Question

In the given figure, $AB$ and $DE$ are perpendicular to $BC$.

Prove that $\Delta ABC~\Delta DEC$, If $AB=6cm,DE=4cm,AC=15cm$, then calculate $CD$, Find the ratio of $Area\left(\Delta ABC\right):Area\left(\Delta DEC\right)$.

Answer 1

Step 1: Proof for $\Delta ABC~\Delta DEC$:

$\angle ACB=\angle DCE$ (Common Angle)

$\angle ABC=\angle DEC=90°$ (Given)

From the above statements, we come to the conclusion that:

$\Delta ABC~\Delta DEC$ (By AA axiom)

[AA axiom: In two triangles, if two pairs of corresponding angles are congruent, then the triangles are similar.]

Step 2: Find the value of $CD$:

We know that,$AB=6cm,DE=4cm,AC=15cm$

We proved that $\Delta ABC~\Delta DEC$ and we know the property of similar triangles so

$\frac{AB}{DE}=\frac{AC}{CD}$

$\frac{6}{4}=\frac{15}{CD}$

$6\left(CD\right)=60$

$CD=10cm$

Step 3: Find the ratio of $Area\left(\Delta ABC\right):Area\left(\Delta DEC\right)$:

We proved above that $\Delta ABC~\Delta DEC$ and we know the property of similar triangles: "The ratio of the areas of two similar triangles is equal to the square of the ratio of any pair of their corresponding sides".

So putting the theorem into the equation and data we have we get the equation as,

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

$\frac{Area\left(\Delta ABC\right)}{Area\left(\Delta DEC\right)}=\frac{6^2}{4^2}$

$\frac{Area\left(\Delta ABC\right)}{Area\left(\Delta DEC\right)}=\frac{36}{16}=\frac{9}{4}$

So, we get the ratio of $Area\left(\Delta ABC\right): Area\left(\Delta DEC\right)= 9:4$.

Thus, we proved that $\Delta ABC~\Delta DEC$, the value of $CD$ is $10cm$ and the ratio of $Area\left(\Delta ABC\right): Area\left(\Delta DEC\right)= 9:4$.