What is the area of the shaded region above, if ABCD

Question

ABCD.png What is the area of the shaded region above, if ABCD is a square and line segment AB is a diameter of the circle with center O? (1) The radius of the circle with center O is 4. (2) The area of triangle ADC is 32. OE To find area of shaded region, you need to find area of triangle ACD, and subtract area of unshaded region inside this triangle from it. To find area of unshaded region, you need to first draw in a line perpendicular to diameter AB of circle from center O to diagonal AC. Call this point E. This point is also one of endpoints of arc AFE. ABCD2.png Arc AFE and radii OA and OE bound a sector of circle O: this sector is one quarter of entire area of circle. Now, if you take area of small triangle AEO, and subtract this area from area of sector OAFE, then you have area of unshaded portion of triangle ADC. So, to solve this question, you need to find area of triangle ADC, one quarter of area of circle O, and area of small triangle AEO. To find these three areas, you need radius of circle, or length of one side of square. (1): This is exactly what we need; once we have radius of circle we can find diameter of circle which is same as a side of square and also same as each leg of right triangle ADC. So you can, with this information, solve for different areas necessary, and then solve for area of shaded region. Sufficient. (2): able to find length of height and width (which are equal to each other) of triangle ADC. Let's say that both base and height of triangle ADC = x. 32 = (½)x^2 → x = 8 You have leg lengths (which are equal) of isosceles right triangle ADC; this means that you also have diameter of circle → radius. Sufficient

Bunuel · Answer

https://gmatclub.com/forum/download/file.php?id=21975 
 What is the area of the shaded region above, if ABCD is a square and line segment AB is a diameter of the circle with center O?

Actually there is no need for any calculations. From both statements we have everything "fixed", only one case possible where we know everything possible.

(1) The radius of the circle with center O is 4 --> diameter = side = 8. We have "fixed" circle and square. We can find anything there. Sufficient.

(2) The area of triangle ADC is 32 --> 1/2*AD^2=32 --> AD=8 --> AD = diameter = 8. The same info as above. Sufficient.

Answer: D.

b00gigi · Answer

Bunuel, Can you please explain why if you only have the radius you can find anything. How would you get the AF part?

I knew that Statement 1 and 2 gave the same information, but couldn't figure out how to find the white portion in TriangleADC

Thanks

Bunuel · Answer

That's the beauty of DS: we don't need to solve if we know that we CAN solve. We can solve because we know the radius of the circle (so we know everything about it) and we know the side of the square (so we know everything about it). If still interested in solving you can check under the spoiler in the initial post.

Cadaver · Answer

D--

One subtle hint....both the options A and B pointing towards same conclusion.....so can be either D or E....

Like the solution provided in spoiler....either is enough

Can,Will · Answer

gmatphobia  
how do we figure out the angle AOE (o is extended to point E)?

gmatophobia · Answer

Hey AnujL Not sure if the question was targetted at me (as you tagged another user with a similar username

) My apologies that I couldn't follow your question well. Can you please elaborate on your question? Does E lie on the line AB?

Can,Will · Answer

gmatophobia sorry for the wrong tag lately.

gmatophobia · Answer

AnujL  - Thank you for sharing an image.

I have used the image of the question to answer your query.

We know that ABCD is square and AC is a diagonal of that square.

\angle AEB = 90  (AB is the diameter of the circle)

Hence the line EB is perpendicular to the AC. Therefore we conclude that EB is a part of the other diagonal of the square. ( as diagonals in a square are perpendicular to each other).

Also, diagonals in a square are equal in length and bisect each other, hence we can conclude that AE = EB and that AEB is an isosceles triangle.

A line dropped from E to AB at point O (i.e. line OE) will bisect the line AB (as OA = AB = radius). As  	riangle AEB   is isosceles, we can conclude that OE will be perpendicular to AB.

Thus,  \angle AOE = 90

Note: This is a DS question. We don't need to find the actual value of the angle. If we can establish the information is sufficient, we are good to go.

Hope this helps!

What is the area of the shaded region above, if ABCD

Prep Toolkit

Top 5 GMAT Debrief Videos

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