Source: Official Guide Revised GRE 1st Ed. Geometry Exercises; #9

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1

Ashlesha Kshirsagar

Here is the Easiest Way :

since Theorem says- In two similar triangles, the ratio of their areas is the square of the ratio of their sides.

area of CDE/area of ADG=(CD/AD)^2

Here in figure suppose CD=x,then AD=3x(Already mentioned in question AB=BC=CD)

so (area of CDE)/(area of ADG)=(x/3x)^2

Setting x to a value of 1, 42(/area of ADG)=1/9 area of ADG=378

ANSWER:378

Jul 28, 2018 • Comment

David Recine, Magoosh Tutor

I like the way you laid out the steps there, Sudharson. :)

Jul 29, 2018 • Reply

Haaris Raja

Could you also use sides GD and EG since they also look equal? The bases only make the difference.

Aug 6, 2018 • Reply

Sam Kinsman, Magoosh Tutor

I think you meant to say GD and ED. And the answer is yes, you could. ED will be one third of GD. However, we can't just make this assumption because the diagram makes it look that way. Instead, we know this (that ED will be one third of GD) because these triangles are all similar. So the ratio of their hypotenuse sides has to be equal to the ratio of their heights.

Aug 9, 2018 • Reply

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Emily Marsh

I don't understand this at all. I understand up to get the equation of AG = 3*CE but do not understand how to get the area of the whole triangle. Please help.

Nov 11, 2017 • Comment

Chandan Troughia

Since you know AG = 3CE and DG = 3DE ...eq(1),

Now the area of the top most triangle(CDE)

=1/2(EC x DE) = 42 ...eq(2) (given)

And Now, consider the area of the entire triangle(ADG) = 1/2(AG x DG)

= 1/2(3EC x 3DE)-------- replace the values using (eq(1))

= 9/2(EC x DE)

= 9(1/2(EC x DE)) ---eq(3)--- in this 1/2(EC x DE) is equal to 42(given) i.e. eq(2) replace it in the eq(3) with the value 42.

= 9 (42) = 378.

I hope this is helpful.

Nov 18, 2017 • Reply

Sam Kinsman, Magoosh Tutor

Hi Emily,

We know that AG = 3*CE. Also, since the triangle are similar, we know that DC/DE=AD/DG. The prompt tells us that AB=BC=CD, so AD is 3*CD, so we have DC/DE=3*DC/DG. Cross multiply and cancel, to get DG=3*DE.

We know that the area of a triangle is (base) * (height) * (1/2). So the area of ADG=AG*DG*(1/2)

Now we can replace AG with 3*CE, and DG with 3*DE, to get:

= 1/2*3CE*3DE

= 9 * (1/2) * CE * DC

= 9 * (triangle DEC)

= 9 * 42 = 378

I hope this clarifies how we got to answer! :-)

Nov 18, 2017 • Reply

Sam Kinsman, Magoosh Tutor

Hi Ali,

First, we know that the three triangles are similar (DCE, DBF, and DAG). And since AB = BC = CD, we know that DA = 3*DC.

Now, let's think about triangles DCE and DAG. We know that the hypotenuse of DAG (which is DA) is 3 times as big as the the hypotenuse of DCD (which is DC). We know this because DA = 3*DC.

This tells us that every side of the triangle DAG must be three times as large as the corresponding side in triangle DCE. Therefore, it must be the case that AG is three times as large as CE. So we can say that AG = 3*CE.

Apr 11, 2018 • Reply

Andrew Chu

Hi Sam,

I hope you are doing well. I understand that DA=3*DC. I get that part. What I don't understand is that how are you assuming that the entire DAG must be 3 times greater. Please explain. Thanks.

Apr 17, 2018 • Reply

Sam Kinsman, Magoosh Tutor

Hi Andrew,

From the diagram, we can see that lines CE and AG are parallel. That means that DCE and DAG must be similar triangles (they have three sets of corresponding and equal angles).

So DCE and DAG are similar. Now, the hypotenuse of DCE is DC, and the hypotenuse of DAG is DA. We know that DA is three times as long as DC. Therefore, it must be the case that every side of DAG is three times as long as every side of DCE. Therefore, we know that DAG must be three times as big as DCE.

Apr 19, 2018 • Reply

Andrew Chu

AHHH that makes a lot of sense. It falls under the principal of similar triangles, which adheres to having either congruent angles or the lengths corresponding to the same ratio correct?.

Apr 20, 2018 • Reply

How did we get 9 after this stage: "= 1/2(3EC x 3DE)-"?

I can see how you get that by multiplying 3 times 3, but I thought I couldn't multiply coefficients with different variables.

May 15, 2018 • Reply

David Recine, Magoosh Tutor

You can't multiply coefficients with different variables. However, you can remove the coefficient number and detach it from the variables by using the commutative property of multiplication. Recall that the commutative property states that you can multiply a group of numbers (and/or variables) in any order.

So this:

(1/2)*3CE*3DE

Can be reordered as this:

3*3*(1/2)*CE*DE

And then the re-ordered pair of 3s can be multiplied together into a single 9, which leads to this:

9 * (1/2) * CE * DC

I hope that helps. :)

May 18, 2018 • Reply

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Sanchit Bogra

Since CE||BF||AG,

We can conclude that the triangles are similar.

So,

DC/CE =AD/AG

=) DC/CE=3*DC/AG ( Because AB=BC=CD)

=) AG= 3*CE

Similarly,

DC/DE=AD/DG

=)DC/DE=3*DC/DG( Because AB=BC=CD)

=)DG=3*DE

Now area of triangle ADG=1/2*AG*DG

=1/2*3CE*3DE

=9(area of triangle CDE)

=9*42 i.e 378 (Ans.)

Aug 7, 2016 • Comment

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Suzanne ashcraft

I found this explantion when researching a solution: http://math.stackexchange.com/questions/192298/geometry-triangle-question

Sep 21, 2015 • Comment

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