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E is the midpoint of AC in right triangle ABC shown above. If the area
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22 Sep 2016, 04:05
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60% (03:08) correct 40% (02:59) wrong based on 106 sessions
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E is the midpoint of AC in right triangle ABC shown above. If the area
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22 Sep 2016, 05:13
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. \(6\sqrt{\frac{2}{3}}\) 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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E is the midpoint of AC in right triangle ABC shown above. If the area
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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 454590, 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 454590 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
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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. \(6\sqrt{\frac{2}{3}}\) \(24=\frac{1}{2}*base*height\) we know that base and height are same( it is an isosles triangle) \(SIDE= \sqrt{48}\) \(AE=EC=\sqrt{48}/2\) Area of ABE= \(\frac{1}{2}*\sqrt{48}*\sqrt{48}/2=12\) \(AEB=BDE\) Answer C: 12



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Re: E is the midpoint of AC in right triangle ABC shown above. If the area
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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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Re: E is the midpoint of AC in right triangle ABC shown above. If the area
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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
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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 = 123 = 9



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E is the midpoint of AC in right triangle ABC shown above. If the area
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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
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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.
Correct Answer is D.



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