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

Question

https://gmatclub.com/forum/download/file.php?id=61017 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 Project PS Butler Subscribe to get Daily Email - Click Here | Subscribe via RSS - RSS 1.png

sumitkrocks · Answer

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

asishron29181 · Answer

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

TarunKumar1234 · Answer

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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Official Answer

D

Scuven · Answer

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

RajatJ79 · Answer

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.

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

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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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