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In the diagram above, BD = 5, CD = 10, and AE = 40. What is

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by Mike@Magoosh » Thu Dec 18, 2014 2:29 pm
Image
In the diagram above, BD = 5, CD = 10, and AE = 40. What is the area of the shaded region?
(A) 175
(B) 350
(C) 775
(D) 1150
(E) 1575

For a set of practice problems on Similar Geometric Figures, as well as the OA & OE for this particular problem, see this blog.


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Last edited by Mike@Magoosh on Mon Dec 22, 2014 5:28 pm, edited 1 time in total.
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by Mathsbuddy » Fri Dec 19, 2014 2:40 am
BDC and AEC are similar triangles.
AE is 40 and BD is 5, so the larger triangle is 8 times the area of the smaller one.
The area of BDC is 5 x 10/2 = 25
Therefore AEC has area 25 x 8 = 200
200-25 = 175
Answer A
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by Mike@Magoosh » Fri Dec 19, 2014 8:20 am
Mathsbuddy wrote:BDC and AEC are similar triangles.
AE is 40 and BD is 5, so the larger triangle is 8 times the area of the smaller one.
The area of BDC is 5 x 10/2 = 25
Therefore AEC has area 25 x 8 = 200
200-25 = 175
Answer A
Dear Mathsbuddy,
My friend, remember that when lengths are scaled up by, say, a scale factor of k = 8, area is NOT scaled by the same factor. Think of a square, 1 x 1, with area of 1. If we triple each side, to a 3 x 3 square, we don't get a new area of three, but rather an area of 9. When lengths are multiplied by scale factor k, areas are multiplied by scale factor (k^2). If volumes are relevant in the problem, volumes would be multiplied by (k^3).
In this problem, choice (A) was a trap answer, for those who used the same scale factor for area that was used for length.
Does all this make sense?
Mike :-)
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by MartyMurray » Fri Dec 19, 2014 10:11 am
Mike@Magoosh wrote:Image
In the diagram above, BD = 5, CD = 10, and AE = 40. What is the area of the shaded region?
(A) 175
(B) 350
(C) 775
(D) 1150
(E) 1575
BCD and ACE both have angle C and a right angle. So the third angles of each, angle B and angle E are also the same and the two triangles are similar triangles.

From triangle BCD we see that the two non hypotenuse sides of the triangles are in the ratio 5:10 or 1:2.

The shorter side of triangle BCD, the side which is opposite angle C, corresponds to AE, which is also opposite angle C.

The longer side of triangle BCD corresponds to AC.

So the ratio of the lengths of AE and AC is 1:2.

So given that the length of side AE = 40, length of side AC of triangle ACE is 80.

Area ACE is bh/2 = (80)(40)/2 = 1600

Area BCD = (5)(10)/2 = 25

Area ACE - Area BCD = 1575.

Choose E.
Mike@Magoosh wrote:For a set of practice problems on Similar Geometric Figures, as well as the OA & OE for this particular problem, see:
https://magoosh.com/gmat/2014/gmat-pract ... c-figures/
By the way Mike, I followed the link to the Magoosh site and did not see that OA and OE to this question.
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by Matt@VeritasPrep » Mon Dec 22, 2014 12:31 pm
Here's an easier visual. The key idea is to call some angle x°, then to find the other angles in terms of x (as shown). Once you do, you'll see the triangle similarity and be set.

Image
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by Mike@Magoosh » Mon Dec 22, 2014 5:31 pm
Marty Murray wrote:By the way Mike, I followed the link to the Magoosh site and did not see that OA and OE to this question.
Dear Marty,
I'm sorry --- the BTG system automatically curtailed the URL, and the curtailed URL misdirected readers to another blog. Not the most adept design. I corrected the link in the post at the top of this thread, and you can find it here.
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by Mike@Magoosh » Mon Dec 22, 2014 5:37 pm
Matt@VeritasPrep wrote:Here's an easier visual. The key idea is to call some angle x°, then to find the other angles in terms of x (as shown). Once you do, you'll see the triangle similarity and be set.
Matt,
(1) Don't you think it's more powerful to teach students the incredible shortcut of AA similarity, rather than the laborious method of finding every angle in the diagram? Doesn't the latter take 5-10x longer in a problem such as this?
(2) I notice that your "easier" diagram was even less to scale than mine. In my diagram, the angles were all precise, but the smaller triangle was scaled considerably bigger to be clear visually. You apparently changed the angles entirely --- were you aware of this difference?
Mike :-)
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