Question

In the figure above, triangles $ABC$ and $CDE$ are equilateral and line segment $\overline{AE}$ has length $25$. What is the sum of the perimeters of the two triangles?

• A. $50$
• B. $75$
• C. $100$
• D. $80$

Answer 1

The correct option is B $75$

Given

$△$$ABC$ and $△$$CDE$ are equilateral.

$\overline{AE}$ $=$ $25$

To find perimeters of the two triangles,

Let us consider the lengths of $\overline{AC}$ and $\overline{CE}$ to be ${}^{\prime }{x}^{\prime }$ and ${}^{\prime }y$ respectively.

As $△$$ABC$ is equilateral, $\overline{AC}$ $=$ $\overline{AB}$ $=$ $\overline{BC}$ $=$ $x$

As $△$$CDE$ is equilateral, $\overline{CE}$ $=$ $\overline{CD}$ $=$ $\overline{DE}$ $=$ $y$

From the figure,

$\overline{AE}$ $=$ $\overline{AC}$ $+$ $\overline{CE}$

$25$ $=$ $x$ $+$ $y$

$x$ $+$ $y$ $=$ $25$

Perimeter of the triangle is the sum of all sides of the triangle.

For $△$$ABC$,

Perimeter of $△$$ABC$ $=$ $\overline{AC}$ $+$ $\overline{AB}$ $+$ $\overline{BC}$

$=$ $x$ $+$ $x$ $+$ $x$

$=$ $3x$

For $△$$CDE$,

Perimeter of $△$$CDE$ $=$ $\overline{CE}$ $+$ $\overline{CD}$ $+$ $\overline{DE}$

$=$ $y$ $+$ $y$ $+$ $y$

$=$ $3y$

Now,

Perimeter of two triangles $=$ Perimeter of $△$$ABC$ $+$ Perimeter of $△$$CDE$

$=$ $3x$ $+$ $3y$

$=$ $3$ $\times$ $(x$ $+$ $y)$

$=$ $3$ $\times$ $25$ $\left(fromabove\right)$

$=$ $75$

Therefore, Perimeter of the two triangles is ${}^{\prime }{75}^{\prime }$ units.