# Ex.6.4 Q7 Triangles Solution - NCERT Maths Class 10

## Question

Prove that the area of an equilateral triangle described on one side of a square is Equal to half the area of the equilateral triangle described on one of its diagonals.

## Text Solution

**Reasoning:**

As we know that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

**Steps:**

\(\Delta ABE\,\,\) is described on the side \(AB\) of the square \(ABCD\)

\(\Delta DBF\) is described on the diagonal \(BD\) of the square \(ABCD\)

Since \(\Delta ABE\,\) and \(\,\Delta DBF\) are equilateral triangles

\(\Delta ABE \sim \Delta DBF\) [each angle in an equilateral triangle measures \({{60}^ \circ}\) ]

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

\[\begin{align}\frac{{Area\,\,of\,\,\Delta ABE}}{{Area\,\,of\,\,\Delta DBF}}&= \frac{{{{(AB)}^2}}}{{{{(DB)}^2}}} \\\frac{{Area\,\,of\,\,\Delta ABE}}{{Area\,\,of\,\,\Delta DBF}}&= \frac{{{{\left( {AB} \right)}^2}}}{{{{\left( {\sqrt 2 AB} \right)}^2}}}\end{align}\] [\(\because\) diagonal of a square is \(\,\sqrt 2\times \) side]

\[\begin{align}\frac{{Area\,of\,\Delta ABE}}{{Area\,of\,\Delta DBF}}&= \frac{{A{B^2}}}{{2A{B^2}}}\\\frac{{Area\,of\,\Delta ABE}}{{Area\,of\,\Delta DBF}} &= \frac{1}{2}\\ \Rightarrow \quad \text{Area of }\Delta ABE &= \frac{1}{2} \text{Area of }\Delta DBF\end{align}\]

Tick the correct answer and justify.