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Rule

# Area of a Kite

The area of a kite is one-half the product of the length of the diagonals.

### Proof

Consider a kite and draw its two diagonals which are perpendicular to each other.

The diagonals divide the kite into two big triangles – namely, $\triangle ABC$ and $\triangle ACD.$

The area of the kite equals the sum of the areas of the two triangles. $\begin{gathered} A_{ABCD} = A_{\triangle ABC} + A_{\triangle ACD} \end{gathered}$ By taking $\overline{AC}$ as the base and $\overline{BE}$ as the height, the area of $\triangle ABC$ can be calculated as follows. $\begin{gathered} A_{\triangle ABC} = \dfrac{1}{2}AC\cdot BE \end{gathered}$ Similarly, the area of $\triangle ACD$ can be found taking $\overline{AC}$ as the base and $\overline{ED}$ as the height. $\begin{gathered} A_{\triangle ACD} = \dfrac{1}{2}AC\cdot ED \end{gathered}$ Next, substitute the areas of the triangles into the area of the kite.
$A_{ABCD} = A_{\triangle ABC} + A_{\triangle ACD}$
$A_{ABCD} = {\color{#0000FF}{\dfrac{1}{2}AC\cdot BE}} + {\color{#009600}{\dfrac{1}{2}AC\cdot ED}}$
$A_{ABCD} = \dfrac{1}{2}AC(BE+ED)$
By the Segment Addition Postulate, $BE+ED$ can be written as $BD.$
$A_{ABCD} = \dfrac{1}{2}AC(BE+ED)$
$A_{ABCD} = \dfrac{1}{2}AC\cdot BD$
$A_{ABCD} = \dfrac{1}{2}{\color{#0000FF}{d_{1}}}{\color{darkorange}d_{2}}$