Expand menu menu_open Minimize Start chapters Home History history History expand_more
{{ item.displayTitle }}
navigate_next
No history yet!
Progress & Statistics equalizer Progress expand_more
Student
navigate_next
Teacher
navigate_next
{{ filterOption.label }}
{{ item.displayTitle }}
{{ item.subject.displayTitle }}
arrow_forward
No results
{{ searchError }}
search
menu_open
{{ courseTrack.displayTitle }}
{{ statistics.percent }}% Sign in to view progress
{{ printedBook.courseTrack.name }} {{ printedBook.name }}
search Use offline Tools apps
Login account_circle menu_open

Area of a Kite

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, ABC\triangle ABC and ACD.\triangle ACD.

The area of the kite equals the sum of the areas of the two triangles. AABCD=AABC+AACD\begin{gathered} A_{ABCD} = A_{\triangle ABC} + A_{\triangle ACD} \end{gathered} By taking AC\overline{AC} as the base and BE\overline{BE} as the height, the area of ABC\triangle ABC can be calculated as follows. AABC=12ACBE\begin{gathered} A_{\triangle ABC} = \dfrac{1}{2}AC\cdot BE \end{gathered} Similarly, the area of ACD\triangle ACD can be found taking AC\overline{AC} as the base and ED\overline{ED} as the height. AACD=12ACED\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.
AABCD=AABC+AACDA_{ABCD} = A_{\triangle ABC} + A_{\triangle ACD}
AABCD=12ACBE+12ACEDA_{ABCD} = {\color{#0000FF}{\dfrac{1}{2}AC\cdot BE}} + {\color{#009600}{\dfrac{1}{2}AC\cdot ED}}
AABCD=12AC(BE+ED)A_{ABCD} = \dfrac{1}{2}AC(BE+ED)
By the Segment Addition Postulate, BE+EDBE+ED can be written as BD.BD.
AABCD=12AC(BE+ED)A_{ABCD} = \dfrac{1}{2}AC(BE+ED)
AABCD=12ACBDA_{ABCD} = \dfrac{1}{2}AC\cdot BD
AABCD=12d1d2A_{ABCD} = \dfrac{1}{2}{\color{#0000FF}{d_{1}}}{\color{darkorange}d_{2}}