Chapter 11.4, Problem 6PSB

To determine

To calculate: The area of kite ABCD

Answer to Problem 6PSB

Area of kite ABCD is $85$ square units.

Explanation of Solution

Given information : $\angle BAD$ is a right angle, $BD=10$ and $BC=13$

Formula used :

Area of kite = $\frac{1}{2}\times d_1\times d_2$ (where $d_1$ and $d_2$ are diagonals of the kite)

Pythagoras Theorem is given as follows:

  ${(Hypotenuse)}^2={(Perpendicular)}^2+{(Base)}^2$ .

Calculation: Let diagonals AC and BD intersects at point O.

Now, $OB=\frac{BD}{2}$ (Longer diagonal of kite bisects the shorter diagonal into two halves)

  $\begin{array}{l}\Rightarrow OB=\frac{10}{2}\\ \Rightarrow OB=5\end{array}$

Now, In $\Delta BOC$ by Pythagoras theorem,

  ${(BC)}^2={(OB)}^2+{(OC)}^2$

  $\begin{array}{l}$ \Rightarrow {(13)}^2={(5)}^2+O{C}^2$\Rightarrow 169=25+O{C}^2\\ \Rightarrow O{C}^2=169-25\\ \Rightarrow O{C}^2=144\\ \Rightarrow OC=\sqrt{144}\\ \Rightarrow OC=12\end{array}$

Now, as AB = AB,

  $\Rightarrow \Delta ABD$ is an isosceles triangle.

As, $\angle BAD=90°$ ,

  $\Rightarrow \angle OAB=45°$ (Altitude of isosceles triangle bisects the angle at the vertex)

Now, in $\Delta OAB$ ,

  $\angle OAB=45°$ and $\angle AOB=90°$ (Diagonals of kite are perpendicular to each other)

  $\Rightarrow OA=5$ (Perpendicular and Base are equal of $90°,45°,45°$ triangle)

Now, $AC=OA+OC=5+12=17$

  $\therefore$ Diagonals are $17$ and $10$ .

Area of kite = $\frac{1}{2}\times d_1\times d_2$

  $\begin{array}{l}\Rightarrow \frac{1}{2}\times AC\times BD\\ \Rightarrow \frac{1}{2}\times 17\times 10\\ \Rightarrow \frac{170}{2}\\ \Rightarrow 85\end{array}$