find area of kite given side length

Question

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!

Answer 1

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.

Answer 2

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 $AC\cdot CD$.

Answer 3

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).