How much area in square centimeters is blue in Nacho's trademark of an isosceles triangle inscribed in a circle with a radius of 5 cm?

How much area in square centimeters is blue in Nacho's trademark of an isosceles triangle inscribed in a circle with a radius of 5 cm?

• A. 50.0 cm²
• B. 53.5 cm²
• C. 55.0 cm²
• D. 60.0 cm²

Answer: (B)

To determine the area of the blue section in Nacho's trademark, we first need to establish the total area of the isosceles triangle inscribed in a circle with a radius of 5 cm. The first step is to find the area of the circle, which can be calculated using the formula for the area of a circle, \( A = \pi r^2 \). Since the radius \( r \) is 5 cm, the area of the circle becomes: \[ A = \pi (5)^2 = 25\pi \text{ cm}^2 \approx 78.5 \text{ cm}^2. \] Next, we need to calculate the area of the isosceles triangle inscribed within this circle. For an isosceles triangle inscribed in a circle, we can use the formula: \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height}. \] In the case of an isosceles triangle inscribed in a circle, its height can be determined based on the geometry of the situation. The height extends from the center of the circle perpendicular to the base.

To determine the area of the blue section in Nacho's trademark, we first need to establish the total area of the isosceles triangle inscribed in a circle with a radius of 5 cm.

The first step is to find the area of the circle, which can be calculated using the formula for the area of a circle, ( A = \pi r^2 ). Since the radius ( r ) is 5 cm, the area of the circle becomes:

[

A = \pi (5)^2 = 25\pi \text{ cm}^2 \approx 78.5 \text{ cm}^2.

]

Next, we need to calculate the area of the isosceles triangle inscribed within this circle. For an isosceles triangle inscribed in a circle, we can use the formula:

[

\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}.

]

In the case of an isosceles triangle inscribed in a circle, its height can be determined based on the geometry of the situation. The height extends from the center of the circle perpendicular to the base.