How exactly would one generalize the area of a kite inscribed within a circle? Through a lot of calculation, which I do think was actually way more complicated then required, using various trigonometric functions, chords, amongst other things, I found that the area in a unit circle would be $\sqrt{3}$. But how exactly would this be generalized?

I am thinking that one would draw a line $r$ to the centre of a circle from the edge, and draw a line across the entire circle. Draw a line starting at the point a of the original line, and go down perpendicular to the central line. Then connect all the points which hit the edges to eachother. Creating a kite.

This should allow one to find an isosceles triangle with the hypotenuses being $r$, with a height of $r$ thus the base can be $r\sqrt{3}$. This should be the side of a larger triangle, which is actually an equilateral, which would have area calculated by $\dfrac{\sqrt{3}}r r(\sqrt{3})^2$, or $\sqrt{3} × 3r^2$.

The second can be found by finding how two other cords, can make up another equilateral, showing that the hypotenuses of the bottom triangles are actually equal to r, thus it is an isosceles with the side lengths of $r$ and base of $r \sqrt{3}$.

However, we need to find the height with $(\sqrt{3}/2) × (r \sqrt{3})$ or $3r/2$, making the height of the smaller triangle $r/2$, thus the area is $\sqrt{3} r^2$.

Thus the generalized formula should be $r^2\sqrt{3}$

This should be correct, but I am open to people pointing out me messing up or overcomplicating this formula.

Noticed the small gap, that is not supposed to be there, I just suck at editing images, imagine it touches the central line.

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    $\begingroup$ You need to add a diagram to illustrate your argument. Otherwise it is difficult to follow sufficiently to find your error. There is an error since the area of a kite inscribed in a circle can be anywhere from zero up to the area of a square inscribed in a circle. $\endgroup$ Commented Dec 25, 2021 at 1:15
  • $\begingroup$ I expected this, exact parametres can be hard to describe. I am gonna add an example $\endgroup$
    – Zoey
    Commented Dec 25, 2021 at 1:16
  • $\begingroup$ It appears that you are working with a kite of specific proportions. $\endgroup$ Commented Dec 25, 2021 at 1:17
  • $\begingroup$ I am, yeah, that is why I attempted to explain the proportions. I just failed. I have added an image. $\endgroup$
    – Zoey
    Commented Dec 25, 2021 at 1:21
  • $\begingroup$ You are working specifically with a kite with one edge equal to $r$. In that case, the line drawn to the center will form an equlateral triangle. The height of the equilateral will be $h=(\sqrt{3}/2)r$. But this is also the height of half the kite, which has a base of length $2r$. So the area of half the kite will be $bh/2$ or $\sqrt{3}r^2/2$. But that is just the area of half the kite, which will be $\sqrt{3}r^2$. In general, however, a kite does not have to have one side of length $r$. $\endgroup$ Commented Dec 25, 2021 at 1:46

2 Answers 2


Let $ABCD$ be the cyclic kite in question, with diagonals $AC$ and $BD$ and $X$ their intersection. Let $O$ be the centre of the circle and $r$ its radius.

A kite inscribed in a circle

Lemma 1: One diagonal of a kite is the perpendicular bisector of the other diagonal. Without loss of generality, assume $AC$ bisects $BD$.

Lemma 2: The perpendicular bisector of a chord of a circle is a diameter of the circle.

Hence $AC$ is a diameter of the circle and $|AC|=2r$.

Lemma 3 (Thales’ theorem): A triangle constructed on a diameter of its circumcircle is a right triangle, with the right angle at the vertex not on the diameter.

Hence $\triangle ABC$ and $\triangle ADC$ are right triangles with $AC$ the hypotenuse of each. They are congruent by SSS ($|AB|=|AD|$ and $|BC|=|DC|$ by definition of a kite; $AC$ common), each having half the kite’s area.

Consider $\triangle ABC$. The altitude $BX$ is constrained such that $0<|BX|\le r$ (length $0$ is the degenerate case where $A$ and $B$ coincide, and length $r$ is the case where $|AC|=|BD|$ and $ABCD$ is a square).

Hence the area $\mathrm{A}_{ABC}$ is constrained to $0<\mathrm{A}_{ABC}\le r^2$, and since it’s half of $ABCD$, we have bounds on the area $\mathrm{A}_{ABCD}$:

$$0<\mathrm{A}_{ABCD}\le 2r^2$$

For the special case in the question, where $|AB|=r$, $\triangle ABO$ is equilateral (since $AO$ and $BO$ are also radii) and its altitude is $|BX|=\frac{\sqrt 3}{2}r$. Hence $\mathrm{A}_{ABC}=\frac{\sqrt 3}{2}r^2$ and:

$$\mathrm{A}_{ABCD}=\sqrt 3 r^2$$

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    $\begingroup$ Thanks again for showing me I did it right. And yeah, I knew about the 2r<sup>2</sup> situation, which is a square.. $\endgroup$
    – Zoey
    Commented Dec 25, 2021 at 16:15

If the height of the smaller part is $h$, where $h \le r$, the height of the upper part is $2r-h$.

The distance $d$ from the center to the crossbar is $r-h$, so, if $c$ is half the length of the crossbar, $r^2 =c^2+(r-h)^2 =c^2+r^2-2rh+h^2 $ so $c^2=2rh-h^2 $.

The area of the lower half is $2(ch/2) =ch $ and the area of the upper half is $2(c(2r-h))/2 =c(2r-h) $ so the combined area is $ch+c(2r-h) =2rc =2r\sqrt{2rh-h^2} $.


If the crossbar is in the middle, $c=h=r$ so the area is $2r^2$ which is correct since each triangle has base $2r$ and altitude $r$.

If the crossbar is at the bottom, $h=0$ so the area is also $0$ which is correct since the kite is a stick.

Note that

$\begin{array}\\ 2r(2r-h)-(2r-h)^2 &=4r^2-2rh-(4r^2-4rh+h^2)\\ &=2rh-h^2\\ \end{array} $

so the expression for the area is the same when $h$ (the lower height) is replaced by $2r-h$ (the upper height).

  • $\begingroup$ Another confirmation that my math was right, yes. Also another example of a far easier way I could do it. Lol. $\endgroup$
    – Zoey
    Commented Dec 25, 2021 at 2:28

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