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

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Bunuel
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Re: If BE || CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of [#permalink] New post  20 May 2014, 00:04
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russ9 wrote:
Bunuel wrote:
russ9 wrote:

If BE || CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of trapezoid BEDC?
A. 12
B. 18
C. 24
D. 30
E. 48

Triangles ABE and ACD are similar (they share the same angle CAD and also as BE is parallel to CD then angles by BE and CD are equal, so all 3 angles of these triangles are equal so they are similar triangles).

Property of similar triangles: ratio of corresponding sides are the same: \(\frac{AB}{AC}=\frac{BE}{CD}\) --> \(\frac{3}{6}=\frac{BE}{10}\) --> \(BE=5\) and \(AD=2AE=8\).

So in triangle ABE sides are \(AB=3\), \(AE=4\) and \(BE=5\): we have 3-4-5 right triangle ABE (with right angle CAD, as hypotenuse is \(BE=5\)) and 6-8-10 right angle triangle ACD.

Now, the \(area_{BEDC}=area_{ACD}-area_{ABE}\) --> \(area_{BEDC}=\frac{6*8}{2}-\frac{3*4}{2}=18\).

Answer: B.

Also discussed here: trapezium-area-99966.html

Hope it helps.


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.


Similar Triangles are triangles in which the three angles are identical. It is only necessary to determine that two sets of angles are identical in order to conclude that two triangles are similar; the third set will be identical because all of the angles of a triangle always sum to 180º.

So, for our question, triangles ABE and ACD are similar because their three angles are identical: they share the same angle CAD and also as BE is parallel to CD then angles by BE and CD are equal, so all 3 angles of these triangles are equal, which means that they are similar.
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Re: If BE || CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of [#permalink] New post  31 Jul 2014, 22:46
Bunuel wrote:
Attachment:
Trapezoid.GIF
If BE || CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of trapezoid BEDC?
A. 12
B. 18
C. 24
D. 30
E. 48

Triangles ABE and ACD are similar (they share the same angle CAD and also as BE is parallel to CD then angles by BE and CD are equal, so all 3 angles of these triangles are equal so they are similar triangles).

Property of similar triangles: ratio of corresponding sides are the same: \(\frac{AB}{AC}=\frac{BE}{CD}\) --> \(\frac{3}{6}=\frac{BE}{10}\) --> \(BE=5\) and \(AD=2AE=8\).

So in triangle ABE sides are \(AB=3\), \(AE=4\) and \(BE=5\): we have 3-4-5 right triangle ABE (with right angle CAD, as hypotenuse is \(BE=5\)) and 6-8-10 right angle triangle ACD.

Now, the \(area_{BEDC}=area_{ACD}-area_{ABE}\) --> \(area_{BEDC}=\frac{6*8}{2}-\frac{3*4}{2}=18\).

Answer: B.

Also discussed here: trapezium-area-99966.html

Hope it helps.


HI Bunuel,

Can we directly use here prop BE || CD . so can we say BE = 1/2(CD) = 1/2*10 = 5

Thanks
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Re: If BE || CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of [#permalink] New post  01 Aug 2014, 00:13
PathFinder007 wrote:
Bunuel wrote:
Attachment:
Trapezoid.GIF
If BE || CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of trapezoid BEDC?
A. 12
B. 18
C. 24
D. 30
E. 48

Triangles ABE and ACD are similar (they share the same angle CAD and also as BE is parallel to CD then angles by BE and CD are equal, so all 3 angles of these triangles are equal so they are similar triangles).

Property of similar triangles: ratio of corresponding sides are the same: \(\frac{AB}{AC}=\frac{BE}{CD}\) --> \(\frac{3}{6}=\frac{BE}{10}\) --> \(BE=5\) and \(AD=2AE=8\).

So in triangle ABE sides are \(AB=3\), \(AE=4\) and \(BE=5\): we have 3-4-5 right triangle ABE (with right angle CAD, as hypotenuse is \(BE=5\)) and 6-8-10 right angle triangle ACD.

Now, the \(area_{BEDC}=area_{ACD}-area_{ABE}\) --> \(area_{BEDC}=\frac{6*8}{2}-\frac{3*4}{2}=18\).

Answer: B.

Also discussed here: trapezium-area-99966.html

Hope it helps.


HI Bunuel,

Can we directly use here prop BE || CD . so can we say BE = 1/2(CD) = 1/2*10 = 5

Thanks


BC=AB=3,meaning B is the mid point of AB and since BE|| CD means E is the mid point of AD and hence by Mid point theorem ..BE=5
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Re: If BE || CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of [#permalink] New post  07 Sep 2014, 18:35
How can we determine the triangle heights? I've read through the above and haven't been able to grasp it. I'm sure its very simple, so thanks in advance, and excuse my oversight.
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Re: If BE || CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of [#permalink] New post  07 Sep 2014, 21:58
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JackSparr0w wrote:
How can we determine the triangle heights? I've read through the above and haven't been able to grasp it. I'm sure its very simple, so thanks in advance, and excuse my oversight.


No problem!!

Attachment:
Untitled.jpg
Untitled.jpg [ 24.18 KiB | Viewed 4604 times ]

Notice that Triangle BAE is right angled at A and Angle A=90..
So BA is perpendicular to AE...so Area of ABE will be 1/2*3*4=6
Area of Bigger Triangle CAD (right angled at A) =1/2*6*8=24

So area of Trapezium is Area of CAD -Area of BAE= 24-6=18
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Re: If BE || CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of [#permalink] New post  06 Jun 2016, 08:27
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enigma123 wrote:
Attachment:
Trapezoid.GIF
If BE || CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of trapezoid BEDC?

A. 12
B. 18
C. 24
D. 30
E. 48


Consider the triangle ABE,
The two sides are 3 and 4. Hence the third side would be 5 (3 - 4 - 5 right triangle)
Area = 1/2 * 3 * 4 = 6

In triangle ABC,
The sides of this triangle are twice the sides of triangle ABE, hence area would be 2^2 times.
Area of ABC = 4*6 = 24

Area od the trapezoid = 24 - 6 = 18

Correct Option: B
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If BE || CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of [#permalink] New post  15 Oct 2017, 02:40
Bunuel wrote:


If BE || CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of trapezoid BEDC?
A. 12
B. 18
C. 24
D. 30
E. 48

Triangles ABE and ACD are similar (they share the same angle CAD and also as BE is parallel to CD then angles by BE and CD are equal, so all 3 angles of these triangles are equal so they are similar triangles).

Property of similar triangles: ratio of corresponding sides are the same: \(\frac{AB}{AC}=\frac{BE}{CD}\) --> \(\frac{3}{6}=\frac{BE}{10}\) --> \(BE=5\) and \(AD=2AE=8\).

So in triangle ABE sides are \(AB=3\), \(AE=4\) and \(BE=5\): we have 3-4-5 right triangle ABE (with right angle CAD, as hypotenuse is \(BE=5\)) and 6-8-10 right angle triangle ACD.

Now, the \(area_{BEDC}=area_{ACD}-area_{ABE}\) --> \(area_{BEDC}=\frac{6*8}{2}-\frac{3*4}{2}=18\).

Answer: B.

Also discussed here: https://gmatclub.com/forum/trapezium-area-99966.html

Hope it helps.


Hi Bunuel, Could you help to explain the area? Why we could use the side to calculate the area of triangle?

While formula for area of triangle = (1/2)*(Base)*(Height)
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Re: If BE || CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of [#permalink] New post  15 Oct 2017, 02:52
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hazelnut wrote:
Bunuel wrote:


If BE || CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of trapezoid BEDC?
A. 12
B. 18
C. 24
D. 30
E. 48

Triangles ABE and ACD are similar (they share the same angle CAD and also as BE is parallel to CD then angles by BE and CD are equal, so all 3 angles of these triangles are equal so they are similar triangles).

Property of similar triangles: ratio of corresponding sides are the same: \(\frac{AB}{AC}=\frac{BE}{CD}\) --> \(\frac{3}{6}=\frac{BE}{10}\) --> \(BE=5\) and \(AD=2AE=8\).

So in triangle ABE sides are \(AB=3\), \(AE=4\) and \(BE=5\): we have 3-4-5 right triangle ABE (with right angle CAD, as hypotenuse is \(BE=5\)) and 6-8-10 right angle triangle ACD.

Now, the \(area_{BEDC}=area_{ACD}-area_{ABE}\) --> \(area_{BEDC}=\frac{6*8}{2}-\frac{3*4}{2}=18\).

Answer: B.

Also discussed here: https://gmatclub.com/forum/trapezium-area-99966.html

Hope it helps.


Hi Bunuel, Could you help to explain the area? Why we could use the side to calculate the area of triangle?

While formula for area of triangle = (1/2)*(Base)*(Height)


ABE and ACD are right triangles: area = 1/2*(leg1)(leg2), which is the same as (1/2)*(Base)*(Height), because legs in a right triangle are perpendicular to each other (so we can consider one of them to be base and another to be the height).
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Re: If BE || CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of [#permalink] New post  24 Apr 2018, 07:12
Bunuel wrote:
pgmat wrote:
Quote:
So in triangle ABE sides are AB=3, AE=4 and BE=5: we have 3-4-5 right triangle ABE (with right angle CAD, as hypotenuse is BE=5) and 6-8-10 right angle triangle ACD.


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


Yes, any triangle whose sides are in the ratio 3:4:5 is a right triangle. Such triangles that have their sides in the ratio of whole numbers are called Pythagorean Triples. There are an infinite number of them, and this is just the smallest. If you multiply the sides by any number, the result will still be a right triangle whose sides are in the ratio 3:4:5. For example 6, 8, and 10.

A Pythagorean triple consists of three positive integers \(a\), \(b\), and \(c\), such that \(a^2 + b^2 = c^2\). Such a triple is commonly written \((a, b, c)\), and a well-known example is \((3, 4, 5)\). If \((a, b, c)\) is a Pythagorean triple, then so is \((ka, kb, kc)\) for any positive integer \(k\).

For more on this check Triangles chapter of Math Book: https://gmatclub.com/forum/math-triangles-87197.html

Hope it helps.


Hi Bunuel,

Why are we considering the height as 8 and 4 can u pls tell me ?

How can 8 and 4 be the height?
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If BE || CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of [#permalink] New post  24 Apr 2018, 11:58
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zanaik89 wrote:
Hi Bunuel,

Why are we considering the height as 8 and 4 can u pls tell me ?

How can 8 and 4 be the height?


A height (altitude) of a triangle is the perpendicular from a vertex to the opposite side. Because legs (non-hypotenuse sides) in a right triangle are perpendicular to each other we can consider one of them to be the base and another to be the height.
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Re: If BE || CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of [#permalink] New post  21 Jul 2021, 10:14
Can we not directly use formula of trapezium rather than finding area of smaller triangle and bigger triangle and then subtracting?
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Re: If BE || CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of [#permalink] New post  21 Jul 2021, 11:30
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barolia wrote:
Can we not directly use formula of trapezium rather than finding area of smaller triangle and bigger triangle and then subtracting?


The area of a trapezoid is (a + b)/2*h, where a and b is the lengths of the parallel sides and h is the length of the height. To apply it directly we are missing the lengths of BE and the height.
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