Question

The centre of a circle inside a triangle is at distance of 625cm. From eaxh of the vertices

Answer 1

Here is how you could construct this situation: You could start by drawing the circle, with radius 3502=1753502=175 around any point. Then draw a larger circle with radius 625625 around that same center. All three corners must lie on the larger circles, and two edges must be tangents to the smaller one. So pick an arbitrary point AA on the larger circle, draw tangents to the smaller circle, and the points where these tangents intersect the larger circle are the points BB and CC.

But I take it that you want to know the number exactly, not measure some construction. So think about the image you obtained. For reasons of symmetry, AE=ECAE=EC. Since you know AFAF and you know FEFE and since ∠AEF=90°∠AEF=90°, you can compute AEAE and therefore ACAC.

AC=26252−1752−−−−−−−−−√AC=26252−1752

Furthermore, if you define α=∠FAEα=∠FAEthen you know sinα=175625=725sin⁡α=175625=725. From that you can deduce the vector from AA to CC:

AC−→−=AC(cosαsinα)=(AGGC)AC→=AC(cos⁡αsin⁡α)=(AGGC)

From that you get the area of the triangle as

2⋅12⋅AG⋅GC=AC2sinαcosα=4(6252−1752)1756251−(175625)2−−−−−−−−−−√=387072