A right(-angled) triangle with a horizontal and a vertical side. The vertical side and the hypotenuse are labeled/labelled three and five, respectively. The horizontal side has a semicircle that is tangent to the hypotenuse.

A semicircle is inscribed in a right(-angled) triangle.

What is the exact area of the semicircle?

Solution

The right(-angled) triangle is a Pythagorean triple: (3, 4, 5).

The semicircle is tangent to the hypotenuse of the right(-angled) triangle and secant to its base.

Then, the tangent line segments have the same length, 3.

Let r be the radius of the semicircle. Then, we can draw r tangent to the hypotenuse:

A right/right-angled triangle with a horizontal and a vertical side. The horizontal side and the vertical side are labeled/labelled four and three, respectively. The horizontal side has a semicircle that is tangent to the hypotenuse. The radius of the semicircle is perpendicular to the hypotenuse and it is labeled/labelled r. The hypotenuse at the left of the radius is labeled/labelled two and at the right of the radius is labeled/labelled three.

As the length of the base of the triangle is 4, we have:

A right(-angled) triangle with a horizontal and a vertical side. The vertical side is labeled/labelled three. A double arrow representing the length of the horizontal side is labeled/labelled four. The horizontal side has a semicircle that is tangent to the hypotenuse. The radius of the semicircle is perpendicular to the hypotenuse and it is labeled/labelled r. The hypotenuse at the left of the radius is labeled/labelled two and at the right of the radius is labeled/labelled three. A double arrow representing the distance from the left vertex to the semicircle is labeled/labelled four minus two r.

Using the tangent-secant theorem,

2^2=4 \left(4-2r \right)

4=4 \left(4-2r \right)

1=4-2r

2r=4-1

2r=3

r=\frac{3}{2}

So, the area of the semicircle is

\frac{\pi \times \left( \frac{3}{2} \right)^2}{2}

\frac{\frac{9\pi}{4}}{2}

\frac{9\pi}{8}

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