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Re: DS Geometry Problem -Right Traiangle- Median [#permalink]
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.
Triangle ABC is right angled at B. BD, the median to [#permalink]
12 Sep 2015, 14:47
hiteshahire22 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 : \(\sqrt{2}\)
2. The hypotenuse AC is 10 units
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Re: Triangle ABC is right angled at B. BD, the median to [#permalink]
14 Sep 2015, 02:41
2
This post received KUDOS
Expert's post
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.
Re: Triangle ABC is right angled at B. BD, the median to [#permalink]
23 Sep 2015, 15:52
1
This post was BOOKMARKED
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.
Re: Triangle ABC is right angled at B. BD, the median to [#permalink]
23 Sep 2015, 15:58
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 _________________
Brent Hanneson - Founder of GMAT Prep Now, a free & comprehensive GMAT course with: - over 500 videos (35 hours of instruction) - over 800 practice questions - 2 full-length practice tests and other bonus offers - http://www.gmatprepnow.com/
Re: Triangle ABC is right angled at B. BD, the median to [#permalink]
24 Sep 2015, 03:53
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
Re: Triangle ABC is right angled at B. BD, the median to [#permalink]
24 Sep 2015, 04:31
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.
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