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1.

From the given figure, prove that:AP + BQ + CR = BP + CQ + ARAlso show that:AP + BQ + CR = (1/2) × Perimeter of ΔABC.

Answer»

Since from B, BQ and BP are the tangents to the circle

Therefore, BQ = BP ………..(i)

Similarly, we can prove that

AP = AR …………..(ii)

and CR = CQ ………(iii)

Adding,

AP + BQ + CR = BP + CQ + AR ………(iv)

Adding AP + BQ + CR to both sides

2(AP + BQ + CR) = AP + PQ + CQ + QB + AR + CR

2(AP + BQ + CR) = AB + BC + CA

Therefore, AP + BQ + CR = (1/2) × (AB + BC + CA)

AP + BQ + CR = (1/2) × perimeter of triangle ABC
Discussion
2.

In the following figure, a circle is inscribed in the quadrilateral ABCD.If BC = 38 cm, QB = 27 cm, DC = 25 cm and that AD is perpendicular to DC, find the radius of the circle.

Answer»

From the figure we see that BQ = BR = 27 cm (since length of the tangent segments from an external point are equal)

As BC = 38 cm

CR = CB − BR = 38 − 27

= 11 cm

Again,

CR = CS = 11cm (length of tangent segments from an external point are equal)

Now, as DC = 25 cm

∴ DS = DC − SC

= 25 − 11

= 14 cm

Now, in quadrilateral DSOP,

∠PDS = 90° (given)

∠OSD = 90°, ∠OPD = 90° (since tangent is perpendicular to the radius through the point of contact)

DSOP is a parallelogram

OP ∥ SD and  PD ∥ OS

Now, as OP = OS (radii of the same circle)

OPDS is a square. ∴ DS = OP = 14cm

∴ radius of the circle = 14 cm
Discussion