### The perimeter of a quadrilateral is $56 \space cm$. If the first three sides of a quadrilateral, taken in order are $17 \space cm, 16 \space cm$, and $15 \space cm$ respectively, and the angle between fourth side and the third side is a right angle, find the area of the quadrilateral.

Area: $180 \space cm^2$

Step by Step Explanation:
1. The following picture shows the quadrilateral $ABCD$, The perimeter of the quadrilateral $ABCD = 56 \space cm$
\begin{align} \text { Therefore }, & AB + BC + CD + DA = 56 \\ \implies & 17 + 16 + 15 + DA = 56 \\ \implies & 48 + DA = 56 \\ \implies & DA = 56 - 48 \\ \implies & DA = 8 \space cm \end{align}
2. Let's draw a line $AC$.
$\sqrt { DA^2 + DC^2 }$
The $\triangle ACD$ is the right angled triangle.
\begin{align} \text { Therefore, } AC^2 & = DA^2 + DC^2 \\ \implies AC & = \sqrt { DA^2 + DC^2 } \\ & = \sqrt { (8)^2 + (15)^2 } \\ & = 17 \space cm \end{align}
3. \begin{align} \text { The area of the right angled triangle } \triangle ACD & = \dfrac { DA \times DC } { 2 } \\ & = \dfrac { 8 \times 15 } { 2 } \\ & = 60 \space cm^2 \end{align}
4. Now, we can see that, this quadrilateral consists of the triangles $\triangle ACD$ and $\triangle ABC$.
The area of the $\triangle ABC$ can be calculated using Heron's formula since all sides of the triangle are known.
\begin{align} S & = \dfrac { AB + BC + CA } {2} \\ & = \dfrac { 17 + 16 + 17 }{2} \\ & = 25 \space cm \end{align}.
\begin{align} \text { The area of the } \triangle ABC & = \sqrt { S(S - AB)(S - BC)(S - CA) } \\ & = \sqrt { 25(25 - 17)(25 - 16)(25 - 17) } \\ & = 120 \space cm^2 \end{align}
5. \begin{align} \text { The area of the quadrilateral } ABCD & = Area(\triangle ACD) + Area(\triangle ABC) \\ & = 60 + 120 \\ & = 180 \space cm^2 \end{align} 