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

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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 : √2

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

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Re: Triangle ABC is right angled at B. BD, the median to hypotenuse AC, is [#permalink]

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(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: Triangle ABC is right angled at B. BD, the median to hypotenuse AC, is [#permalink]

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New post 04 Jun 2011, 09:54
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: Triangle ABC is right angled at B. BD, the median to hypotenuse AC, is [#permalink]

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New post 06 Jun 2011, 03:37
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: Triangle ABC is right angled at B. BD, the median to hypotenuse AC, is [#permalink]

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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem.
Remember equal number of variables and independent equations ensures a solution.

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

Transforming the original condition and the question, we have the below image.

In case of the triangle we have 3 variables and 2 equations (ang B=90 degree , BD=5) thus we need 1 more equation to match the number of variables and equations. Therefore there is high probability that D is the answer.
In case of 1), from 1:1:root2, we have root2*AC=5 and thus the length of AC. Then we have AB=BC=5root2, therefore the area is (1/2)(5root2)^2=25. The condition is sufficient.
In case of 2), we have AC=10 but no information regarding the length of AB,BC. Therefore the condition is not sufficient.
The answer is A.

Normally for cases where we need 1 more equation, such as original conditions with 1 variable, or 2 variables and 1 equation, or 3 variables and 2 equations, we have 1 equation each in both 1) and 2). Therefore D has a high chance of being the answer, which is why we attempt to solve the question using 1) and 2) separately. Here, there is 59 % chance that D is the answer, while A or B has 38% chance. There is 3% chance that C or E is the answer for the case. Since D is most likely to be the answer according to DS definition, we solve the question assuming D would be our answer hence using 1) and 2) separately. Obviously there may be cases where the answer is A, B, C or E.
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Re: Triangle ABC is right angled at B. BD, the median to hypotenuse AC, is [#permalink]

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Can someone explain why my reasoning behind St2 being sufficient is incorrect??

So we're given the line between the corner at B to the midpoint of AC, which, if you flip the triangle, is the height of the triangle. Now all you need to solve for the area is the base, which is given by the statement.
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Re: Triangle ABC is right angled at B. BD, the median to hypotenuse AC, is [#permalink]

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New post 23 Sep 2015, 16:58
ugur wrote:
Can someone explain why my reasoning behind St2 being sufficient is incorrect??

So we're given the line between the corner at B to the midpoint of AC, which, if you flip the triangle, is the height of the triangle. Now all you need to solve for the area is the base, which is given by the statement.

We can't be certain that BD is the height of the triangle.
In fact, the only time it will be the height is when we have a 45-45-90 triangle.

Cheers,
Brent
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Re: Triangle ABC is right angled at B. BD, the median to hypotenuse AC, is [#permalink]

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New post 24 Sep 2015, 04:53
GMATPrepNow wrote:
ugur wrote:
Can someone explain why my reasoning behind St2 being sufficient is incorrect??

So we're given the line between the corner at B to the midpoint of AC, which, if you flip the triangle, is the height of the triangle. Now all you need to solve for the area is the base, which is given by the statement.

We can't be certain that BD is the height of the triangle.
In fact, the only time it will be the height is when we have a 45-45-90 triangle.

Cheers,
Brent


Why, we can prove and,in fact, we get 45-45-90 triangle ABC
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Re: Triangle ABC is right angled at B. BD, the median to hypotenuse AC, is [#permalink]

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New post 24 Sep 2015, 05:31
kiskinen wrote:
GMATPrepNow wrote:
ugur wrote:
Can someone explain why my reasoning behind St2 being sufficient is incorrect??

So we're given the line between the corner at B to the midpoint of AC, which, if you flip the triangle, is the height of the triangle. Now all you need to solve for the area is the base, which is given by the statement.

We can't be certain that BD is the height of the triangle.
In fact, the only time it will be the height is when we have a 45-45-90 triangle.

Cheers,
Brent


Why, we can prove and,in fact, we get 45-45-90 triangle ABC


I dont know what you are trying to say but GMATPrepNow is correct. Statement 2 does not mention that the median is also perpendicular to the hypotenuse AC.

If you are asking how can we say that ABC is 45-45-90 triangle per statement 1, you need to remember that for any right angled triangle having sides in the ratio \(1:1:sqrt{2}\), then it MUST be 45-45-90 triangle.

This can be seen as (this proof is easiest using trigonometry but I will follow the standard GMAT treatment): lets say the sides are \(x,x, x*\sqrt{2}\). You know that angles opposite equal sides are equal. Thus, if B is the right angle such that AB=BC=x and Ac = \(x*\sqrt{2}\), then \(\angle{BAC} = \angle{BCA}=Z\)

Now, per the property of any triangle, \(\angle{ABC} + \angle{BAC}+\angle{BCA} = 180\) ---> 90+Z+Z=180 ---> Z=45 --->\(\angle{BAC}=\angle{BCA} = 45\)

Thus triangle ABC is a 45-45-90 triangle.

We can arrive at this result using statement 2 and hence statement 2 is NOT sufficient to find the area of the triangle.

Hope this helps.
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Re: Triangle ABC is right angled at B. BD, the median to hypotenuse AC, is [#permalink]

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New post 10 Jun 2016, 03:15
subhashghosh wrote:
(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



triangle ABD ~ triangle ACB ???????????? Why . It doesnt follow any of the rule.

Congruence of triangles Two triangles are congruent if their corresponding sides are equal in length and their corresponding angles are equal in size.

1. SAS (Side-Angle-Side): If two pairs of sides of two triangles are equal in length, and the included angles are equal in measurement, then the triangles are congruent.

2. SSS (Side-Side-Side): If three pairs of sides of two triangles are equal in length, then the triangles are congruent.

3. ASA (Angle-Side-Angle): If two pairs of angles of two triangles are equal in measurement, and the included sides are equal in length, then the triangles are congruent.
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Re: Triangle ABC is right angled at B. BD, the median to hypotenuse AC, is [#permalink]

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New post 10 Jun 2016, 06:32
abrakadabra21 wrote:
subhashghosh wrote:
(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



triangle ABD ~ triangle ACB ???????????? Why . It doesnt follow any of the rule.

Congruence of triangles Two triangles are congruent if their corresponding sides are equal in length and their corresponding angles are equal in size.

1. SAS (Side-Angle-Side): If two pairs of sides of two triangles are equal in length, and the included angles are equal in measurement, then the triangles are congruent.

2. SSS (Side-Side-Side): If three pairs of sides of two triangles are equal in length, then the triangles are congruent.

3. ASA (Angle-Side-Angle): If two pairs of angles of two triangles are equal in measurement, and the included sides are equal in length, then the triangles are congruent.


If you read carefully, ~ is a symbol used for triangle SIMILARITY and not CONGRUENCY.
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Re: Triangle ABC is right angled at B. BD, the median to hypotenuse AC, is [#permalink]

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New post 10 Jun 2016, 07:06
soaringAlone wrote:
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


Statement 1 tells ... triangle is isosceles .

1:1:sqrt 2

Thus can calculate the area .

Statement 2 tells AC 10 and thus we can tell AD =DC=5... BUT NOTHING ABOUT THE AREA .

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Re: Triangle ABC is right angled at B. BD, the median to hypotenuse AC, is [#permalink]

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Re: Triangle ABC is right angled at B. BD, the median to hypotenuse AC, is   [#permalink] 28 Dec 2017, 05:27
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