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^@

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}^@

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