Question

Line l 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 l are equal to 7 and 11 respectively, and AK=13. Round the answer to the closest integer.

Answer 1

Let $F\in l, AF\perp l; T\in l, BT\perp l; N\in BT, AN\perp BT;$ O is the center of the circle ω.

$AF=7, AK=13, BT=11.$

$\triangle AKF: KF=\sqrt{AK^2-AF^2}=\sqrt{13^2-7^2}=\sqrt{120}\approx 10.95.$

$\angle AKF=arcsin\frac{AF}{AK}=arcsin\frac{7}{13}.$

$\triangle AOK: \angle AOK=2\angle AKF=2arcsin\frac{7}{13}.$

$OB=OA=OK=\frac{AK}{2sin\frac{\angle AOK}{2}}=\frac{13}{2\cdot \frac{7}{13}}=\frac{169}{14}.$

$\triangle KBT: \angle BKT=arcsin\frac{BT}{BK}=arcsin\frac{11}{BK}.$

$\triangle OBK: \angle BOK=2\angle BKT=2arcsin\frac{11}{BK}.$

$BK=2OKsin\frac{\angle BOK}{2}=2\cdot \frac{ 169}{14}\cdot \frac{ 11}{BK},$

$BK=\sqrt{\frac{169\cdot11}{7}}\approx16.30.$

$\triangle KBT: TK=\sqrt{BK^2-BT^2}=\sqrt{16.3^2-11^2}\approx 12.03.$

$AN=TF=TK+KF=12.03+10.95=22.98.$

$AF=NT=7, BN=BT-NT=11-7=4.$

$\triangle ANB: AB=\sqrt{AN^2+BN^2}=\sqrt{22.98^2+4^2}\approx 23.33.$

And by the Heron's formula, $S=102.35\approx 102.$