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## Comment on

Assumptions and Estimation on the GMAT## Do we have answer to the

## The question is an example of

The question is an example of how students might make a good guess if they can’t solve the question.

Having said that, the correct answer is D (70 degrees). Here’s why:

Angle BAD = 20 degrees since this inscribed angle “holds” the same arc as angle BOD, and we know that the central angle must be twice the inscribed angle.

Since AB || CD, angle ADC = 20 degrees.

Finally, since angle ECD in an inscribed angle “holding” the diameter (ED), angle ECD = 90 degrees.

At this point, we know 2 of the 3 angles in a triangle (20 degrees and 90 degrees), so the third angle (x) must equal 70 degrees.

## If the sides of a triangle

I. 28

II. 36

III. 42

A I only

B II only

C I and II only

D I and III only

E I, II, and III

## We have to test each

We have to test each statement separately.

I. 28

This cannot be the perimeter, since sides x and y already have a sum of 30. So, the perimeter cannot be less than 30.

This allows us to eliminate A, C, D and E, since they all suggest that 28 IS a possible perimeter.

Answer: B

## Hi Brent,

In this question,

http://www.beatthegmat.com/mgmat-geometry-t285465.html

I have two concerns with regards to Mitch's asnwer :

1. "The degree measurement of an inscribed angle = 1/2 the degree measurement of the intercepted arc."

I didn't came across this principle in our concept file that dealt with inscribed angles.

2. How did you assume CB to equal EB? Which is the principle applicable for the same?

## Question link: http://www

Question link: http://www.beatthegmat.com/mgmat-geometry-t285465.html

1. Mitch uses different words to describe the rule I discuss at 2:54 of the following video: https://www.gmatprepnow.com/module/gmat-geometry/video/880

I would typically say "The central angle COE = 120 degrees," however Mitch says "Arc CAE = 120 degrees"

ASIDE: I don't believe I've ever seen an official GMAT question refer to an arc as having a measurement in degrees.

2. Since AB is the diameter, it "cuts" the circle into two IDENTICAL semicircles. Lines CB and FB create 30 degrees with AB.

Since CB and FB are lines within two IDENTICAL semicircles, then they must have equal length.

## Hi Brent,

In the question https://gmatclub.com/forum/the-shaded-region-in-the-figure-above-represents-a-rectangular-frame-135095.html

I dont understand the wording. The question says that the frame has the same area as the picture and begins with saying the frame has sides 18X15 then in the solution we are supposed to divide the area by half?

I understood your answer but if I were to compute in the conventional way then how is the area divided into two equal parts?

I found it confusing to divided it by two when the wording says the areas are the same.

Thank you,

Ari Banerjee

## Question link: https:/

Question link: https://gmatclub.com/forum/the-shaded-region-in-the-figure-above-represe...

Here's another way to put it: If you examine the image at https://imgur.com/t6zGIOR, we can say that the area of the red portion (the frame) is equal to the area of the blue portion (the picture)

18 x 15 = 270

So, the area of the red portion = 135, and the area of the blue portion = 135

Does that help?

Cheers,

Brent

## https://gmatclub.com/forum

the diagonal divides the square into two parts. So the area of of one triangle cannot be greater than 144/2 whch is 72. SO the D and E are eliminated.

The shaded area is clearly greater than half of the triangle, so area is greater than 36.

I'm stuck here. how do I know if its 45 or 60?

## Question link: https:/

Question link: https://gmatclub.com/forum/the-quadrilateral-abcd-is-a-square-with-sides...

In my diagram (at https://gmatclub.com/forum/the-quadrilateral-abcd-is-a-square-with-sides...) the blue region has area 72. Notice that 3/4 of that blue region is shaded.

3/4 of 72 = 54, so the answer must be greater than 54.

Does that help?

Cheers,

Brent