangle.

$\therefore Ar\left(\Delta BGF\right)=\frac{1}{4}Ar\left(\Delta ABC\right)\Rightarrow Ar\left(\Delta ABC\right)=4Ar\left(\Delta BGF\right)=4\times 4=16c{m}^{2}Similarly,Ar\left(\Delta DHE\right)=\frac{1}{4}Ar\left(\Delta ADC\right)\Rightarrow Ar\left(\Delta ADC\right)=4\times Ar\left(\Delta DHE\right)=4\times 4[\because Ar(\Delta DHE)=4c{m}^2]=16c{m}^2\therefore Ar\left(ABCD\right)=Ar\left(\Delta ABC\right)+Ar\left(\Delta ADC\right)=16+16=32c{m}^2$

Hence, the correct answer is option (a).