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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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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 5867 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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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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Bunuel


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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Bunuel


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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Bunuel
pgmat
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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zanaik89

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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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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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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Can we not directly use formula of trapezium rather than finding area of smaller triangle and bigger triangle and then subtracting?

Yes.

Any side of a triangle can be deemed the base.
Each base has a corresponding height.
No matter which base-height combination is used, the area of the triangle must be the same value.
Implication:
Regardless of which side of a triangle is deemed the base, bh must always yield the same product.

Triangle ABE:
If AB=3 is considered the base, then AE=4 is the corresponding height.
If BE=5 is considered the base, let AF = the corresponding height.
Since each base-height combination must yield the same product, we get:
\(5AF = 3*4\)
\(AF = \frac{12}{5}\)

Triangle ACD:
If AC=6 is considered the base, then AD=8 is the corresponding height.
If CD=10 is considered the base, let AG = the corresponding height.
Since each base-height combination must yield the same product, we get:
\(10AG = 6*8\)
\(AG = \frac{48}{10} = \frac{24}{5}\)

Trapezoid BEDC:
height \(= AG - AF = \frac{24}{5} - \frac{12}{5} = \frac{12}{5}\)
area = (average of the two bases)(height) \(= \frac{BE+CD}{2} * \frac{12}{5} = \frac{5+10}{2} * \frac{12}{5} = 18\)
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These questions are not part of the new gmat focus edition are they?
enigma123

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

Attachment:
Trapezoid.GIF
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frostingontoast
These questions are not part of the new gmat focus edition are they?


No, you can ignore this question.
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