Question

In the given figure, $\angle B=\angle E,\angle ACD=\angle BCE,AB=10.4$ cm and DE = 7.8 cm. Find the ratio of areas of the $\Delta ABC$ and $\Delta DEC$.

• A. 4 : 3
• B. 16 : 9
• C. 9 : 16
• D. 3 : 4

Answer 1

The correct option is B

16 : 9

In the figure DE =7.8 cm, AB =10.4 cm

$\angle ACD=\angle BCE$ (given)

Adding $\angle DCB$ both sides,

$\angle ACD+\angle DCB=\angle DCB+\angle BCE$

$\Rightarrow \angle ACB=\angle DCE$

Now in $\Delta ABC$ and $\Delta DEC$,

$\angle B=\angle E$ (given)

$\angle ACB=\angle DCE$ (proved)

$\therefore \Delta ABC\sim \Delta DEC$ (by AA similarity criterion)

$\therefore \frac{area\left(\Delta ABC\right)}{area\left(\Delta DEC\right)}=\frac{A{B}^2}{D{E}^2}$

(The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.)

$=\frac{(10.4{)}^2}{(7.8{)}^2}={\left(\frac{104\times 10}{10\times 78}\right)}^2={\left(\frac{4}{3}\right)}^2=\frac{16}{9}$

$\therefore$ $Ar\left(\Delta ABC\right):Ar\left(\Delta DEC\right)=16:9$