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Area: Do you see a correct answer choice here?

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03 Feb 2011, 07:43
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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
[Reveal] Spoiler: OA

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03 Feb 2011, 07:54
gmatpapa wrote:
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.
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Current Student
Status: Up again.
Joined: 31 Oct 2010
Posts: 541
Concentration: Strategy, Operations
GMAT 1: 710 Q48 V40
GMAT 2: 740 Q49 V42
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Kudos [?]: 397 [0], given: 75

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03 Feb 2011, 08:02
Bunuel wrote:
gmatpapa wrote:
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

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03 Feb 2011, 08:07
gmatpapa wrote:
Bunuel wrote:
gmatpapa wrote:
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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Joined: 31 Oct 2010
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03 Feb 2011, 08:15
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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05 Feb 2011, 04:03
i did a bit different - x^2+x^2 = 10^2
x=squrt2*5
x^2 = 2*25 = 50
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05 Feb 2011, 05:01
144144 wrote:
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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05 Feb 2011, 09:23
ye, i noticed it after i posted. thanks.
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Re: Area: Do you see a correct answer choice here?   [#permalink] 05 Feb 2011, 09:23
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