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^@

Step by Step Explanation:
  1. The following picture shows the quadrilateral ^@ABCD,^@
  2. 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}^@
  3. ^@\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}^@
  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 { 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}^@
  5. ^@\begin{align} \text { The area of the quadrilateral } ABCD & = Area(\triangle ACD) + Area(\triangle ABC) \\ & = 30 + 60 \\ & = 90 \space cm^2 \end{align}^@

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