Question

In ΔAВС, АВ = АC = 25 cm and BC = 14 cm. Find the radius of the circle circumscribing the triangle.

Answer 1

Given: In ΔABC, AB = AC = 25 cm, BC = 14 cm.

Step-wise calculation:

1. Let M be the midpoint of BC.   ...(Draw AM ⟂ BC as in the isosceles-triangle example) 

So, BM = CM

= `14/2`

= 7 cm

2. Compute the height AM:

`AM = sqrt(AB^2 - BM^2)`

= `sqrt(25^2 - 7^2)`

= `sqrt(625 - 49)`

= `sqrt(576)`

= 24 cm

3. Area Δ = `1/2` × base BC × height AM 

= `1/2 xx 14 xx 24`

= 168 cm²

4. Use the circumradius formula

`R = (a xx b xx c)/(4Δ)`

With a = 25, b = 25, c = 14:

`R = (25 xx 25 xx 14)/(4 xx 168)` 

= `8750/672` 

= `625/48 cm` 

Equivalently, for an isosceles triangle this simplifies to

`R = s^2/(2h)`

= `25^2/(2 xx 24)`

= `625/48`

The circumradius R = `625/48` cm ≈ 13.020833... cm.