# find area of kite given side length

Circle with two tangent lines

Above is the picture in question. A circle is given, center (-2,4) and a point outside the circle (0,10) is shown. Asked to calculate the area of the quadrilateral ABCD, I figured that this kite has sides 2 (the radii of the circle) and 6 (the difference between (0,4) and (0,10)). How can I calculate the area of the kite?

(I tried calculating the diagonals and got rad40 for one of the diagonals, but cannot figure out how to calculate the other)

Thanks!

Note that, $$AB=AD=6$$ units (tangents from a point to a circle are equal in length), and $$BC=DC=2$$(radius) units. Join $$AC$$. $$\triangle ACD$$ is congruent to triangle $$\triangle ABC$$ (tangents make $$90^0$$ with the radius, AB=AD, BC=DC, hence they're congruent by the RHS criteria). Hence,

total area $$= 2*1/2*CD*AD=2*6=12$$ unit square.

• This is very clear. Would there be a way of calculating the length of the other diagonal? – bagrut1 Jun 6 at 14:47
• @DaniellaLejtman You can mark it correct if it answers your question. And yes, that can also be done. Consider $\triangle BCD$ and $\triangle ABD$ and find BD. It's a bit tedious. AC is easy to find out since $\triangle ACD$ is a right-angled triangle. Now, $Area= AC.BD/2$ – Tapi Jun 6 at 14:59
• It'd actually be easier to calculate BD once you find out the area. Use the formula $A=(1/4)b\sqrt{4a^2−b^2}$ for $\triangle BCD$ and $\triangle ABD$. @DaniellaLejtman They have the same value of $b$ (which we're finding out) but different $a$ values, i.e. $2$ and $6$. Now add the areas of those two triangles and equate it with the area of $ABCD$ (which is $12$). It'd take a lot of time if you have to calculate the diagonal $BD$. – Tapi Jun 7 at 14:33

Since $$\overline{AB}$$ is tangent to the circle, $$m\angle B = 90°$$, so the quadrilateral consists of two right-triangles. Both triangles share a common hypotenuse $$\overline{AC}$$ and have a congruent leg, as both are the radius of the circle. Hence, the two right-triangles are congruent, so $$A_{kite} = 2A_{\triangle} = 2\cdot\frac{1}{2}bh = bh$$. From the diagram, $$b = 0-(-2) = 2$$ and $$h = 10-4 = 6$$, so $$A_{kite} = 2\cdot 6 = 12$$ (in square units).

Since $$AB=AD$$ and $$BC=CD$$, and the radii are perpendicular to the tangent lines, the triangles $$ABC$$ and $$ACD$$ are equal. The area of $$ACD$$ is $$(AC\cdot CD)/2$$. Then the area of quadrilateral is $$AC\cdot CD$$.

• Should be AD⋅𝐶𝐷 – Ido Sarig Jun 15 at 18:55