Chapter 11.3, Problem 54SPR

To determine

To find: Lengths of Diagonals of a Kite.

Answer to Problem 54SPR

The lengths of diagonals are:

AC = 27.42 inches

BD = 13.71 inches

Explanation of Solution

Figure:

  

Solution:

Split the Kite into two Triangles i.e. Δ ABD & Δ CBD.

Now,

Let the length of the diagonals be as follows:

**emsp; $\begin{array}{l}AC=2x\\ BD=x\end{array}$

  $\begin{array}{l}\text{Area Of Triangle = }\frac{\text{1}}{\text{2}}\text{×Base×Height}\\ \text{Thus,}\\ \text{Ar}\Delta ADB=\frac{1}{2}\times \text{BD×AO}\\ \text{Ar}\Delta CBD=\frac{1}{2}\times \text{BD×CO}\\ \text{Ar}\Delta ABD+\text{Ar}\Delta CBD=\text{Area Of Kite ABCD}\\ \Rightarrow \frac{\text{1}}{\text{2}}\text{BD×AO + }\frac{\text{1}}{\text{2}}\text{×BD×CO}=188\\ \Rightarrow \frac{\text{1}}{\text{2}}\text{×BD×}\left(\text{AO+OC}\right)=188\\ \Rightarrow \frac{1}{2}\times x\times 2x=188\\ \Rightarrow x^2=188\\ \Rightarrow x=\sqrt{188}\\ \Rightarrow x=13.71\text{ inches}\end{array}$

Therefore,

  $\begin{array}{l}\text{AC = 2×}x\\ \text{AC = 2×13}\text{.71}\\ \text{AC = 27}\text{.42 inches}\\ \text{\&}\\ \text{BD = }x\\ \text{BD = 13}\text{.71 inches}\end{array}$

The lengths of diagonals are:

AC = 27.42 inches

BD = 13.71 inches