# Find the Area a Kite Inscribed in a Circle

The question is as follows:

$A$, $B$, $C$ and $D$ lie in alphabetical order on a circle so that ABCD forms a kite. $AB = DA = 8 cm$ and $BC = CD = 13 cm$. Find the area of the kite $ABCD$.

I thought that I might be able to use $AC$ as the diameter of the circle, therefore, $\angle ADC$ would be a right angle. Using that I thought that I can use the formula for finding the area of a triangle using sine ($\frac{1}{2}ab \times \sin C$) to find the area of the two congruent triangles. However, the answer that I get does not match with the correct answer. What am I doing incorrectly? Any help will be greatly appreciated!

• Hint...you have a height and a corresponding base for the triangles, so you don't need the sine formula Feb 5 '18 at 18:42
• $ADC$ is a right triangle and you know both legs. So, what is the area of $\triangle{ADC}$? Feb 5 '18 at 18:47
• I got the area of $\triangle {ADC}$ to be 52 square centimetres. So the area of the other triangle would also be the same, right? If so, then the area of the kite would be 104 square centimetres, however, that's not the answer. According to the answer key, the area is 85.4 square centimetres. Feb 5 '18 at 19:01
• the area is 85.4 $\;8\color{red}{4.5}\,$ would be suspiciously equal to $\,13^2 / 2\,$, but I don't see how that could relate to the problem as stated, either.
– dxiv
Feb 6 '18 at 4:07

An overkill solution:

We know that the area of kite is $ef/2$ where $e,f$ are diagonals. By Ptolomey theorem we have:

$$2A =ef = ac+bd = 2ac = 2\cdot 104$$ so $A= 104$.

I leave it as an exercise to find why ($M$ is the point where the perpendicular diagonals of the kite cut)

$$\frac{1}{BM^2}= \frac{1}{8^2} + \frac{1}{13^2}$$

is correct ( compare proportionate sides of similar triangles )

Now as you have done for total kite area

$$(AC \cdot BD/2 \cdot \frac12) \cdot 2$$

$AC$ is hypotenuse etc.