Area of △ P R G \triangle PRG

Geometry Level pending

ABC is an equilateral triangle with side = 2 3 2\sqrt3 cm.

A circular arc with radius equal to the inradius of △ A B C \triangle ABC is drawn with centre at A that cuts the sides, AB and AC at points P and Q, respectively.

Similarly, another circular arc with radius equal to the circumradius of △ A B C \triangle ABC is drawn with centre at C that cuts sides, CA and CB at points R and S, respectively.

Find the area of △ P R G \triangle PRG , where G is the centroid of △ A B C \triangle ABC .

3 3 − 4 2 \frac{3\sqrt3 - 4}{2} 3 − 1 2 \frac{\sqrt3 - 1}{2} 2 3 − 1 2 \frac{2\sqrt3 - 1}{2} 3 3 − 3 2 \frac{3\sqrt3 - 3}{2}

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1 solution

Inradius of △ A B C \triangle ABC = 2 3 2 3 \frac{2\sqrt3}{2\sqrt3} = 1 cm

Circumradius of △ A B C \triangle ABC = 2 3 3 \frac{2\sqrt3}{\sqrt3} = 2 cm

Join CG and extend it to intersect the side AB at D.

Area of △ A D C \triangle ADC = 1 2 \frac{1}{2} x △ A B C \triangle ABC = 1 2 \frac{1}{2} x 3 4 \frac{\sqrt3}{4} x ( 2 3 ) 2 (2\sqrt3)^{2} = 3 3 2 \frac{3\sqrt3}{2} c m 2 cm^{2} ... (a)

As AP = 1 cm, PD = 3 − 1 \sqrt3 - 1 cm

As CR = 2 cm, AR = 2 3 − 2 2\sqrt3 - 2 cm = 2 ( 3 − 1 ) 2(\sqrt3 - 1) cm

Now, Area of △ A P R \triangle APR = 1 2 \frac{1}{2} x AP x AR x sin ⁡ 6 0 o \sin60^{o} = 3 ( 3 − 1 ) 2 \frac{\sqrt3(\sqrt3 - 1)}{2} c m 2 cm^{2} ... (b)

Area of △ C R G \triangle CRG = 1 2 \frac{1}{2} x CR x CG x sin ⁡ 3 0 o \sin30^{o} = 1 c m 2 cm^{2} ... (c)

Area of △ P D G \triangle PDG = 1 2 \frac{1}{2} x PD x DG = 3 − 1 2 \frac{\sqrt3 - 1}{2} c m 2 cm^{2} ... (d)

Therefore, Area of △ P R G \triangle PRG = (a) - (b) - (c) - (d)

= 3 3 2 \frac{3\sqrt3}{2} - 3 ( 3 − 1 ) 2 \frac{\sqrt3(\sqrt3 - 1)}{2} - 1 - 3 − 1 2 \frac{\sqrt3 - 1}{2}

= 3 3 − 4 2 \frac{3\sqrt3 - 4}{2} c m 2 cm^{2}

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