Australia
The sides of a quadrilateral, taken in order are ^@ 13 \space cm, 10 \space cm, 12 \space cm,^@ and ^@5 \space cm^@ respectively. The angle contained by the last two sides is a right angle. Find the area of the quadrilateral.
Answer:
Area: ^@90 \space cm^2^@
- The following picture shows the quadrilateral ^@ABCD,^@
- Let's draw a line ^@AC^@.
^@\sqrt { AD^2 + DC^2 }^@
The ^@\triangle ACD^@ is the right-angled triangle.
^@ \begin{align} \text { Therefore, } AC^2 & = AD^2 + DC^2 \\ \implies AC & = \sqrt { AD^2 + DC^2 } \\ & = \sqrt { (5)^2 + (12)^2 } \\ & = 13 \space cm \end{align}^@ - ^@\begin{align} \text { The area of the right-angled triangle } \triangle ACD & = \dfrac { AD \times DC } { 2 } \\\ & = \dfrac { 5 \times 12 } { 2 } \\ & = 30 \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 { 13 + 10 + 13 }{2}\\ & = 18 \space cm. \end{align}^@
^@\begin{align} \text { The area of the } \triangle ABC & = \sqrt { S(S - AB)(S - BC)(S - CA) } \\ & = \sqrt { 18(18 - 13)(18 - 10)(18 - 13) } \\ & = 60 \space cm^2 \end{align}^@ - ^@\begin{align} \text { The area of the quadrilateral } ABCD & = Area(\triangle ACD) + Area(\triangle ABC) \\ & = 30 + 60 \\ & = 90 \space cm^2 \end{align}^@
