The centre of a circle inside a triangle is at a distance of 625 cm. from each of the vertices of the triangle. If the diameter of the circle is 350 cm. and the circle is touching only two sides of the triangle, find the area of the triangle.

If a point is equidistant from all 3 vertices, it has to be the circumcentre. The given circle with centre S is concentric and touches two sides.

As S is equidistant from 2 of the sides (say AB and AC), => It lies on angle bisector of $$\angle A$$.

=> $$\triangle ABC$$ is isosceles with AB = AC

Radius of the circle = RS = SQ = 175 cm and SA = SB = SC = 625 cm

=> $$AR = \sqrt{625^2 - 175^2} = 600$$

Let SP = x

=> $$(BP)^2 = (BA)^2 - (AP)^2 = (BS)^2 - (SP)^2$$

=> $$1200^2 - (625 + x)^2 = 625^2 - x^2$$

=>$$1200^2 - 625^2 - x^2 - 2*625x = 625^2 - x^2$$

=> $$1200^2 - 2 * 625^2 = 1250x$$

=> $$x = \frac{658750}{1250} = 527$$

=> $$BP = \sqrt{625^2 - 527^2} = 336$$

$$\therefore$$ ar $$(\triangle ABC)$$ = $$\triangle ASB + \triangle ASC + \triangle SBC$$

= $$(600 \times 175) + (600 \times 175) + (527 \times 336)$$

= $$105000 + 105000 + 177072 = 387072$$