Kite

Example 1:

In kite $\mathit{\text{P}}\mathit{\text{Q}}\mathit{\text{R}}\mathit{\text{S}}$, $\mathit{\text{Q}}\mathit{\text{S}}$ is the main diagonal. If $\mathit{\text{P}}\mathit{\text{R}}=10$ units and $\mathit{\text{Q}}\mathit{\text{O}}=4$ units what is the length $\mathit{\text{P}}\mathit{\text{Q}}$?

Here, $\mathit{\text{Q}}\mathit{\text{S}}$ is the perpendicular bisector of $\mathit{\text{P}}\mathit{\text{R}}$. Then, $\mathit{\text{m}}\angle \mathit{\text{Q}}\mathit{\text{O}}\mathit{\text{P}}=90°$ and $\mathit{\text{P}}\mathit{\text{O}}=\frac{1}{2}\mathit{\text{P}}\mathit{\text{R}}=\mathit{\text{R}}\mathit{\text{O}}$.

$\mathit{\text{P}}\mathit{\text{O}}=\frac{1}{2}\left(10\right)$

$=5$

$\Delta \mathit{\text{P}}\mathit{\text{O}}\mathit{\text{Q}}$ is a right triangle, $\mathit{\text{P}}\mathit{\text{O}}=5$ units, $\mathit{\text{Q}}\mathit{\text{O}}=4$ units. Use the Pythagorean Theorem to find the length of the hypotenuse.

$\mathit{\text{P}}\mathit{\text{Q}}=\sqrt{{\left(\mathit{\text{P}}\mathit{\text{O}}\right)}^2+{\left(\mathit{\text{Q}}\mathit{\text{O}}\right)}^2}$

$=\sqrt{5^2+4^2}$

$=\sqrt{25+16}$

$=\sqrt{41}$

Example 2:

In kite $\mathit{\text{L}}\mathit{\text{M}}\mathit{\text{N}}\mathit{\text{O}}$, the length of the main diagonal $\mathit{\text{M}}\mathit{\text{O}}$ is $12$ units and that of the cross diagonal $\mathit{\text{L}}\mathit{\text{N}}$ is $5$ units. What is the area of the kite $\mathit{\text{L}}\mathit{\text{M}}\mathit{\text{N}}\mathit{\text{O}}$?

$\mathit{\text{A}}=\frac{1}{2}\left(\mathit{\text{M}}\mathit{\text{O}}\right)\left(\mathit{\text{L}}\mathit{\text{N}}\right)$

$=\frac{1}{2}\left(12\right)\left(5\right)$

$=\frac{1}{2}\left(60\right)$

$=30$

Therefore, the area of the kite $\mathit{\text{L}}\mathit{\text{M}}\mathit{\text{N}}\mathit{\text{O}}$ is $30$ square units.