Please may someone explain why the height of the parallelogram in this question is sqrt(2)/2 and not 1/sqrt(2)?
[Problem text removed by editor]
If the height of the parallelogram P is perpendicular to the base forming a 90 degree angle then, given statement (1), it will form an internal 45-45-90 triangle with hypotenuse of 1.
Therefore if x sqrt(2) = 1 then surely x = 1/sqrt(2). The height of P forms one of the legs of this 45-45-90 triangle and so should also equal 1/sqrt(2) and not sqrt(2)/1.
Indeed, using a^2+b^2=c^2 and inserting 1/sqrt(2) as a and b we get 1 as the hypotenuse: (1/sqrt(2))^2+(1/sqrt(2))^2 = 1
Am I doing something wrong here?
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- FrancisP624
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Re: OG 2020 DS 317
Please read the forum guidelines before posting; unfortunately we can't post OG problems here due to copyright.
I'm not exactly sure what your question about this problem is. Remember that in DS, we don't actually have to calculate the numerical answer to the question, but simply need to say whether we have enough information to do so. For that reason, this is how I'd think about the problem:
Statement (1) tells me one angle of the parallelogram, from which I can deduce the other angles. This, along with the side lengths given in the question stem, "fixes" the parallelogram, i.e. there's only one size of parallelogram with these features. Consequently I can deduce the area. The same applies to statement (2): if I know the height (that's what 'altitude' means) and the side lengths, the parallelogram is fixed.
It looks like you're doing an extra task by actually trying to work out the area. This can be a useful bonus challenge, so long as you remember that you don't need to do this if you were solving the problem under test conditions. I agree that the height of the parallelogram is 1/sqroot(2), given that the hypotenuse of the 45-45-90 triangle is 1. In DS problems the statements are consistent, so that's perhaps the presentation of statement (2) as sqroot(2)/2 is confusing. However, remember that 2 = sqroot(2) x sqroot(2), so that by dividing top and bottom by sqroot(2), you can show that 1/sqroot(2) = sqroot(2) / 2
I'm not exactly sure what your question about this problem is. Remember that in DS, we don't actually have to calculate the numerical answer to the question, but simply need to say whether we have enough information to do so. For that reason, this is how I'd think about the problem:
Statement (1) tells me one angle of the parallelogram, from which I can deduce the other angles. This, along with the side lengths given in the question stem, "fixes" the parallelogram, i.e. there's only one size of parallelogram with these features. Consequently I can deduce the area. The same applies to statement (2): if I know the height (that's what 'altitude' means) and the side lengths, the parallelogram is fixed.
It looks like you're doing an extra task by actually trying to work out the area. This can be a useful bonus challenge, so long as you remember that you don't need to do this if you were solving the problem under test conditions. I agree that the height of the parallelogram is 1/sqroot(2), given that the hypotenuse of the 45-45-90 triangle is 1. In DS problems the statements are consistent, so that's perhaps the presentation of statement (2) as sqroot(2)/2 is confusing. However, remember that 2 = sqroot(2) x sqroot(2), so that by dividing top and bottom by sqroot(2), you can show that 1/sqroot(2) = sqroot(2) / 2
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