Question

CRITICAL THINKING Find the area between the three congruent tangent circles. The radius of each circle is 6 inches. Round your answer to the nearest hundredth. Three congruent circles touch each other. The centers are marked as points A, B, and C. Equilateral triangle A B C is formed and the region inside triangle, but not in a circle, is shaded. The area is about a square inches.

Answer 1

Answer:

$Area = 5.84in^2$

Step-by-step explanation:

Given

$Circles = 3$

$r = 6in$ --- The radius of each

Required

The area between them

See attachment for illustration of the question. (figure 1)

First, calculate the height of the equilateral triangle formed by the 3 radii (See figure 2)

Using Pythagoras theorem, we have:

$12^2 = h^2 + 6^2$

$144 = h^2 + 36$

Collect like terms

$h^2 = 144 - 36$

$h^2 = 108$

Take square roots

$h = \sqrt{108$

Expand

$h = \sqrt{36 * 3$

Split

$h = \sqrt{36} * \sqrt{3$

$h = 6\sqrt{3$

Now, the area of the equilateral triangle can be calculated using:

$A = \frac{1}{2}bh$

Where

$h = 6\sqrt{3$

$b = 2r = 2 * 6 = 12$

$A = \frac{1}{2} * 12 * 6\sqrt 3$

$A = 6 * 6\sqrt 3$

$A_1 = 36\sqrt 3$

Next, is to calculate the area of the sector formed by 2 radii in each circle (figure 3).

Since the radii formed an equilateral triangle, then the central angle will be 60. So:

$A = \frac{\theta}{360} * \pi r^2$

$A = \frac{60}{360} * \pi 6^2$

$A = \frac{60}{360} * \pi * 36$

$A = \frac{60}{10} * \pi$

$A = 6\pi$

For the three circles, the area is:

$A_2 = 3 * 6\pi$

$A_2 = 18\pi$

Subtract the areas of the sectors (A2) from the area of the equilateral triangle (A1), to get the area between them.

$A = A_1 - A_2$

$A = 36\sqrt 3 - 18\pi$

$A = 36*1.7321 - 18 * 3.14$

$A = 5.8356$

Approximate

$A = 5.84in^2$