Question #142444
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.
1
Expert's answer
2020-11-16T19:14:47-0500


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.


DKB=12KB=KAB\angle DKB=\dfrac{1}{2}\stackrel \frown {KB}=\angle KAB

CKA=12KA=KBA\angle CKA=\dfrac{1}{2}\stackrel \frown {KA}=\angle KBA

KAHBKD=>AKBK=KHBD\triangle KAH\sim \triangle BKD=>\dfrac{AK}{BK}=\dfrac{KH}{BD}

KBHAKC=>BKAK=KHAC\triangle KBH\sim \triangle AKC=>\dfrac{BK}{AK}=\dfrac{KH}{AC}

KHBD=ACKH=>KH=ACBD\dfrac{KH}{BD}=\dfrac{AC}{KH}=>KH=\sqrt{AC\cdot BD}


KH=69=36KH=\sqrt{6\cdot 9}=3\sqrt{6}

BK=AKKHAC=10366=56BK=\dfrac{AK\cdot KH}{AC}=\dfrac{10\cdot 3\sqrt{6}}{6}=5\sqrt{6}

By the Pythagorean Theorem


BH=BK2KH2=(56)2(36)2=46BH=\sqrt{BK^2-KH^2}=\sqrt{(5\sqrt{6})^2-(3\sqrt{6})^2}=4\sqrt{6}

AH=AK2KH2=(10)2(36)2=46AH=\sqrt{AK^2-KH^2}=\sqrt{(10)^2-(3\sqrt{6})^2}=\sqrt{46}


SAKB=12(AH+BH)KH=S_{\triangle AKB}=\dfrac{1}{2}(AH+BH)KH=

=12(46+46)36=69+3644=\dfrac{1}{2}(\sqrt{46}+4\sqrt{6})3\sqrt{6}=\sqrt{69}+36\approx44

The area of triangle AKB is 44 square units.



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