A regular pentagon has all sides and angles equal. If the shaded pentagon is enclosed by squares and triangles, as shown, what is the size of angle $x$ ?
(A) $75^{\circ}$
(D) $60^{\circ}$
(B) $108^{\circ}$
(C) $90^{\circ}$
(E) $72^{\circ}$
![](https://cdn.mathpix.com/cropped/2024_04_20_6027bc27089ed4fc493cg-104.jpg?height=538&width=244&top_left_y=2107&top_left_x=1445)
A regular pentagon has all sides and angles equal. If the shaded pentagon is enclosed by squares and triangles, as shown, what is the size of angle $x$ ?
(A) $75^{\circ}$
(B) $108^{\circ}$
(C) $90^{\circ}$
(D) $60^{\circ}$
(E) $72^{\circ}$

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-472.jpg?height=304&width=315&top_left_y=1542&top_left_x=1382)

## Solution

Since a pentagon can be divided into three triangles, the sum of the angles in the pentagon is $3 \times 180^{\circ}=540^{\circ}$. Since the angles in a regular pentagon are all equal, each one is $540^{\circ} \div 5=108^{\circ}$. The sum of the angles of each vertex of the pentagon is $x+90^{\circ}+90^{\circ}+108^{\circ}=360^{\circ}$.

Therefore $x=72^{\circ}$.

ANSWER: (E)