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# E is the midpoint of AC in right triangle ABC shown above. If the area

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Math Expert
Joined: 02 Sep 2009
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E is the midpoint of AC in right triangle ABC shown above. If the area  [#permalink]

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22 Sep 2016, 04:05
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65% (hard)

Question Stats:

60% (03:08) correct 40% (02:56) wrong based on 108 sessions

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E is the midpoint of AC in right triangle ABC shown above. If the area of ΔABC is 24, what is the area of ΔBED?

A. $$9\sqrt{3}$$
B. $$9\sqrt{2}$$
C. 12
D. 9
E. $$6\sqrt{\frac{2}{3}}$$

Attachment:

T7878.png [ 4.38 KiB | Viewed 2473 times ]

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E is the midpoint of AC in right triangle ABC shown above. If the area  [#permalink]

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22 Sep 2016, 05:13
1
Bunuel wrote:

E is the midpoint of AC in right triangle ABC shown above. If the area of ΔABC is 24, what is the area of ΔBED?

A. $$9\sqrt{3}$$
B. $$9\sqrt{2}$$
C. 12
D. 9
E. $$6\sqrt{\frac{2}{3}}$$

Attachment:
T7878.png

This is how I tried.

Since we are given that E is the midpoint of AC then AE and EC are same.

We know that when the two bases are same the areas of triangles are also same..i.e.e 24 = 12+12 for two triangles.

Now BEC triangle is 12..

In BEC = BED + EDC triangles = 12.. we can happily discard option C because BED has to be lesser than 12...

From options we can say that BED area given then we can apply our logic.

A) 9 *1.7 = 15.3 which is greater than 12...not correct
B) 9*1.4 = 12.6 > 12...not correct
D) can be
E) 6*1.4/1.7 ~ 4.6... we can say from the shaded region BED is greater than EDC and this can't be the option.

IMO option D.
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GMAT 1: 750 Q49 V44
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E is the midpoint of AC in right triangle ABC shown above. If the area  [#permalink]

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22 Sep 2016, 07:06
I have D as well:

1) Area of total triangle is 24. BE is the median of side AC and divides the triangle in 2 triangles of area 12.
2) Since the triangle ABC is right and isosceles with angles 45-45-90, and we know that area is 24, we know the length of the sides is 4*sqrt(3)
3) Since BE splits AC in half, we know that the length of EC is 2*sqrt(3), and as we have again a 45-45-90 triangle, ED is 2*sqrt(3)/sqrt(2), simplify to sqrt (6)
4) Area of EDC is therefore (sqrt(6)^2)/2 = 3
5) Area of BED is 12 -3 = 9
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Re: E is the midpoint of AC in right triangle ABC shown above. If the area  [#permalink]

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23 Sep 2016, 05:04
Bunuel wrote:

E is the midpoint of AC in right triangle ABC shown above. If the area of ΔABC is 24, what is the area of ΔBED?

A. $$9\sqrt{3}$$
B. $$9\sqrt{2}$$
C. 12
D. 9
E. $$6\sqrt{\frac{2}{3}}$$

Attachment:
T7878.png

$$24=\frac{1}{2}*base*height$$ we know that base and height are same( it is an isosles triangle)

$$SIDE= \sqrt{48}$$

$$AE=EC=\sqrt{48}/2$$

Area of ABE= $$\frac{1}{2}*\sqrt{48}*\sqrt{48}/2=12$$

$$AEB=BDE$$

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Schools: Judge"18 (A)
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Re: E is the midpoint of AC in right triangle ABC shown above. If the area  [#permalink]

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23 Sep 2016, 08:56
Azamaka, are you sure that AEB=BED? since BE is the median of AC, it should split the are of triangle ABC in half, so AEB=BCE not BED - or is there something I am missing?
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Posts: 74
Re: E is the midpoint of AC in right triangle ABC shown above. If the area  [#permalink]

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23 Sep 2016, 10:09
Since BE is the median, Area of triangle ABE = Area of triangle EBC = 1/2 * Area of triangle ABC = 12.

Also, since Angle C = 45, AB = AC.

But, 1/2 * AB*AC = 24 => AC = \sqrt{48} = 4\sqrt{3}

AE = AC/2 = 2\sqrt{3}

Now, triangle EDC is again right angled. so area of triangle EDC = 1/2 * ED * DC .

But ED = DC and ED^2 + DC^2 = 12 => ED = \sqrt{6}

So area of triangle EDC = 1/2 *\sqrt{6} *\sqrt{6} = 3

Now area of triangle BCD = Area of triangle EBC - Area of triangle EDC => 12 -3 = 9.

D.
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E is the midpoint of AC in right triangle ABC shown above. If the area  [#permalink]

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26 Sep 2016, 01:49
ABC area = 0.5 AC^2 ( SINCE AB=AC) = 24 , THUS AC = 4 SQRT3 , EC = 0.5 AC = 2 SQRT3

EDC IS 45 45 90 , THUS sides are sqrt6:sqrt6: 2sqrt3 , area of EDC = 0.5(SQRT6)^2 = 3

BE IS MEDIAN THUS BEC area is 0.5*24 = 12 , area of BED = area of BEC - are of EDC = 12-3 = 9
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E is the midpoint of AC in right triangle ABC shown above. If the area  [#permalink]

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26 Sep 2016, 02:23
Bunuel wrote:

E is the midpoint of AC in right triangle ABC shown above. If the area of ΔABC is 24, what is the area of ΔBED?

After getting ΔBEC to be 12 units, I assumed for a moment ED to be of 1 unit. I did this only to work out the ratio.
ED=DC=1. So, EC = sqrt2. AC=2(sqrt2).
Because angle C is 45, AB=AC=2(sqrt2). Hypotenuse BC = 4 = BD+DC (DC =1, as mentioned above)
BD=3. Finally, the ration of ΔBDE and ΔEDC will be in the ratio if 3:1 because they have a common height of ED.
Divide 12 in the ratio and 9 is the answer for the shaded region.

Hope it helps
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Re: E is the midpoint of AC in right triangle ABC shown above. If the area  [#permalink]

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26 Sep 2016, 03:08
It is given that, Angle ACB = 45 Degrees. So, Angle ABC = 45 Degrees and thus AB=AC
Let AC = 2x, so AB = 2x

Also given that area of ABC = 24
=> (1/2)*AB*AC = 24
=> AB*AC = 48
=> (2x) (2x) = 48
=> x = 2 sqrt(3)

and BC [= sqrt (AB^2 + AC^2)] = 4 sqrt(6)

Also, EC is half of AC. Thus, EC = x = sqrt(3)
Since, Angle DCE = 45 = Angle DEC, thus DE = DC = x/sqrt(2) = sqrt(6)
BD = BC - DC = 4 sqrt(6) - sqrt(6) = 3 sqrt(6)

Hence, Area of Triangle BDE = (1/2)*BD*DE = (1/2)*3 sqrt(6)*sqrt(6) = 9.

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Re: E is the midpoint of AC in right triangle ABC shown above. If the area  [#permalink]

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15 Dec 2017, 14:07
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Re: E is the midpoint of AC in right triangle ABC shown above. If the area   [#permalink] 15 Dec 2017, 14:07
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