Answer to Question #142444 in Geometry for Peace College

Question

Line l touches the circle w at point K. Points A and B are chosen on w in such a way that they are
situated on opposite sides with respect to the diameter of w that passes through point K. Find the area of triangle AKB if the distances from points A and B to line l are equal to 6 and
9 respectively, and AK = 10. Round the answer to the closest integer.

Answer 1

The alternate segment theorem states that an angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment.

$$\angle DKB=\dfrac{1}{2}\stackrel \frown {KB}=\angle KAB$$

$$\angle CKA=\dfrac{1}{2}\stackrel \frown {KA}=\angle KBA$$

$$\triangle KAH\sim \triangle BKD=>\dfrac{AK}{BK}=\dfrac{KH}{BD}$$

$$\triangle KBH\sim \triangle AKC=>\dfrac{BK}{AK}=\dfrac{KH}{AC}$$

$$\dfrac{KH}{BD}=\dfrac{AC}{KH}=>KH=\sqrt{AC\cdot BD}$$

$$KH=\sqrt{6\cdot 9}=3\sqrt{6}$$

$$BK=\dfrac{AK\cdot KH}{AC}=\dfrac{10\cdot 3\sqrt{6}}{6}=5\sqrt{6}$$

By the Pythagorean Theorem

$$BH=\sqrt{BK^2-KH^2}=\sqrt{(5\sqrt{6})^2-(3\sqrt{6})^2}=4\sqrt{6}$$

$AH=\sqrt{AK^2-KH^2}=\sqrt{(10)^2-(3\sqrt{6})^2}=\sqrt{46}$

$$S_{\triangle AKB}=\dfrac{1}{2}(AH+BH)KH=$$

$$=\dfrac{1}{2}(\sqrt{46}+4\sqrt{6})3\sqrt{6}=\sqrt{69}+36\approx44$$

The area of triangle AKB is 44 square units.