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Question:

In the figure given below, find the distance PQ.

Explanation:

BC² = AB² + AC² = 15² + 20² = 625

BC = 25

Let BD = x, so DC = 25 − x

In ΔABD,

BD² + AD² = AB²

x² + AD² = 225 ... (i)

In ΔADC,

AD² + (25 − x)² = 400

∴ AD² + 625 − 50x + x² = 400

∴ (AD² + x²) + 625 − 50x = 400 ... (ii)

Substituting (i) in (ii),

225 + 625 − 50x = 400

∴ x = 9

BD = 9 and DC = 16

We use the formula A = r × S, where A is the area of a triangle, r is the inradius and S is the semi-perimeter to find the radii of the two circles.

The radius of the circle inscribed in ∆ABD = 3 m

The radius of the circle inscribed in ∆ADC = 4 m

∴ PQ = radius of ∆ABC + radius of ∆ADC = 7 m

Hence, option (a).