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Re: Trapezium area [#permalink]
28 Aug 2010, 10:10

Expert's post

jananijayakumar wrote:

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

Attachment:

Untitled.jpg

I guess it's BE||CD (BE is parallel to CD).

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

Re: Trapezium area [#permalink]
28 Aug 2010, 11:30

Bunuel - In this case it turned out to be a rt angled triangle so the area calc was easy.. Lets say the bigger triangle were 5 7 9 in sides and the smaller was correspondingly half in size inside of it.. In that case, what would be the easiest way to calculate the trapezium area.. I guess does GMAT require us to remember the formula of the area of a triangle (not the .5 base height) but the one with the sides? Is there an easy way to calculate the area where the sides are given for any triangle? _________________

Re: Trapezium area [#permalink]
28 Aug 2010, 11:48

Expert's post

mainhoon wrote:

Bunuel - In this case it turned out to be a rt angled triangle so the area calc was easy.. Lets say the bigger triangle were 5 7 9 in sides and the smaller was correspondingly half in size inside of it.. In that case, what would be the easiest way to calculate the trapezium area.. I guess does GMAT require us to remember the formula of the area of a triangle (not the .5 base height) but the one with the sides? Is there an easy way to calculate the area where the sides are given for any triangle?

If it were not right triangle we could calculate the area of triangles with Heron's Formula (as we still would knew the lengths of all sides), though I've never seen a GMAT question requiring it. _________________

Re: Trapezium area [#permalink]
13 Sep 2010, 06:18

Bunuel wrote:

jananijayakumar wrote:

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

Attachment:

Untitled.jpg

I guess it's BE||CD (BE is parallel to CD).

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.

Hope it's clear.

Bunuel... If you say that BC = AB = 3, AE = 4 then how can BE||CD (BE is parallel to CD) true. I am not able to visualize such a figure!!!

Re: Trapezium area [#permalink]
13 Sep 2010, 07:42

Expert's post

utin wrote:

Bunuel wrote:

jananijayakumar wrote:

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

Attachment:

Untitled.jpg

I guess it's BE||CD (BE is parallel to CD).

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.

Hope it's clear.

Bunuel... If you say that BC = AB = 3, AE = 4 then how can BE||CD (BE is parallel to CD) true. I am not able to visualize such a figure!!!

Why not? Consider triangle CAD, where CA=6 and AD=8. Now draw midsegment BE of a triangle (a line segment joining the midpoints of two sides of a triangle). Properties of a midsegment:

• The midsegment is always parallel to the third side of the triangle. • The midsegment is always half the length of the third side.

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