The trapezoid shown has a height of length $12 \mathrm{~cm}$, a base of length $16 \mathrm{~cm}$, and an area of $162 \mathrm{~cm}^{2}$. The perimeter of the trapezoid is
(A) $51 \mathrm{~cm}$
(B) $52 \mathrm{~cm}$
(C) $49 . \overline{6} \mathrm{~cm}$
(D) $50 \mathrm{~cm}$
(E) $56 \mathrm{~cm}$

![](https://cdn.mathpix.com/cropped/2024_04_20_6027bc27089ed4fc493cg-053.jpg?height=298&width=442&top_left_y=1033&top_left_x=1275)
We first label the trapezoid $A B C D$ as shown in the diagram below.

Since $A D$ is the perpendicular height of the trapezoid, then $A B$ and $D C$ are parallel.

The area of the trapezoid is $\frac{A D}{2} \times(A B+D C)$ or $\frac{12}{2} \times(A B+16)$ or $6 \times(A B+16)$.

Since the area of the trapezoid is 162 , then $6 \times(A B+16)=162$ and $A B+16=\frac{162}{6}$ or $A B+16=27$, so $A B=11$.

Construct a perpendicular from $B$ to $E$ on $D C$.

Since $A B$ is parallel to $D E$ and both $A D$ and $B E$ are perpendicular to $D E$, then $A B E D$ is a rectangle.

Thus, $D E=A B=11, B E=A D=12$, and $E C=D C-D E=16-11=5$.

Since $\angle B E C=90^{\circ}$, then $\triangle B E C$ is a right-angled triangle.

Thus by the Pythagorean Theorem, $B C^{2}=B E^{2}+E C^{2}$ or $B C^{2}=12^{2}+5^{2}$ or $B C^{2}=169$ so
$B C=13$ (since $B C>0$ ).

The perimeter of the trapezoid is $A B+B C+C D+D A=11+13+16+12=52 \mathrm{~cm}$.

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-253.jpg?height=399&width=523&top_left_y=283&top_left_x=324)

ANSWER: (B)