Question

In the given figure, DE is parallel to BC and AD = 1cm, BD = 2cm. What is the ratio of the area of ABC to the area of ADE?

Answer 1

Answer:

9:1

Step-by-step explanation:

answer for this question is 9:1

Answer 2

Given:

Two triangles ABC and ADE are given. in which AD=1 cm and BD =2 cm.

To Find:

The ratio of the area of ABC to the area of ADE?

Step-by-step explanation:

• In triangle ABC and ADE  side, DE of triangle ADE is parallel to side BC of triangle ABC.

        So,  $\angle ADE=\angle ABC$

               $\angle AED=\angle ACB$

• Angle  A is common in both triangles.

        So,    $\angle A= \angle A (common)$

• By the Base proportionality theorem,

         $\frac{AB}{AD} =\frac{AC}{AE}$

         $\frac{1+2}{1} =\frac{AC}{AE}$  

          $\frac{AC}{AE} =\frac{3}{1}$

        triangle ABC and ADE are similar.

• By Area theorem. The ratio of areas of two similar triangles is equal to the squares of the ratio of their corresponding sides.

        $\frac{Area \triangle ABC}{Area \triangle ADC} =(\frac{AB}{AD} )^2$

        $\frac{Area \triangle ABC}{Area \triangle ADC} =(\frac{3}{1} )^2$

        $\frac{Area \triangle ABC}{Area \triangle ADC} =(\frac{9}{1} )$

So, the ratio of the area of ABC to the area of ADE is 9:1