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.
Answer:
Area: ^@180 \space cm^2^@
- 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}^@ - 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}^@ - ^@ \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} ^@
- 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}^@ - ^@\begin{align} \text { The area of the quadrilateral } ABCD & = Area(\triangle ACD) + Area(\triangle ABC) \\ & = 60 + 120 \\ & = 180 \space cm^2 \end{align}^@
