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Gemetry question

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by GMATGuruNY » Sun Aug 16, 2015 2:40 am
Test an EASY CASE.

In a 45-45-90 triangle, the sides are in the following ratio:
1 : 1 : √2.

Let ∆ABD, ∆BCD, and ∆ABC each be a 45-45-90 triangle, as follows:
Image
In the figure above:
BD=1, BC=√2, and AB=√2.
Test these values in the answer choices.
Only B works:

B: 1/(BD)² = 1/(BC)² + 1/(AB)²
1/1² = 1/(√2)² + 1/(√2)²
1 = 1/2 + 1/2
1 = 1.

The correct answer is B.
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by lucas211 » Mon Aug 17, 2015 4:05 am
Thank you for the explanation :-)
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by sandipgumtya » Mon Aug 17, 2015 8:30 pm
GMATGuruNY wrote:Test an EASY CASE.

In a 45-45-90 triangle, the sides are in the following ratio:
1 : 1 : √2.

Let ∆ABD, ∆BCD, and ∆ABC each be a 45-45-90 triangle, as follows:
Image
In the figure above:
BD=1, BC=√2, and AB=√2.
Test these values in the answer choices.
Only B works:

B: 1/(BD)² = 1/(BC)² + 1/(AB)²
1/1² = 1/(√2)² + 1/(√2)²
1 = 1/2 + 1/2
1 = 1.

The correct answer is B.

Hi Mitch,
How did u get this to be a 45-45-90 triangle?Cant it be done without assuming this.Moreover in this case is it right that BD=1/2AC.Pl explain.
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by GMATGuruNY » Tue Aug 18, 2015 11:14 am
Let ∆ABD, ∆BCD, and ∆ABC each be a 45-45-90 triangle, as follows:
Image
Hi Mitch,
How did u get this to be a 45-45-90 triangle?Cant it be done without assuming this.Moreover in this case is it right that BD=1/2AC.Pl explain.
In a must-be-true problem, the correct answer choice must be true for ANY CASE THAT SATISFIES THE GIVEN CONSTRAINTS.
The figure in the posted problem indicates the following constraint:
∆ABD, ∆BCD, and ∆ABC must all be right triangles.
I tested an easy case that satisfies this constraint.
While ∆ABD, ∆BCD, and ∆ABC do not HAVE to be 45-45-90 triangles, they all COULD be 45-45-90 triangles.
Thus, the correct answer choice must be true for the lengths shown in my post above.
Since only B is true when BD=1, BC=√2, and AB=√2, the correct answer is B.
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by sandipgumtya » Wed Aug 19, 2015 6:04 am
thanks mitch,got ur point.bt pl clarify my one doubt:for any right triangle is the altitude drawn to the hypotaneous is equal to half of hypotaneous?
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by GMATGuruNY » Wed Aug 19, 2015 6:44 am
sandipgumtya wrote:thanks mitch,got ur point.bt pl clarify my one doubt:for any right triangle is the altitude drawn to the hypotaneous is equal to half of hypotaneous?
An altitude drawn to the hypotenuse of a 45-45-90 triangle will be equal to half the hypotenuse.
An altitude drawn to the hypotenuse of any OTHER type of right triangle will NOT be equal to the half the hypotenuse.

Other problems involving an altitude drawn to the hypotenuse of a right triangle:
https://www.beatthegmat.com/data-suffici ... 85748.html (second post)
https://www.beatthegmat.com/inscribed-tr ... 74152.html
https://www.beatthegmat.com/length-of-th ... 71979.html
https://www.beatthegmat.com/geo-question ... nta-14-649
https://www.beatthegmat.com/geomatry-t273588.html
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