In the figure above, ABC is a right isosceles triangle, where AB = AC

Question

Competition Mode Question https://gmatclub.com/forum/download/file.php?id=54316 In the figure above, ABC is a right isosceles triangle, where AB = AC = 6. If BQC is a semi-circle, with diameter BC and BPC is an arc of a circle centered at A, what is the area of the shaded region? A. 9 B. 9π - 18 C. 18 D. 9π E. 10π Are You Up For the Challenge: 700 Level Questions area.png

freedom128 · Answer

BQC is a semi-circle, with diameter BC
--> Area of BCQ = π*(6√2)^2 /8 = 9π

Area of ABPC= π*(6)^2 /4 = 9π
--> Area of BPC = 9π - 6*6/2 = 9π - 18

Shaded region = Area of BCQ - Area of BPC
= 9π - (9π - 18)
= 18

FINAL ANSWER IS (C)

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ArunSharma12 · Answer

In the figure above, ABC is a right isosceles triangle, where AB = AC = 6. If BQC is a semi-circle, with diameter BC and BPC is an arc of a circle centered at A, what is the area of the shaded region?

A. 9
B. 9π - 18
C. 18
D. 9π
E. 10π

Area of the segment BPC =  \frac{\pi6^2}{4} -\frac{6^2}{2}

Area of semi-circle BQC =  \frac{\pi}{2}	imes (\frac{6}{\sqrt{2}})^2

Area of shaded region =  \frac{\pi}{2}	imes (\frac{6}{\sqrt{2}})^2  - ( \frac{\pi6^2}{4} -\frac{6^2}{2} ) =  \frac{6^2}{2} 	imes (\frac{\pi}{2} - \frac{\pi}{2} + 1)  = 18
Ans: C

GMATWhizTeam · Answer

Solution 
 • Area of the shaded region = Area of the semicircle BCQ + Area of the triangle ABC – Area of the sector ABPC
• It is given that that  AB = AC = 6 
 o So,  BC =6\sqrt{2} 
  Radius of semi-circle BCQ  = \frac{ 6\sqrt{2} }{ 2} = 3\sqrt{2}   
• So area of semicircle BCQ  = \frac{1}{2}*\pi *( 3\sqrt{2})^2 = 9*\pi 
• Area of the triangle ABC  = \frac{1}{2}*base*height = \frac{1}{2}*6*6 = 18 
• Now area of the sector ABPC  = \frac{90}{360} * \pi* 6*6 = 9\pi  
• So area of the shaded region  = 9*\pi + 18 -9*\pi = 18  
Thus, the correct answer is  Option C.

ankitgoswami · Answer

Area of semicircle BQC = 9π

Area of arc BPC = area of BPCA - area of triangle ABC
 = 9π - 18

So, area of shaded region = 9π - 9π + 18
 = 18

So C is the ans.

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exc4libur · Answer

shaded: area semi - area slice
area slice: area circle * 1/4 - area iso-triangle
area iso-triangle (x:x:xV2 = 6:6:6V2): 6*6/2=18
area circle * 1/4: pi.r^2 * 1/4 = pi(6)^2/4 = 9pi
area slice: 9pi-18
area semi: pi(6V2/2)^2/2 = pi(18)/2 = 9pi
shaded: 9pi-(9pi-18) = 18

ans (C)

GMATinsight · Answer

Please find the video solution as attached

Answer: Option C

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https://www.youtube.com/watch?v=CopmqAYg8mM

pabpinor · Answer

Ok, I do not see how to solve this quickly, only a standard hard work calculation works.

We have a right isosceles triangle with sides 6 and hypotenuse being \sqrt{72}=6\sqrt{2}.

Then we have a semicircle with center at our right angle, so with radius being 6. And another semicircle with center at the middle of our hypotenuse, so its radius being 3\sqrt{2}.

We need to get the area between the two semicircles, so we need to subtract the smaller from the larger one.

The smaller circle area would be π *6^2=36π, the area to subtract would be a fourth of that minus the triangle area: 36π /4-6^2=9π -18.

The larger circle area would be π *3\sqrt{2}^2=18π.

Then we have to substrate to half of that circle area the coinciding area of the smaller circle, so:
1/2*18π =9π. 9π -9π +18=18.

Therefore the right answer is   C  .

lacktutor · Answer

Since BC is diameter of BQC semi-circle, Radius of it should be equal to  \frac{√ (36+36)}{2}= \frac{6√2}{2}= 3√2

The area of BQC semi-circle =  \frac{π(3√2)^{2}}{2}= \frac{18π}{2}= 9π

The area of BPC arc=  \frac{πr^{2}*90}{360}- \frac{6*6}{2}= 9π- 18

---> The shaded region =  9π - (9π- 18 )= 18

Answer (C).

monikakumar · Answer

In the figure above, ABC is a right isosceles triangle, where AB = AC = 6. If BQC is a semi-circle, with diameter BC and BPC is an arc of a circle centered at A, what is the area of the shaded region?

A. 9
B. 9π - 18
C. 18
D. 9π
E. 10π
area of big circle with radius 3*2^1/2 = 18π
half will be 9π
area of triangle=18
1/4th of area of small circle with centred at A=9π
so 9π-(9π-18)=18
Ans C

unraveled · Answer

Image
In the figure above, ABC is a right isosceles triangle, where AB = AC = 6. If BQC is a semi-circle, with diameter BC and BPC is an arc of a circle centered at A, what is the area of the shaded region?

A. 9
B. 9π - 18
C. 18
D. 9π
E. 10π

Area of triangle ABC =  \frac{1}{2}*6*6  = 18
Area of segment of circle with center A =  \frac{π*6*6}{4}  = 9π
Area of sector BPC = 9π - 18
Area of semicircle BQC =  \frac{π*(6\sqrt{2})^2}{4*2}  = 9π

Shaded Area = 9π - 9π + 18 = 18

Answer C.

Fdambro294 · Answer

I guess the winner of this one never got announced? lol

To Find the Area of the Shaded Region, the Strategy used is the following:
Shaded Area =

(Area of Semi-Circle BQC with BC as the Diameter) - (Sector Area of the 1/4 portion of a Circle with Radius 6) + (then ADD Back the over-subtraction of the Area of the Right Triangle that was Subtracted out as part of the Sector Area, but should NOT have been Subtracted out)

(1/2) *  ^2 (pi)

-

(90/360) * (6)^2 (pi)

+

(1/2) * (6) * (6)

=

9(pi) - 9(pi) + (1/2)(6)(6) = 18

18 = Area of Shaded Region

(C)

SylvesterF · Answer

Area of shaded region = Area (semi-circle BQC) - Area (region BPC)

For Area (semi-circle BQC) 
Area (semi-circle BQC) =  \pi  r^2  / 2
Diameter = BC = 6 \sqrt {2} ... radius = 3 \sqrt {2} 
  	riangle ABC is an isoceles  	riangle  i.e.  45^{\circ} - 45^{\circ} - 90^{\circ} , with sides in the ratio 1:1: \sqrt {2}  
Area (semi-circle BQC) = 9 \pi

For Area (region BPC) 
Area (sector BPC) =  	heta  /  360^{\circ}  x  \pi  r^2  ( \angle BAC =  90^{\circ} )
Area (sector BPC) =  90^{\circ}  /  360^{\circ}  x  \pi  6^2 ... (BPC is an arc of a circle centered at A)
Area (sector BPC) = 9 \pi 
 	herefore   Area of region BPC = Area (sector BPC) - A( 	riangle ABC) = 9 \pi  - 18

Area of shaded region = 9 \pi  - (9 \pi  - 18) = 18

Option C

In the figure above, ABC is a right isosceles triangle, where AB = AC

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GMAT Problem Solving (PS) Questions

In the figure above, ABC is a right isosceles triangle, where AB = AC

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards