A circle of diameter $8$ inches is inscribed in a triangle ABC where $\angle A B C=90^{\circ}$. If $B C=10$ inches then the area of the triangle in square inches is

We have a right triangle $ABC$ with:

Right angle at $B$ ($\angle ABC = 90°$)

An inscribed circle with diameter $8$ inches (so radius $= 4$ inches)

$BC = 10$ inches

We need to find the area of triangle $ABC$

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Before using complex formulas, let's check if this is a special right triangle using Pythagorean triplets.

The most common triplets with $10$ are:

$6-8-10$ (scaled version of $3-4-5$)

$10-24-26$ (scaled version of $5-12-13$)

For a right triangle with an inscribed circle:

$\text{Inradius} = \frac{\text{leg}_1 + \text{leg}_2 - \text{hypotenuse}}{2}$

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Let's test the $10-24-26$ triplet:

$\text{leg}_1 = 10$, $\text{leg}_2 = 24$, $\text{hypotenuse} = 26$

$\text{Inradius} = \frac{10 + 24 - 26}{2} = \frac{8}{2} = 4$

This matches our given radius of $4$ inches!

So our triangle has sides: $10, 24, 26$

$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \times 24 = 120$ square inches

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Alternative method if you don't remember the triplets:

Let $AB = h$ (unknown leg)

$AC = \sqrt{h^2 + 100}$ (hypotenuse using Pythagorean theorem)

Using the inradius formula for right triangles:

$4 = \frac{10 + h - \sqrt{h^2 + 100}}{2}$

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Multiply both sides by $2$:

$8 = 10 + h - \sqrt{h^2 + 100}$

Rearrange:

$\sqrt{h^2 + 100} = 10 + h - 8 = h + 2$

Square both sides:

$h^2 + 100 = (h + 2)^2$

$h^2 + 100 = h^2 + 4h + 4$

$100 = 4h + 4$

$96 = 4h$

$h = 24$ inches

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$\text{Area} = \frac{1}{2} \times 10 \times 24 = 120$ square inches

Therefore, the area of triangle $ABC$ is $120$ square inches.