A circle is inscribed in a rhombus with sides of length 4 cm. If the two acute angles in the rhombus each measure $90^{\circ}$, what is the length of the circle's radius? Express your answer in simplest radical form.

bader Mar 30, 2024

#1**0 **

Properties of the Rhombus:

Since the given rhombus has all sides equal (4 cm) and two acute angles of 90 degrees each, it's essentially a square.

Diagonals of the Square:

The diagonals of a square bisect each other at right angles and divide the square into four congruent right triangles. Each diagonal has a length equal to the square root of twice the side length squared (sqrt(2s^2)).

In this case, the side length (s) is 4 cm. Therefore, the length of each diagonal (d) is:

d = sqrt(2 * s^2)

d = sqrt(2 * 4^2 cm^2)

d = sqrt(32) cm

d = 4√2 cm (since the square root of 32 can be simplified to 4 times the square root of 2)

Radius of the Inscribed Circle:

The inscribed circle in a square touches the midpoint of each side. Since the diagonals bisect each other at the center of the square and are perpendicular, they also pass through the center of the inscribed circle.

Therefore, the radius (r) of the inscribed circle is half the length of a diagonal divided by two (since there are two diagonals and the circle touches them both at their midpoints):

r = (d / 2) / 2

r = (4√2 cm) / 2 / 2

r = √2 cm

Answer:

The radius of the inscribed circle is √2 centimeters.

booboo44 Mar 30, 2024