Expert answers to "Either of two unit circles passes through the..."

Kamari Simon

Answered question

2022-02-10

Either of two unit circles passes through the center of the other. How do you prove that the common area is $\frac{2}{3} π - \frac{\sqrt{3}}{2}$ areal units?

Answer & Explanation

meldElafellrbo

Beginner2022-02-11Added 13 answers

Step 1
The common area is enclosed by unit circle arcs subtending

∠ 120^{\circ} = \frac{2}{3} π

radian, at the respective centers.
For this arc, sector area of a unit circle

= \frac{\frac{2}{3} π}{2 π}

(area of unit circle)

= \frac{\frac{2}{3 π}}{2 π} π = \frac{π}{3}

One half of the common area
This sector area less area of the inner triangle, of sides

[1, \sqrt{3}, 1]

= \frac{π}{3} - (\frac{1}{2}) (\sqrt{3}) (\frac{1}{2})

Twice this is the common area

= \frac{2}{3} π - \frac{\sqrt{3}}{2}

areal units.

deliria1t4

Beginner2022-02-12Added 13 answers

Step 1

1) Equilateral triangle area
$A_{e} = \sqrt{\frac{3}{4}} \times 1^{2} = \frac{\sqrt{3}}{4}$
2) yellow area
$A_{y} = \frac{π}{6} - \frac{\sqrt{3}}{4} = \frac{2 π - 3 \sqrt{3}}{12}$
3) Common area
$A_{c} = 2 A_{e} + 4 A_{y}$
$A_{c} = 2 \times \frac{\sqrt{3}}{4} + 4 (\frac{2 π - 3 \sqrt{3}}{12})$
$= \frac{\sqrt{3}}{2} + \frac{2 π - 3 \sqrt{3}}{3}$
$= \frac{3 \sqrt{3} + 4 π - 6 \sqrt{3}}{6}$
$= \frac{4 π - 3 \sqrt{3}}{6} = 2 \frac{π}{3} - \frac{\sqrt{3}}{2}$

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