# Solve question 3 on the attached worksheet. Explain your working.

Images: We often use degrees to measure angles, where a degree is 1/360th of the angular distance around a circle. (So 90 degrees is 1/4 of the angular measure around the circle.)

Radians are another way to measure angles. Here we are using the arc length subtended by the angle. If the radius of the circle is 1 (the unit circle), then the radian measure of an angle is the arc length subtended by that angle. Since the circumference of a circle is `2pi ` units, the radian measure that corresponds to 360 degrees is `2pi ` . Using proportionality we have `180^@=pi, 90^@=pi/2, 60^@=pi/3, 45^@=pi/4, 30^@=pi/6 ` etc... where we drop the word radians as usual.

If s is the arc length and k the area of a sector, we have `s=d/360(2pi*r)=r theta ` where d is the degree measure of the angle, `theta ` is the radian measure, and r the radius of the circle; and `k=d/360(pi r^2)=1/2 r^2 theta `

(The formulas in radians are much simpler than the ones involving degrees.)

For problem 3, the central angle is `120^@=(2pi)/3 "rad" ` . (We have an isosceles triangle with congruent sides of length 1, and altitude of length 1/2 as it is 1/2 of the distance between the two centers.)

Thus the area of the sector (radius=1) is `k=1/2(1)^2((2pi)/3)=pi/3 `

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The area of the segment is the area of the sector minus the area of the triangle: the base of the triangle is `sqrt(3) ` (we have two 30-60-90 right triangles with hypotenuse 1, legs `1/2,sqrt(3)/2 ` : see attachment) and height 1/2 so the area of the triangle is `sqrt(3)/4 ` . The area of the segment is then `pi/3-sqrt(3)/4 ` .

By symmetry, the shared area is twice the area of the segment or `2(pi/3-sqrt(3)/4) ` . The area of one circle is `A=pi r^2=pi ` . The proportion of the area of one triangle shared by the two circles is `(2(pi/3-sqrt(3)/4))/pi~~.3910 ` as expected.

(The area shared is `2(pi/3-sqrt(3)/4)~~1.2284 ` )

Images:
Approved by eNotes Editorial Team There are two questions here. I am answering the first of the questions (number 3 on the worksheet) as the policy is one question per post.

The length AB is equal to the radius of the circles (which have the same radius), as are the lengths AC and AD, BC and BD, so we have

AB = AC = AD = BC = BD = 1

This implies that both the triangles ABC and ABD are equilateral triangles with equal lengths of sides 1 and equal angles of 180/3 = 60 degrees.

Angle CAD is equal to angles CAB + BAD = 120 degrees. The area of the circle segment from C to D is then a third of the area of the circle centered at A, since CAD = 120 = 360/3 (a third of the full 360 degrees of the circle).

The area of the circle centered at A is given by the formula for the area of a circle

Area =  `pi r^2 `

where r is the radius. Here we have `r=1 `, so that the area is...

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equal to `pi ` . The area of the circle segment from C to D (call this area S) is hence a third of this. So we have that the area `S = pi/3 `.

The shaded area we are required to calculate is the area of the circle segment (S) minus the area of the triangle ACD (call this T). Note that the area of the triangle ACD is equal to that of the triangles ABC and ABD since it is made up of half of each of those and they have the same area. Those triangles are equilateral with sides equal to 1 so that their area is half base x height = `0.5 times sqrt(1^2 - 0.5^2) = 0.5 times sqrt(3)/2 = sqrt(3)/4 `

Therefore the area of the triangle ACD is  `T = sqrt(3)/4 ` and the shaded area we are required to calculate is given by

`S - T = pi/3 - sqrt(3)/4 `

Finally the area common to both circles is equal to twice the shaded area, so is equal to 2(S - T). If this is indeed 39% of the area of either circle, this being 0.39pi, it will be approximately equal to 1.23. This is the case since

` `` 2(S-T) = 2 (pi/3 - sqrt(3)/4) = 2 times 0.6141848 approx 1.23 approx 0.39pi `

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