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The big triangle is a 30:60:90 and so is the small, white one.
Given \(A_{x}\)= Area of the big triangle and \(A_{y}\)=area of the small triangle
we have that
\(A_{x}=x\cdot\sqrt{3}x=x^2\cdot\sqrt{3}\)
and \(A_{y}=y\cdot\sqrt{3}y=y^2\cdot\sqrt{3}\)

The shaded area equals the difference: \(\sqrt{3}(x^2-y^2)\)

B is the correct answer
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ngie
The big triangle is a 30:60:90 and so is the small, white one.
Given \(A_{x}\)= Area of the big triangle and \(A_{y}\)=area of the small triangle
we have that
\(A_{x}=x\cdot\sqrt{3}x=x^2\cdot\sqrt{3}\)
and \(A_{y}=y\cdot\sqrt{3}y=y^2\cdot\sqrt{3}\)

The shaded area equals the difference: \(\sqrt{3}(x^2-y^2)\)

B is the correct answer

Hi ngie,

You have the right conceptual idea, but you have to be very careful with your work.

The formula for area of a triangle is (1/2)(Base)(Height).

Your calculations don't use the correct formula.

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Rich
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The area of un-shaded region = y^2/x^2 * The area of the big triangle => The area of shaded region = 1 - y^2/x^2

The area of the big triangle = 1/2*x^2*sqrt(3) => The area of shaded one is 1/2*sqrt(3)*(x^2 - y^2)

=> ANSWER: E
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Attachment:
nestedtr_q1.png
In terms of x and y, what is the area of the shaded region? Note: Figure not drawn to scale

A. x^2 - y^2
B. \(\sqrt{3}(x^2 - y^2)\)
C. \(\sqrt{3}x^2 - y^2\)
D. (x^2 -y^2)/2
E. \(\frac{\sqrt{3}(x^2 - y^2)}{2}\)

Kudos for a correct solution.

Both triangles are 30 60 90 triangles since they are similar triangles.
I just subtracted the larger triangle from the smaller one.
(1/2)(x^2)(x*sqrt(3)) - (1/2)(y)(y*sqrt(3))
\(\frac{\sqrt{3}(x^2 - y^2)}{2}\)

Answer: E
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EMPOWERgmatRichC
Hi ngie,

You have the right conceptual idea, but you have to be very careful with your work.

The formula for area of a triangle is (1/2)(Base)(Height).

Your calculations don't use the correct formula.

GMAT assassins aren't born, they're made,
Rich

You are right! Thank you for pointing it out!
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Shaded region is a trapezoid

Attachment:
nestedtr_explanation.png
nestedtr_explanation.png [ 13.55 KiB | Viewed 16362 times ]

There is a direct formula to calculate the area once we have the height of trapezoid \(= \sqrt{3}(x-y)\)

Area \(= \frac{1}{2} (x+y) \sqrt{3}(x-y) = \frac{\sqrt{3} (x^2 - y^2)}{2}\)

Answer = E
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How are we sure that we can use the 30 60 90 proportions? We only know that its a right triangle. Plus 2x, x ratio. does it mean that we must have a 30 60 90?

Thanks!
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How are we sure that we can use the 30 60 90 proportions? We only know that its a right triangle. Plus 2x, x ratio. does it mean that we must have a 30 60 90?

Thanks!

MUST KNOW FOR THE GMAT:
• A right triangle where the angles are 30°, 60°, and 90°.

This is one of the 'standard' triangles you should be able recognize on sight. A fact you should commit to memory is: The sides are always in the ratio \(1 : \sqrt{3}: 2\).
Notice that the smallest side (1) is opposite the smallest angle (30°), and the longest side (2) is opposite the largest angle (90°).

• A right triangle where the angles are 45°, 45°, and 90°.

This is one of the 'standard' triangles you should be able recognize on sight. A fact you should also commit to memory is: The sides are always in the ratio \(1 : 1 : \sqrt{2}\). With the \(\sqrt{2}\) being the hypotenuse (longest side). This can be derived from Pythagoras' Theorem. Because the base angles are the same (both 45°) the two legs are equal and so the triangle is also isosceles.
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Bunuel niks18 gmatbusters amanvermagmat pushpitkc

Is my below understanding correct?

For the larger triangle, I know base x and hypotenuse (2x) hence I could find height using ratio 30:60:90 (angles)
vs sides (1:\(\sqrt{3}\): 2)
But in smaller triangle (i.e with base y) only three angles and base y is known, Hence we derived height as \(\sqrt{3y}\)
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Bunuel niks18 gmatbusters amanvermagmat pushpitkc

Is my below understanding correct?

For the larger triangle, I know base x and hypotenuse (2x) hence I could find height using ratio 30:60:90 (angles)
vs sides (1:\(\sqrt{3}\): 2)
But in smaller triangle (i.e with base y) only three angles and base y is known, Hence we derived height as \(\sqrt{3y}\)

Hello

If your understanding is that smaller triangle is also similar to the larger triangle and thus the three angles of this smaller right angle triangle will be 30-60-90 and thus height of this smaller triangle will be √3*y,..

.. then it's correct :)

Posted from my mobile device
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your question :
Is my below understanding correct?

For the larger triangle, I know base x and hypotenuse (2x) hence I could find height using ratio 30:60:90 (angles)
No, . In fact the the sides x and 2x let you know that the triangle is 30:60:90 triangle.

All 30-60-90-degree triangles have sides with the same basic ratio.
So only a single side is required to find the value of all sides.



A 30-60-90-degree right triangle.
Attachment:
369580.image1.jpg
369580.image1.jpg [ 7.21 KiB | Viewed 12062 times ]



adkikani


Is my below understanding correct?

For the larger triangle, I know base x and hypotenuse (2x) hence I could find height using ratio 30:60:90 (angles)
vs sides (1:\(\sqrt{3}\): 2)
But in smaller triangle (i.e with base y) only three angles and base y is known, Hence we derived height as \(\sqrt{3y}\)
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In the is question the ratio of lengths of sides are mention in the stem as x:2x and from the figure it’s mentioned that it’s a right angle so we think of a 30-60-90 triangle whose sides are in the ratio x:root3x:2x.
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