Question
There is a circle with a centre T and a radius of 8 cm.
Two tangents KE and KI are drawn from point K, which is 17 cm from the centre. The area of the quadrilateral TEKI is:Solution
In the right-angled triangle TEK: TE = Radius = 8 cm KT = Hypotenuse = 17 cm Using Pythagoras theorem: ⇒ KE2 = KT2 - TE2 ⇒ KE2 = 172 - 82 ⇒ KE2 = 289 - 64 ⇒ KE2 = 225 ⇒ KE = √225 ⇒ KE = 15 cm Area of Triangle TEK: ⇒ Area = (1/2) × TE × KE ⇒ Area = (1/2) × 8 × 15 ⇒ Area = 60 sq.cm Since TEKI is symmetrical: Area of Quadrilateral TEKI = 2 × Area of Triangle TEK ⇒ Area of TEKI = 2 × 60 ⇒ Area of TEKI = 120 sq.cm The area of the quadrilateral TEKI is 120 sq.cm.
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