Question

angle d is a circumscribed angle of circle o. what is the perimeter of kite obde? 17 units 23 units 27 units 60 units

Explanation:

Step1: Recall tangent - secant properties

Tangents drawn from an external point to a circle are equal. So, $BD = ED = 5$ and $OB=OE$ (radii of the circle). In right - triangle $OBC$, using the Pythagorean theorem, if $OC = 15$ and $BC = 8$, then $OB=\sqrt{15^{2}-8^{2}}=\sqrt{225 - 64}=\sqrt{161}
eq17,23,27,60$. But we can also use the fact that for a kite $OBDE$, the perimeter $P=2(BD + OB)$. Since tangents from point $D$ to the circle are equal, we know that if we consider the right - triangle formed by the radius and the tangent, and assume the radius is $r$. In right - triangle $OBC$, $r=\sqrt{15^{2}-8^{2}} = 17$. The perimeter of the kite $OBDE$ is $P = 2(5+17)=2\times12 = 27$ (units).

Step2: Calculate the perimeter

The perimeter of a kite with two adjacent side - lengths $a$ and $b$ is $P = 2(a + b)$. Here $a = 5$ and $b = 17$, so $P=2(5 + 17)=27$ units.

Answer:

27 units