Chapter 11.4, Problem 4PSB

To determine

To calculate: The area of kite

Answer to Problem 4PSB

Area of kite is $336$ square units.

Explanation of Solution

Given information :

The side of Kite ( BC ) = 17

The other side of the Kite ( AB ) = 10

Formula used :

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

(where $d_1$ and $d_2$ are diagonals of the rhombus)

Pythagoras Theorem is given as follows

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

Calculation: Let ABCD be the rhombus and AC and BD be the diagonals of the rhombus intersecting at point O.

  

In $\Delta BOC$ , according to Pythagoras theorem,

  $\begin{array}{l}$ {(OB)}^2+{(OC)}^2={(BC)}^2$$ \Rightarrow {(8)}^2+{(OC)}^2={(17)}^2$\Rightarrow 64+O{C}^2=289\\ \Rightarrow O{C}^2=289-64\\ \Rightarrow O{C}^2=225\\ \Rightarrow OC=\sqrt{225}\\ \Rightarrow OC=15\end{array}$

Similarly, in $\Delta BOA$ , according to Pythagoras theorem,

  $\begin{array}{l}$ {(OB)}^2+{(OA)}^2={(BA)}^2$$ \Rightarrow {(8)}^2+{(OA)}^2={(10)}^2$\Rightarrow 64+O{A}^2=100\\ \Rightarrow O{A}^2=100-64\\ \Rightarrow O{A}^2=36\\ \Rightarrow OA=\sqrt{36}\\ \Rightarrow OA=6\end{array}$

Now, $AC=OC=OA=15+6=21$

And, $BD=2(OB)$ (Longer diagonal of kite bisects the shorter diagonal into two equal halves)

  $\begin{array}{l}\Rightarrow BD=2(8)\\ \Rightarrow BD=16\end{array}$

  $\therefore$ Diagonals are $21$ and $16$ .

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

  $\begin{array}{l}\Rightarrow \frac{1}{2}\times AC\times BD\\ \Rightarrow \frac{1}{2}\times 21\times 16\\ \Rightarrow \frac{336}{2}\\ \Rightarrow 168\end{array}$