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If D is the midpoint of AC and arcs DF and DE are centered at C and A

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Bunuel
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If D is the midpoint of AC and arcs DF and DE are centered at C and A [#permalink] New post   11 May 2021, 01:30
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If D is the midpoint of AC and arcs DF and DE are centered at C and A respectively, and If AB = BC = 2, what is the area of the shaded region?


A. \(2-2\pi\)

B. \(\frac{\pi}{6}\)

C. \(0.5\)

D. \(2-\frac{\pi}{2}\)

E. \(4-\pi\)



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If D is the midpoint of AC and arcs DF and DE are centered at C and A [#permalink] New post   11 May 2021, 03:35
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Bunuel wrote:

If D is the midpoint of AC and arcs DF and DE are centered at C and A respectively, and If AB = BC = 2, what is the area of the shaded region?


A. \(2-2\pi\)

B. \(\frac{\pi}{6}\)

C. \(0.5\)

D. \(2-\frac{\pi}{2}\)

E. \(4-\pi\)


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Area of right angled triangle ABC=\(\frac{ 1}{2 }* 2 * 2 \)

AC =\( \sqrt{(2^2 +2^2)}\) = 2\(\sqrt{2}\)
CD = AD = AC/2 = \(\sqrt{2}\)
Area of symmetrical region AED and CFD = 1/8 * area of circle of radius CD/ or AD =1/8 *Pi* \(\sqrt{2}^2\)
Area of shaded region = Area of triangle ABC- 2* Area of region AED

= 2 - 2*\(\frac{1}{8}\)*pi*2
=2 - \(\frac{Pi}{2}\)

D is the answer IMO
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Re: If D is the midpoint of AC and arcs DF and DE are centered at C and A [#permalink] New post   11 May 2021, 03:54
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Given AB = BC = 2
Area of triangle = 0.5 * 2 * 2 = 2
Area of shaded region = Area of triangle - 2 (area of sectors)
2 - 2( 45/360 * pi * r ^2)

Now since it is a right triangle
AC = 2√ 2
Radius of sector= AC/2
= √2
putting the values in the above equation, area of shaded region = (2 - pi/2)

IMO D
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Re: If D is the midpoint of AC and arcs DF and DE are centered at C and A [#permalink] New post   11 May 2021, 07:30
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If D is the midpoint of AC and arcs DF and DE are centered at C and A respectively, and If AB = BC = 2, what is the area of the shaded region?

AC = \(\sqrt{2^2 +2^2} = \sqrt{8} = 2 \sqrt{2}\)
AD = AC/2 = \(\frac{2\sqrt{2}}{2}\) = \(\sqrt{2}\)

Grey area = \(\frac{1}{2}*2*2 - \frac{pie*(\sqrt{2})^2 }{360} *45 *2\)
= \(2- \frac{pie}{2}.\)

So, I think D. :)
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Re: If D is the midpoint of AC and arcs DF and DE are centered at C and A [#permalink] New post   11 May 2021, 14:09
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AB=BC=2

We can also see that the triangle is 90-45-45.

We recall that the area of a circular sector is pi greek *radius^2 *(angle of the sector/360)

We see that the shadowed area can be calculated as = area triangle - area 2 circular sectors

Thus area triangle is 1/2(base * height )=2

Area of the TWO circular sectors = 2* pi greek * (90/360) =0,5 pi greek (90 is the sum of the two angles 45+45, we calculate here the total area of the two instead of one after the other to save time)

Area of the shadowed part = 2 - 1/2 pi greek thus D
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Re: If D is the midpoint of AC and arcs DF and DE are centered at C and A [#permalink] New post   11 May 2021, 19:48
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If D is the midpoint of AC and arcs DF and DE are centered at C and A respectively, and If AB = BC = 2, what is the area of the shaded region?

AC = Hypotenuse, using Pythagoras theorem:
AC = \(\sqrt{AB^2 +BC^2} = \sqrt{2^2 +2^2} = \sqrt{8} = 2 \sqrt{2}\)
As D is mid-point, AD = \(\frac{AC}{2}\) = \(\frac{2\sqrt{2}}{2}\) = \(\sqrt{2}\)

Area of shaded region = Area of Right angled Triangle - 2 * Area of arc = \(\frac{1}{2}*2*2\) - \( \frac{\pi*(\sqrt{2})^2 *45}{360} * 2 \)
= \(2- \frac{\pi}{2}.\)

Answer should be Option D.
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Re: If D is the midpoint of AC and arcs DF and DE are centered at C and A [#permalink]
11 May 2021, 19:48
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