Question

Find the area of the shaded portion intersecting between the two circles. Show all work for full credit.

Answer 1

Solution:

Join other end of chord from the center of two circles.

As these two circles are concentric circles .

And the triangles are equilateral triangle.

Area of sector

= $\frac{\pi r^2 \times (\alpha )}{360 degrees}\\\\ \frac{\pi \times 16 \times 60}{360}\\\\ \frac{8 \pi}{3}$

Area of equilateral triangle having side 4 cm=$\frac{\sqrt3}{4}\times(side)^2$ = $\frac{\sqrt3}{4} \times {16}= 4 \sqrt3$

Area of region I = Area of sector - Area of equilateral triangle having side 4 cm

                          = $\frac{8\pi }{3} - 4 \sqrt 3=8.389 - 6.928= 1.461$ cm²

Area of colored region = 2 × Area of region I

                                       = 2 × 1.461

                                        = 2.922 cm² (Approx)

Answer 2

So let's try to find the part of it in one circle then multiply it by 2.
if the vertical line of shaded reason is 4, then 2 would form a rt triangle with radius.
thus the cos € = 2/4 = 1/2, so € = arccos (1/2)
€ = 60°, and 2× that = 120°
so now one part of the shaded region is the whole sector (pi×r^2) - the triangle within (base×ht):
[pi(4)^2×120/360] - [2(4cos30)]
= [16pi×1/3] - [2×3.46] = 16.76 - 6.93 = 9.83
now double that for both parts of the shaded region: 9.83×2 = 19.65 sq. units