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Triangle ABC is right angled at B. BD, the median to

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Triangle ABC is right angled at B. BD, the median to [#permalink] New post 04 Jun 2011, 08:08
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27% (01:37) correct 73% (00:10) wrong based on 11 sessions
Triangle ABC is right angled at B. BD, the median to hypotenuse AC, is 5 units. What is the area of the triangle?

1. The sides of triangle ABC are in the ratio 1 : 1 : root 2

2. The hypotenuse AC is 10 units
[Reveal] Spoiler: OA

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Re: DS Geometry Problem -Right Traiangle- Median [#permalink] New post 04 Jun 2011, 08:47
(1)
The median is perpendicular to the hypotenuse, as it's an isosceles right triangle.

The ratio of sides is known, and triangle ABD ~ triangle ACB

AB/AC = BD/AB

AB/AC = 1/root(2) = 5/AB

AB = 5root(2)

BC = 5root(2)

So area of triangle ABC = 1/2 * 25 * 2 = 25

Sufficient

(2)

It is not clear if the median is perpendicular to the hypotenuse

The other sides are not known either.

Not Sufficient

Answer - A
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Re: DS Geometry Problem -Right Traiangle- Median [#permalink] New post 04 Jun 2011, 08:54
Median divides the hypotenuse into two equal lengths, so AD = DC and forms 2 isosceles triangles.
Area of triangle = 1/2 * AB*BC

so if BD =5, AD = DC = 5
Hence AC = 10

St - 1 given the ratio of AB to BC to AC
So we can find AB and BC = AC/sqrt(2) = 10/(sqrt(2) = 5 sqrt(2)
So we can find the area of triangle = 1/2 * 5 sqrt(2) * 5 sqrt(2) = 25

St-2 gives us already known information. We cannot determine the value of AB or BC from this. so insufficient.

Hence A.
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Re: DS Geometry Problem -Right Traiangle- Median [#permalink] New post 06 Jun 2011, 02:37
a in triangles CDB and ABC,

5/AB = AB/AC giving AB^2 = 5 * 2^(1/2)
thus AC can be found out and hence the area = 1/2 * AB * BC.
Sufficient.

b AD = CD = 5 each.
However AB and BC have no relative information given. Not sufficient.

Thus A
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Re: DS Geometry Problem -Right Traiangle- Median   [#permalink] 06 Jun 2011, 02:37
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