If BE || CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of

Since triangles ABE and ACD are similar we get BE = 5 and ED = 4
To determine the height --> 3^2 = h^2+a^2 and 4^2 = h^2+b^2
where a+b+5 = 10 so we get a+b= 5
putting this value in the above equation we get b-a = 7/5
From this value we determine the value of h =12/5
since area of trapezium = 1/2*ht*(sum of parallel sides) = 1/2*12/5*15 = 18

This is how I approached:

Since BE is II to CD AND B is mid point of AC therefore BE is Midsegment which means that BE = 1/2 (CD) = 5 and AE = ED = 4,
Observing triangle ACD : its a 6,8,10 Right Triangle & therefore area is 1/2(b/h) = 24
similarly traingle ABE : its a 3,4,5 triangle & therfore area is 1/2 (b/h) = 6

Therefore area of trapezoid is = 24 -6 = 18.

Answer is B

Bunuel, any time we see a triangle with sides 3,4,5, can we assume its a right triangle?

But MGMAT says that if you see the 3-4-5 as the side ratio, it does not implies it is a rt angled triangle???

I doubt that. But if it does then it's a mistake.

That's because converse of the Pythagorean theorem is also true.

For any triangle with sides a, b, c, if a^2 + b^2 = c^2, then the angle between a and b measures 90°.

Since 3^2+4^2=5^2 then any triangle whose sides are in the ratio 3:4:5 is a right triangle.

Hope it's clear.

Hi,

Area of a triangle is directly proportional to square of a side.
or area (ABC) = \(BE^2k\) (where k, is constant of proportionality)
& area (ACD) = \(CD^2k\)
& k = AB/AC=BE/CD=1/2
Thus, area (trap BCDE) = area (ACD) - area (ABC)
=100k-64k=36k
=36(1/2)
=18

Answer is (B)

Regards,

The problem I am having with this question is, why are we assuming that ADC is a triangle at all? It looks like a triangle, but AB and AE also look like they are equivalent lengths, which they are not. So would couldn't BC and ED skew off into different directions while still maintaining the parallel nature of BE and CD? Because I didn't know why we should make these assumptions, I guessed. I thought we were always supposed to assume that the picture is not drawn to scale.

BE || CD and B is the middle of AC, implies that E is the middle of AD, so AD = 8.
Since \(AC^2+AD^2=CD^2 \, \, (6^2+8^2=10^2)\) we can deduce that ACD is right angled triangle
and its area is 6 * 8 / 2 = 24. The area of the small triangle ABE is 1/4 of the area of the large triangle ACD. Triangle ABE is similar to triangle ACD, similarity ratio being 1:2. The corresponding areas are in relation 1:4.

Therefore, the area of the trapezoid BEDC is 3/4 of the area of the large triangle i.e. (3/4) * 24 = 18.

Answer B.

Merging similar topics. Please refer to the solutions above.

As for your doubt, OG13, page150:
Figures: A figure accompanying a problem solving question is intended to provide information useful in solving the problem. Figures are drawn as accurately as possible. Exceptions will be clearly noted. Lines shown as straight are straight, and lines that appear jagged are also straight. The positions of points, angles, regions, etc., exist in the order shown, and angle measures are greater than zero. All figures lie in a plane unless otherwise indicated.

OG13, page 272:
A figure accompanying a data sufficiency problem will conform to the information given in the question but will not necessarily conform to the additional information given in statements (1) and (2).
Lines shown as straight can be assumed to be straight and lines that appear jagged can also be assumed to be straight.
You may assume that the positions of points, angles, regions, and so forth exist in the order shown and that angle measures are greater than zero degrees.
All figures lie in a plane unless otherwise indicated.

Hope it helps.

I got that how it is a right triangle..

I have another question in my mind.. that is...

If BE and CD are parallel, so does that mean BE bisecting AC?? thats y we have assumed that if AB is 3 then BC is 3 too?

No, BE || CD does not imply that BE bisects AC. The fact that BC = AB = 3 is given in the stem.

Hi Bunuel,

It's quite clear from your explanation that once we apply the properties of similar triangles, ED has to equal 4. That being said, one thing that stumped was whether ED should have been 3 or 4? If lines BE and CD are parallel, don't they have to have a uniform distance between the two? If so, why doesn't BC = ED?

EDIT: What is the main indicator that these are parallel? Is it the fact that they both have the same angle BAE or is it because they have parallel lines? Would having just one vs. the other suffice in identifying that these are parallel?

I think you are getting confused because of the figure. The figure shown is not drawn to scale, and it shows AE = AD which is not the case. It is not necessary that distance between two parallel line is always the same. Think of a line drawn perpendicular to both the parallel lines and then a line making 30 degrees with these parallel lines. The lengths would be different.

If two lines are parallel, then they are constant distance apart (definition of parallel lines). But the distance between parallel lines is perpendicular from one to another, thus neither BC nor ED is the distances between BE and CD, hence the difference.

As for the other question: we know that BE is parallel to CD, because we are directly told about it. BE || CD, means BE is parallel to CD (|| implies parallel).

Hope it's clear.

The red part is not correct: if two lines are parallel, then they are constant distance apart (definition of parallel lines).

I think you are getting confused because of the figure. The figure shown is not drawn to scale, and it shows AE = AD which is not the case. It is not necessary that distance between two parallel line is always the same. Think of a line drawn perpendicular to both the parallel lines and then a line making 30 degrees with these parallel lines. The lengths would be different.[/quote]

The red part is not correct: if two lines are parallel, then they are constant distance apart (definition of parallel lines).[/quote]

Yes, you are right. What I was trying to say was the lengths of the various lines drawn between there two parallel lines might not be the same. The distance is the length of perpendicular drawn between two parallel lines which is always the same. My bad!!!

Thanks for pointing it out.

My apologies -- I meant to ask "what is the main indicator that these triangles are similar triangles" as opposed to mistakenly asking "what is the main indicator that these are parallel lines"?

How do you know that these are similar triangles? is it because we have two parallel lines or is it because one triangle is inscribed in the other?

If one was inscribed in the other but we weren't told that the bases were parallel -- we wouldn't be able to conclude that they were similar triangles. Is that correct?

Thanks and sorry for the confusion.