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gmatpapa
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What is the area of a square that has a diagonal of length 10?

A. 5
B. 10
C. 20
D. 40
E. 45

Source: Cliff's GMAT

\(area_{square}=side^2=\frac{diagonal^2}{2}\): so A to be the answer either option A should read 50 instead of 5 or diagonal should be \(\sqrt{10}\) instead of 10.

Yes. Answer quite clearly should be 50.

Does this OE make sense to you:
From the relationship that exists in a right triangle, we see that the sides of the square
must equal the square root of 5. Thus the area of the square is 5

As OE says that \(side=\sqrt{5}\) then there must be the second case I suggested: diagonal should be \(\sqrt{10}\) instead of 10.
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Yep. Quite correct as diagonal = \(sqrt2\)*side. So side will be \(\frac{sqrt10}{sqrt2}\)= \(sqrt 5\). And area will be \(side^2= 5\).
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i did a bit different - x^2+x^2 = 10^2
x=squrt2*5
x^2 = 2*25 = 50
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i did a bit different - x^2+x^2 = 10^2
x=squrt2*5
x^2 = 2*25 = 50

Its essentially the same thing if you look at it. Diagonal= \(sqrt2\)*side is arrived using the same formula that you have used. (Pythagoras theorem)
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ye, i noticed it after i posted. thanks.

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