Parallelogram $A B C D$ is made up of four equilateral triangles of side length 1. The length of diagonal $A C$ is
(A) $\sqrt{5}$
(B) $\sqrt{7}$
(C) 3
(D) $\sqrt{3}$
(E) $\sqrt{10}$

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-144.jpg?height=228&width=444&top_left_y=1003&top_left_x=1296)
Parallelogram $A B C D$ is made up of four equilateral triangles of side length 1 . The length of diagonal $A C$ is
(A) $\sqrt{5}$
(D) $\sqrt{3}$
(B) $\sqrt{7}$
(C) 3

## Solution

From $C$, we draw a line perpendicular to $A D$ extended so that they meet at point $E$ as shown in the diagram.

This construction makes $\triangle C D E$ a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle with $\angle C D E=60^{\circ}$ and $C D=1$. Thus $C E=\frac{\sqrt{3}}{2}$ and $D E=\frac{1}{2}$. Using pythagoras in $\triangle A C E$, we have $A E=\frac{5}{2}$

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-293.jpg?height=266&width=529&top_left_y=211&top_left_x=1432)
and $C E=\frac{\sqrt{3}}{2}, A C=\sqrt{\left(\frac{\sqrt{3}}{2}\right)^{2}+\left(\frac{5}{2}\right)^{2}}=\sqrt{7}$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-293.jpg?height=266&width=556&top_left_y=580&top_left_x=1424)

ANSWER: (B)