The figure shows the top side of a circular medallion made of a circul

Haha. You beat me to it!

Btw.. how do you get the symbol for pi?

Mark \pi with formula button [m].

\(\pi\)

Cheers!

Hi,

The mistake you are making is that you are considering 'r' to be the radius of the medallion without the glass frame.

The question states that the medallion is made of :
1) Circular piece of glass and
2) Metal frame

Also, we are told that the radius 'r' is that of the medallion.

Given the width of the metal frame to be s, we can conclude that the radius of the just the circular piece of glass will be (r-s).

Thus the expression is in fact as follows :

Area of the frame =\(\pi*r^2-\pi*(r-s)^2=\pi(r^2-r^2+2rs-s^2)=\pi*s*(2r-s)\)

Answer : E

radius = r (cm)
width = s (cm)
Area of the metal frame = ? (cm^2)

Think of this as concentric circles. Then the inner circle will have radius = r - s

The outer circle radius = r

The area of metal frame = Pi(r )^2- Pi(r-s)^2
= Pi(r^2 - s^2 +2sr - r^2) = Pi*s(2r - s )

my answer: E

You're given that the width of the medallion is s cm. And that the radius of the medallion is r cm. So the figure looks like this:

Area of the medallion =\(\pi*r^2\)

Area of the frame = Area of the frame + medallion - Area of the medallion \(= \pi*(r+s)^2 - \pi*r^2 = \pi * [(r+s)^2 - (r)^2] = \pi * ( r+s+r) (r+s-r) = \pi*(s)(2r+s)\)

I'm not sure why the OA has a - sign either.

The question is worded somewhat awkwardly, which I believe is causing you to not picture the medallion correctly. We are asked to find the area of the metal frame, the area of which is part of radius \(r\).

Given:

Area of the whole medallion (frame + glass center): \(\pi r^2\)
The width of the frame is \(s\), so the area of just the glass center is: \(\pi r^2 - \pi (r-s)^2\)

Solve:

\(\pi r^2 - \pi (r^2 - 2rs + s^2)\)

\(\pi r^2 - \pi r^2 - \pi s^2 + 2\pi rs\)

\(2\pi rs - \pi s^2\)

Pull out \(\pi s\) and you're left with \(\pi s(2r - s)\)

Area = Area(Bigger Circle) - Area(Smaller Circle)
= \(\pi (r)^2 - \pi (r-s)^2\)
= \(\pi (-s^ + 2rs)\)
= \(\pi s (2r-s)\)

Note that r is the radius of the medallion, or the bigger circle.

Hence, answer (e)

Darn! I thought the smaller circle has r and the bigger had r+s...

Thanks, shrouded1

i was wondering if it was somehow possible to hide answer from the screenshot? thanks!

Done.

Similar questions to practice about rectangles:
a-square-wooden-plaque-has-a-square-brass-inlay-in-the-89215.html
a-border-of-uniform-width-is-placed-around-a-rectangular-144434.html
a-rectangular-yard-is-20-yards-wide-and-40-yards-long-it-is-8453.html
the-figure-above-represents-a-square-plot-measuring-x-feet-o-160568.html

Similar questions to practice about circles:
marty-s-pizza-shop-guarantees-that-their-pizzas-all-have-a-148428.html
a-circular-lawn-with-a-radius-of-5-meters-is-surrounded-by-a-154737.html
the-figure-above-shows-a-circular-flower-bed-with-its-cente-144448.html

Hope it helps.

Does anyone else think question is really ambiguous ? .By Radius of circular medallion I thought it meant radius of small circle. Hence I calculated it this way.
Radius of medallion i.e. small circle = r, radius of metal frame = s
Total radius of medallion + metal frame = r+s
Total area of medallion + metal frame = pi * (r+s)^2
Area of metal frame = (Total area of medallion + metal frame) - area of small circle
pi*(r+s)^2 - pi*r^2. = pi*s(s+2r)

after doing all this calculation I noticed my ans is not matching with any of the answer choices , so I did the calculation again with the correct understanding. . I understand such silly mistake of mine will have worse consequences in actual GMAT Exam. My question is how to avoid such mistakes. Please please please help.

Regards
Megha

Hey pushpitkc

the solution istelf i understand, i just got stuck in agebraic technical details

we know thise formula,
\((a-b)^2 = a^2-2ab+b^2\)

so,

\(pi*r^2- pi(r-s)^2 = pi*r^2- pi (r^2 -2rs+s^2)\) now what

Hey dave13

\(\pi*r^2 - \pi(r-s)^2 = \pi*r^2 - \pi (r^2 - 2rs + s^2) = \pi*r^2 - \pi*r^2 + \pi*2rs - \pi*s^2\)

Now we are left with \(\pi*2rs - \pi*s^2 = s*\pi(2r - s)\)(when we take s common from both the terms)

Hope this helps you!

Here's an alternative response, just plug in numbers!
Lets say that r=5 and s=2.
That means the area of the whole medallion is 25π, and the area of the medallion without the metal frame is 9π (r-s = 5-2 = 3 to find the radius of the inner circle)
That means the area of the metal frame is 16π
To find the area of the metal ring, substitute the values of r and s in the answer choices.
Only Answer E gives you 16π!

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.