Question

A circle of diameter 8 inches is inscribed in a triangle ABC where ∠ABC = 90°. If BC = 10 inches then the area of the triangle in square inches is

Answer 1

Correct Answer: 120

Given that triangle ABC is a right triangle with ∠ABC = 90° and BC = 10 inches, and there's an inscribed circle with diameter 8 inches.

The radius of the circle, \( r \), is \( \frac{8}{2} = 4 \) inches. 

Now, for a right triangle with an inscribed circle, the radius of the inscribed circle can be represented in terms of the triangle's legs (the two shorter sides) as follows: 

\( r \) = \( \frac{a + b - c}{2} \) 

Where: 

- \( a \) and \( b \) are the lengths of the two legs of the triangle 

- \( c \) is the hypotenuse (BC in this case, which is 10 inches) 

Given that \( r = 4 \) inches and \( c = 10 \) inches, we can set up the following equation: 

\( 4 \) = \( \frac{a + b - 10}{2} \)

 From which: 

\( a + b = 18 \) ... equation (1) 

Now, we know that the sum of the lengths of the two legs of the right triangle is equal to the diameter of the inscribed circle. Therefore, \( a + b = 8 \) inches. But this conflicts with equation (1), so let's reconsider. 

The tangent drawn from a point to a circle is perpendicular to the radius through the point of contact. Given the circle is inscribed inside the triangle, the two legs of the triangle can be considered tangents to the circle. 

Thus, the length from the right angle to the circle along leg \( a \) would be \( r \), and the length from the right angle to the circle along leg \( b \) would also be \( r \). This would divide each of the legs of the triangle into two parts. 

Let \( a \) be the side that's adjacent to side \( c \) and \( b \) be the side that's opposite side \( c \). Then, \( a \) can be split into two lengths: 4 inches and \( a - 4 \) inches. Similarly, \( b \) can be split into 4 inches and \( b - 4 \) inches. 

Using the Pythagorean theorem: 

\( a^2 + b^2 = c^2 \) 

Substituting the known value of \( c \) (10 inches): \( a^2 + b^2 = 100 \) ... equation (2) 

Also, given the geometry of the inscribed circle and the right triangle, the two smaller triangles formed are also similar to triangle ABC. Using their properties and the relationships of the sides: 

\( a = 2r + b - 4 \) and \( b = 2r + a - 4 \) 

Substituting the value of \( r \) (4 inches) in the above equations: \( a = b + 4 \) ... equation (3) 

Using equations (2) and (3): 

\( (b + 4)^2 + b^2 = 100 \) 

Expanding and simplifying: 

\( 2b^2 + 8b - 84 = 0 \) 

Solving this quadratic equation for \( b \) gives: 

\( b = 6 \) or \( b = -7 \) 

The negative value doesn't make sense in this context, so we discard it. 

If \( b = 6 \), then \( a = b + 4 = 10 \). 

Now, the area of triangle ABC is: 

\( \frac{1}{2} \times a \times b = \frac{1}{2} \times 10 \times 6 = 30 square inches.