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: β KE2Β = KT2Β - TE2 β KE2Β = 172Β - 82 β KE2Β = 289 - 64 β KE2Β = 225 β KE = β225 β KE = 15 cm Area of Triangle TEK: β Area = (1/2) Γ TE Γ KE β Area = (1/2) Γ 8 Γ 15 β Area = 60 sq.cm Since TEKI is symmetrical: Area of Quadrilateral TEKI = 2 Γ Area of Triangle TEK β Area of TEKI = 2 Γ 60 β Area of TEKI = 120 sq.cm The area of the quadrilateral TEKI is 120 sq.cm.
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