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21 Aug 2021, 13:42
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EMPOWERgmatRichC
You are always so helpful... I have a clarifying question based on reading the Official Guide's answer. To actually calculate the area is it just (1)(1/root(2))? I see the answer as (root(2))/2, if my understanding is correct, why did the Official Guide feel the need to multiple the 1/root(2) times root(2) in the numerator and denominator... is it because you never want an answer with a radical in the bottom? Thank you so much in advance
21 Aug 2021, 16:59
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Hi woohoo921,
Yes - to both of your questions. The area of a parallelogram is (Base)(Height) - and by the "Height", we really mean the "Altitude." As a rule, it's generally frowned upon to have a radical in the denominator of a fraction - so 'canceling it out' (by multiplying by another radical) is often done. I've seen some rare cases in which that extra step was not included, but you'll be far more likely to need to do that extra step on Test Day than not.
GMAT assassins aren't born, they're made,
Rich
Yes - to both of your questions. The area of a parallelogram is (Base)(Height) - and by the "Height", we really mean the "Altitude." As a rule, it's generally frowned upon to have a radical in the denominator of a fraction - so 'canceling it out' (by multiplying by another radical) is often done. I've seen some rare cases in which that extra step was not included, but you'll be far more likely to need to do that extra step on Test Day than not.
GMAT assassins aren't born, they're made,
Rich
05 Feb 2022, 19:49
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question on this. If item 1 had an angle other than 30, 45, or 60 (those being the angles that result in 30-60-90 or 45-45-90 triangles), would 1 alone have still been sufficient? For instance, say 1 is 65 degrees, we wouldn't have DS on this piece right?
a^2 + b^2 = 1 ^ 2
We wouldn't have enough to solve A or B in parth. theorem right?
a^2 + b^2 = 1 ^ 2
We wouldn't have enough to solve A or B in parth. theorem right?
