Question

Find the area of the shaded portion.
6
120

Answer 1

A = 36√3 - 12π

Explanation:

Make 2 identical 30-60-90 triangles out of the "wedge"

For each 30-60-90 triangle:

Since the short leg is 6 the long leg is 6√3

short leg is the base and long leg is the height

a = (1/2)bh

a = (1/2)(6)(6√3)

a = 3(6√3)

a = 18√3

Since there are 2 triangles

a = 36√3

--------------------------------

Then subtract off the area of the 120º sector

a = (120/360)πr^2

a = (1/3)π(6^2)

a = (1/3)π(36)

a = 12π

Shaded area

a = 36√3 - 12π

Answer 2

Explanation:

we have the areas of 2 right-angled triangles minus 1/3 the area of the circle (because the 120 degree segment of the circle is 1/3 of the whole 360 degree circle).

since the answer is asking for a square root, we need to aim for the result of a Pythagoras calculation.

basically, the area of a triangle is

1/2 × b×c × sin(A)

A being the angle between the 2 sides b and c. for any 2 sides of the triangle.

one triangle is from the outmost left point to the center of the circle and up to the 90 degree angle. the other triangle is then down to the second right angle.

we know all angles in such a triangle : 90, 120/2 = 60, 180-90-60 = 30.

remember, the sum of all angles in a triangle is 180 degrees.

so, via the law of sines we can then calculate all sides :

a/sin(A) = b/sin(B) = c/sin(C)

with the angles being opposite of the sides.

6/sin(30) = baseline/sin(90)

6/0.5 = baseline/1

12 = baseline

top side² + 6² = 12²

top side² = 12² - 6² = 144 - 36 = 108

top side = sqrt(108) = sqrt(36×3) = 6×sqrt(3)

we pick then top side and 6 as the 2 sides, so that we can use sin(90)=1 in the formula.

the area of the 2 triangles is then

6×sqrt(3)×6×sin(90) = 36×sqrt(3)

and the circle segment to be deducted is

pi×r²/3 = pi×6²/3 = pi×36/3 = pi×12

so, the complete answer is

A = 36×sqrt(3) - 12×pi