Calculate the gray shaded area for the following figure. The figure consists of an isosceles triangle with two sides of length 5 cm, one angle of 60 degrees, and a sector of a circ... Calculate the gray shaded area for the following figure. The figure consists of an isosceles triangle with two sides of length 5 cm, one angle of 60 degrees, and a sector of a circle. Read more

Steps to Solve

1. Calculate the area of the sector

The area of a sector of a circle is given by the formula:

$A_{sector} = \frac{\theta}{360} \pi r^2$

where $\theta$ is the angle subtended at the center and $r$ is the radius. However, we are not given enough info to calculate the area using that formula, but rather we are told to use the value $462 \text{ cm}^2$ elsewhere

2. Calculate the area of the triangle

Since the triangle is isosceles with two sides of length 5 cm and the angle between them is 60 degrees, this implies that the triangle is actually equilateral, since the base angles must be equal and sum to $180 - 60 = 120$, hence must each be $60$ as well. So all angles are 60 degrees and all sides are 5 cm.

The area of an equilateral triangle is given by the formula: $A_{triangle} = \frac{\sqrt{3}}{4} a^2$ where $a$ is the side length. Substituting $a=5$ cm: $A_{triangle} = \frac{\sqrt{3}}{4} (5)^2 = \frac{\sqrt{3}}{4} (25) = \frac{25\sqrt{3}}{4} \text{ cm}^2 \approx 10.825 \text{ cm}^2 $

3. Calculate the shaded area

The shaded area is the difference between the sector area and the triangle area. So the shaded area will be $A_{shaded} = A_{sector} - A_{triangle}$ $A_{shaded} = 462 - \frac{25\sqrt{3}}{4} \text{ cm}^2 \approx 462 - 10.825 \text{ cm}^2 = 451.175 \text{ cm}^2$

Given the significant difference in magnitude between the two given values: "Area of sector = 462 cm^2" vs. side lengths of 5 cm, combined with the note "(Figure not to scale)", it seem very unlikely (and visually inconsistent) that the sector in question has area 462 $cm^2$. Thus, because some provided information seems erroneous, calculating a final answer is problematic.

Assuming for a second that the sector area can be assumed proportional to the 60 degree angle, the radius of 5 cm means the total area would be

$\pi r^2 = \pi (5^2) = 25\pi \approx 78.54 \text{ cm}^2$, and so the area the 60 degree sector would be $\frac{60}{360} \times 25\pi = \frac{1}{6} \times 25\pi = \frac{25\pi}{6} \approx 13.09 \text{ cm}^2$.

In this case, the area of the shaded region is $13.09 - 10.825 = 2.265 \text{ cm}^2$.