Question

Line 1 touches the circle ω at point K. Points A and B are chosen on ω in such a way that they are situated on opposite sides with respect to the diameter of ω that passes through point K. Find the area of triangle AKB if the distances from points A and B to line 1 are equal to 7 and 11 respectively, and AK=13. Round the answer to the closest integer.

Answer 1

Using the alternate segment theorem:

$\angle B_1KB = \cfrac{1}{2} \frown{KB} = \angle KAB \\ \angle A_1KA = \cfrac{1}{2}\frown{AK} = \angle KBA \\$

$\triangle KAM \sim \triangle BKB_1 \Rightarrow \cfrac{AK}{BK} = \cfrac{KM}{B_1B}\\ \triangle KBH \sim \triangle AKC \Rightarrow \cfrac{BK}{AK} = \cfrac{KM}{AA_1}$

So we have

$\cfrac{KM}{BB_1}=\cfrac{AA_1}{KM} \Rightarrow KM = \sqrt{AA_1 *BB_1}$

$KM = \sqrt{7 *11} =\sqrt{77}$

$BK = \cfrac{AK*B_1B}{KM} = \cfrac{13*11}{\sqrt{77}}=\cfrac{13\sqrt{11}}{\sqrt{7}}$

By Pythagorean Theorem:

$BM = \sqrt{KB^2 - KM^2} = \sqrt{\cfrac{169*11}{7} - 77}= \sqrt{\cfrac{1320}{7}}= 2\sqrt{\cfrac{330}{7}}$

$AM = \sqrt{169-77} = 2\sqrt{23}$

And finaly:

$S = \cfrac{\sqrt{77}}{2}(2\sqrt{23} + \cfrac{2\sqrt{330}}{\sqrt{7}}) = \sqrt{11*167} + \sqrt{11*1320} = \sqrt{1771}+\sqrt{14520} = 162$