Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

In the figure below, if isosceles right angle PQR has an [#permalink]
19 Jun 2004, 08:56

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct
0% (00:00) wrong based on 0 sessions

In the figure below, if isosceles right angle PQR has an area of 4, what is the area of the shaded portion of the fgure? Assume R is the center. PLEASE EXPLAIN!

A. Pi
B. 2Pi
C. 2. sqrt2. pi
D. 4 pi
E. 8 pi

Attachments

GEO.JPG [ 7.69 KiB | Viewed 812 times ]

Last edited by mirhaque on 19 Jun 2004, 09:45, edited 1 time in total.

Since R is center, let RP = RQ = x
Area of PRQ = 1/2*RQ)RP =x^2/2
Area given = 4
Hence x^2/2 = 4 =>x = 2*sqrt2!!
For this Isoceles triangles if we drop a perpendicular from R (RY) to base PQ, it will bisect PQ at Y.
From pythagoras theorem, PQ^2 = PR^2 + RQ^2 =>PQ = 4 => QY =2!!

Hence RY^2 = RQ^2 - QY^2 = {2*sqrt2}^2 -{2}^2 = 4
=>RY = 2 = radius!!
Area of shaded portion is = 1/4*pi*radius^2=4*pi/4 = pi!!

In the figure below, if isosceles right angle PQR has an area of 4, what is the area of the shaded portion of the fgure? Assume R is the center. PLEASE EXPLAIN!

A. Pi B. 2Pi C. 2. sqrt2. pi D. 4 pi E. 8 pi

S(triangle) = 4 => Radius of the circle = 2 (since R*2*R/2 = area = 4). => S(shaded region) = S(circle)/4 = 2^2*pi/4 = pi.

Remember, if you get booged down with calculations you are going to be screwed, and this question tries to lead you down that path.
The triangle has an area of 4, and the shaded portion has an area obviuosly less than that. Only one answer choice fits that criteria.

Remember, if you get booged down with calculations you are going to be screwed, and this question tries to lead you down that path. The triangle has an area of 4, and the shaded portion has an area obviuosly less than that. Only one answer choice fits that criteria.

In the figure below, if isosceles right angle PQR has an area of 4, what is the area of the shaded portion of the fgure? Assume R is the center. PLEASE EXPLAIN!

A. Pi B. 2Pi C. 2. sqrt2. pi D. 4 pi E. 8 pi

S(triangle) = 4 => Radius of the circle = 2 (since R*2*R/2 = area = 4). => S(shaded region) = S(circle)/4 = 2^2*pi/4 = pi.

I completely understood CBRF3's explanation, but your condensed version is not as clear. Just following up in case you got some great shortcut.

In the figure below, if isosceles right angle PQR has an area of 4, what is the area of the shaded portion of the fgure? Assume R is the center. PLEASE EXPLAIN!

A. Pi B. 2Pi C. 2. sqrt2. pi D. 4 pi E. 8 pi

S(triangle) = 4 => Radius of the circle = 2 (since R*2*R/2 = area = 4). => S(shaded region) = S(circle)/4 = 2^2*pi/4 = pi.

I completely understood CBRF3's explanation, but your condensed version is not as clear. Just following up in case you got some great shortcut.

S(circle) is the area of the circle.

We find area of triangle = 4 => we find all sides of this triangle (2*sqrt(2),2*sqrt(2), 4). Then we find its height (from the vertex which is opposite to hipotenuse), which is equal to 2(because hipotenuse = 4, area = 4 => hipotenuse*height/2 = area = 4). This height is the same as the radius of the circle! Then we find area of the circle, given its radius: pi*r^2 = 4*PI. Then we find shaded area, which is one-fourth of the circle area = PI.

In the figure below, if isosceles right angle PQR has an area of 4, what is the area of the shaded portion of the fgure? Assume R is the center. PLEASE EXPLAIN!

A. Pi B. 2Pi C. 2. sqrt2. pi D. 4 pi E. 8 pi

S(triangle) = 4 => Radius of the circle = 2 (since R*2*R/2 = area = 4). => S(shaded region) = S(circle)/4 = 2^2*pi/4 = pi.

I completely understood CBRF3's explanation, but your condensed version is not as clear. Just following up in case you got some great shortcut.

S(circle) is the area of the circle.

We find area of triangle = 4 => we find all sides of this triangle (2*sqrt(2),2*sqrt(2), 4). Then we find its height (from the vertex which is opposite to hipotenuse), which is equal to 2(because hipotenuse = 4, area = 4 => hipotenuse*height/2 = area = 4). This height is the same as the radius of the circle! Then we find area of the circle, given its radius: pi*r^2 = 4*PI. Then we find shaded area, which is one-fourth of the circle area = PI.

in other words, it's the same throught process as presented by CBRF3. thanks for elaboration

Remember, if you get booged down with calculations you are going to be screwed, and this question tries to lead you down that path. The triangle has an area of 4, and the shaded portion has an area obviuosly less than that. Only one answer choice fits that criteria.

good work Maneesh.

Emmanuel, the third option is 2.sqrt2.pi which is > 4

- ash _________________

ash
________________________
I'm crossing the bridge.........

Emmanuel, the third option is 2.sqrt2.pi which is > 4

- ash

No, ashkg, 2*sqrt(PI) < 4 because 4*PI < 16, because PI = 3.14 < 4.

Emmanuel, what you have written is correct, but what is given is 2.srqt2.pi not 2*sqrt(PI). Maneesh, thanks. You are thinking outside the box. It took me several minutes to find the answer using traditional math. _________________