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# Four semicircular arcs of length 2π are joined to make the figure abov

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Math Expert
Joined: 02 Sep 2009
Posts: 58396
Four semicircular arcs of length 2π are joined to make the figure abov  [#permalink]

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30 Aug 2018, 05:11
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Difficulty:

45% (medium)

Question Stats:

68% (01:58) correct 32% (01:33) wrong based on 55 sessions

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Four semicircular arcs of length 2π are joined to make the figure above. What is the area of the enclosed region?

A. $$8\pi$$

B. $$8 + 8\pi$$

C. $$16 + 8\pi$$

D. $$16 + 16\pi$$

E. $$24\pi$$

Attachment:

image019.jpg [ 2.6 KiB | Viewed 837 times ]

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Re: Four semicircular arcs of length 2π are joined to make the figure abov  [#permalink]

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30 Aug 2018, 06:08
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Bunuel wrote:

Four semicircular arcs of length 2π are joined to make the figure above. What is the area of the enclosed region?

A. $$8\pi$$

B. $$8 + 8\pi$$

C. $$16 + 8\pi$$

D. $$16 + 16\pi$$

E. $$24\pi$$

Area of enclosed region=4*Area of semicircle+Area of square (refer affixed diagram)
Given, length of semicircular arc=2π
Or, $$\frac{2πr}{2}=2π$$
Or, r=2

Area of semi-circle=$$\frac{1}{2}*π*r^2=2π$$
Area of square=4*(2r)=16

So , area of enclosed region=4*2π+16=8π+16

Ans. (C)
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semic.JPG [ 22.09 KiB | Viewed 718 times ]

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Re: Four semicircular arcs of length 2π are joined to make the figure abov  [#permalink]

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30 Aug 2018, 06:08
Top Contributor
Bunuel wrote:

Four semicircular arcs of length 2π are joined to make the figure above. What is the area of the enclosed region?

A. $$8\pi$$

B. $$8 + 8\pi$$

C. $$16 + 8\pi$$

D. $$16 + 16\pi$$

E. $$24\pi$$

Attachment:
image019.jpg

If one SEMI-circle has a perimeter of 2π, then an ENTIRE circle must have a circumference of

Circumference = (diameter)(π)
So, if the circumference of the circle is , then the diameter of each circle is 4

This means the square in the middle of the figure must have area 16.

Now we need only find the area of 4 SEMIcircles.
If we combine 2 semicircles we get an entire circle.
So, if we combine 4 semicircles we get 2 entire circles.

Each circle has diameter 4, which means each circle has RADIUS 2
Area of a circle = π(radius²)
So, area of one circle = π(2²) = 4π
So, the area of TWO circles = 8π

So, the TOTAL area = 16 + 8π

Cheers,
Brent
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Re: Four semicircular arcs of length 2π are joined to make the figure abov   [#permalink] 30 Aug 2018, 06:08
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