Problem 6 - Entrance Test

To find the radius of the inscribed circle, we need to use the formula for the area of a triangle and the formula for the radius of the inscribed circle. The area of the triangle can be found using Heron's formula, which is given by A = √(s(s - AB)(s - BC)(s - AC)), where s is the semi-perimeter of the triangle. The semi-perimeter is given by s = (AB + BC + AC)/2 = (5 + 7 + 9)/2 = 10.5. Plugging this value into Heron's formula, we get A = √(10.5(10.5 - 5)(10.5 - 7)(10.5 - 9)) = √(10.5 * 5.5 * 3.5 * 1.5) = √(10.5 * 28.875) = √(303.1875). The area of the triangle is also equal to rs, where r is the radius of the inscribed circle and s is the semi-perimeter. Setting these two expressions for the area equal to each other, we get rs = √(303.1875), so r = √(303.1875)/10.5 = √(28.875/10.5) = √(2.75) = 1.6583, which is close to the value given by the formula r = A/s, and A is given by the formula for area using the height of the triangle. Using height we can calculate it or apply the formula r = (AB + BC - AC)/2 = (5 + 7 - 9)/2, or r = (7 + 9 - 5)/2, or r = (9 + 5 - 7)/2. Since only r = (9 + 5 - 7)/2 = 7/2 = 3.5 is the only equation that satisfies the formula for area of triangle using the inradius and semiperimeter, this option should be considered as possible answer for this problem.