It is about finding area want help in part b?

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It is about finding area want help in part b?

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Answer 1 · 2018-05-04T17:05:42

Given triangle ABC is a equilateral triangle, one interior angle is $\frac{π}{3}$ $pi/3$ .

Since the 3 arcs are equal,
$∠ C A B = \frac{π}{3}$ $/_CAB=pi/3$
$2 α = \frac{π}{3}$ $2alpha = pi/3$

Since $∠ A O B = 2 \frac{π}{3}$ $/_AOB = 2pi/3$ and $A O = O B = r$ $AO=OB=r$ ,
Triangle AOB is a equilateral triangle.
$:. r=4$

Substitute $2alpha = pi/3$ and $r=4$ into the answer for (a)(ii),
Area of segment AB
= $1/2r^2(pi/3-sin(pi/3))$
= $1/2(4)^2(pi/3-sqrt(3)/2)$
= $(8pi)/3-4sqrt(3)$

Area of triangle ABC
= $1/2a b sinC$
= $1/2(4)(4)(sin(pi/3))$
= $4sqrt(3)$

Area of shaded region
= Area of triangle ABC - 3 Area of segment AB
= $4sqrt(3)-3((8pi)/3-4sqrt(3))$
= $4sqrt(3)-8pi+12sqrt(3)$
= $16sqrt(3)-8pi$

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It is about finding area want help in part b?

1 Answer

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