A simple geometry problem

[MathBuster Forum]

ABCD is a rectangle inscribed in a quarter-circle as shown. A is on BF, B is the centre of the quarter-circle, C is on BE, and D is on arc FE.

If AD = 12 cm and CE = 1 cm, determine the length of AF.

[Please explain the solution, instead of just giving the answer.]

[Santhosh Reddy]

Given

AD=12 cm,CE=1 cm

BC=12 cm (AD and BC are opp.sides of Rectangle)

But, BE = BC + CE

            = 12 + 1 = 13 cm

And BE = BD = BF = 13 cm (Radii of circle)

By pythagoras theorem,

                 BD² = AB² + AD²

13² = AB² + 12²

AB=5 cm

AF = BF – AB= 13 – 5 = 8cm

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