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If the length of a diagonal of a square is 2x^(1/2), what is the area

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parkhydel
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If the length of a diagonal of a square is 2x^(1/2), what is the area [#permalink] New post   28 Apr 2020, 07:37
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If the length of a diagonal of a square is \(2\sqrt{x}\), what is the area of the square in terms of x ?


A. \(\sqrt{x}\)

B. \(\sqrt{2x}\)

C. \(2\sqrt{x}\)

D. x

E. 2x


PS60231.02
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E
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BrentGMATPrepNow
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Re: If the length of a diagonal of a square is 2x^(1/2), what is the area [#permalink] New post   28 Apr 2020, 08:35
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parkhydel wrote:
If the length of a diagonal of a square is \(2\sqrt{x}\), what is the area of the square in terms of x ?


A. \(\sqrt{x}\)

B. \(\sqrt{2x}\)

C. \(2\sqrt{x}\)

D. x

E. 2x


PS60231.02


Here's what we're dealing with:


We're dealing with a right triangle
So, we can apply the Pythagorean theorem to write: k² + k² = (2√x)²
Simplify to get: 2k² = 4x
Divide both sides by 2 to get: k² = 2x

Since k = length of one side of the square, it follows that k² = the area of the square
Since k² = 2x, we can conclude that 2x = the area of the square.

Answer: E

Cheers,
Brent
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Re: If the length of a diagonal of a square is 2x^(1/2), what is the area [#permalink] New post   29 Apr 2020, 11:40
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parkhydel wrote:
If the length of a diagonal of a square is \(2\sqrt{x}\), what is the area of the square in terms of x ?


A. \(\sqrt{x}\)

B. \(\sqrt{2x}\)

C. \(2\sqrt{x}\)

D. x

E. 2x


PS60231.02




Well, if we know the relation between the diagonal and the side of a square. The problem becomes fairly simple.

Let us assume that the side of the square is 's'. The relation between the side of a square and its diagonal (d) is 'd= s \(\sqrt{2}\)

Hence,

s \(\sqrt{2}\)= 2 \(\sqrt{x}\)
(These two are equal since we are talking about the length of the same diagonal)

s = \(2 \sqrt{x}\) / \(\sqrt{2}\)

Squaring both the sides (since area of a square is (\(S^2\))

\(S^2\) = \(\frac{4x}{2}\)

\(S^2\) = 2x (Area of Square)

Hope this helps a little
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If the length of a diagonal of a square is 2x^(1/2), what is the area [#permalink] New post   03 May 2020, 06:16
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parkhydel wrote:
If the length of a diagonal of a square is \(2\sqrt{x}\), what is the area of the square in terms of x ?


A. \(\sqrt{x}\)

B. \(\sqrt{2x}\)

C. \(2\sqrt{x}\)

D. x

E. 2x


PS60231.02


Solution:

For a square, diagonal = side√2, so we have:

2√x= side√2

2√(x/2) = side

Thus, the area of the square is [2√(x/2)]^2 = 4(x/2) = 2x.

Alternate Solution:

The area of a square with diagonal length of d is A = d^2/2. Thus, the area of the square in question is (2√x)^2 / 2 = (4x)/2 = 2x.

Answer: E
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Re: If the length of a diagonal of a square is 2x^(1/2), what is the area [#permalink] New post   31 May 2020, 00:11
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The area of a square when diagonal is give = \(\frac{1}{2}\)\(d^{2}\)

So the area now will be \(\frac{1}{2}*(2\sqrt{x})^{2}\)

= \(\frac{1}{2}\) * 4x

= 2x
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Re: If the length of a diagonal of a square is 2x^(1/2), what is the area [#permalink] New post   28 Apr 2020, 08:21
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parkhydel wrote:
If the length of a diagonal of a square is \(2\sqrt{x}\), what is the area of the square in terms of x ?


A. \(\sqrt{x}\)

B. \(\sqrt{2x}\)

C. \(2\sqrt{x}\)

D. x

E. 2x


PS60231.02


Asked: If the length of a diagonal of a square is \(2\sqrt{x}\), what is the area of the square in terms of x ?

Side of a square = \(Length of diagonal / \sqrt{2} =2\sqrt{x}/\sqrt{2} = \sqrt{2x}\)

Area of a square \(= (side)^2 = (\sqrt{2})^2 = 2x\)

IMO E
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Re: If the length of a diagonal of a square is 2x^(1/2), what is the area [#permalink] New post   30 Apr 2020, 08:54
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Since it's a square we get an isoceles right triangle with the diagonal and the sides will be in the ratio
1:1:√2
?:?:2√x
So side of the square is2√x/√2
Square this for Area 4x/2=2x
Hence E

Posted from my mobile device
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If the length of a diagonal of a square is 2x^(1/2), what is the area [#permalink] New post   21 Mar 2021, 00:35
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Solution:

Area of a square with diagonal length of d= d^2/2

d = 2√x

=>Area = (2√x)^2/2

=2x (option e)

Hope this helps :thumbsup:

Devmitra Sen (GMAT Quant Expert)
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Re: If the length of a diagonal of a square is 2x^(1/2), what is the area [#permalink] New post   30 May 2021, 07:09
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parkhydel wrote:
If the length of a diagonal of a square is \(2\sqrt{x}\), what is the area of the square in terms of x ?


A. \(\sqrt{x}\)

B. \(\sqrt{2x}\)

C. \(2\sqrt{x}\)

D. x

E. 2x


PS60231.02


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Re: If the length of a diagonal of a square is 2x^(1/2), what is the area [#permalink] New post   30 Apr 2020, 08:38
given that digonal of square = 2√x
we know that digonal of square = √2 * side of square
so from given info ; side of square = 2√x/2 ; √2x
area of square ( √2x) ^2 ; 2x
OPTION E


parkhydel wrote:
If the length of a diagonal of a square is \(2\sqrt{x}\), what is the area of the square in terms of x ?


A. \(\sqrt{x}\)

B. \(\sqrt{2x}\)

C. \(2\sqrt{x}\)

D. x

E. 2x


PS60231.02
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Re: If the length of a diagonal of a square is 2x^(1/2), what is the area [#permalink] New post   31 May 2020, 00:50
2 ways to approach this que.
1.with direct formula
Area = 1/2 (diagonal)^2

2.as pee info it is square and diagonal will cut right angle to 45:45.
So using 45:45:90 we can find side of square
Area= side^2

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Re: If the length of a diagonal of a square is 2x^(1/2), what is the area [#permalink] New post   31 May 2020, 08:08
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Solution


Given
    • Length of a diagonal of a square = \( 2 \sqrt{(x)}\).

To Find
    • Area of square in terms of x.

Approach and Working out
Let’s say that the square has all of its sides = a units. Area of the square will be \(a^2\).
    • Diagonal of the square can be found out using Pythagoras theorem since all the angles of the square are of 90 degrees each.
      o Therefore \(a^2 + a^2 = diagonal ^2\)
        o Therefore \(diagonal = \sqrt{(2a^2)}\)
      .
Now, diagonal of the square \(= 2\sqrt{(x)} = \sqrt{(2a^2)}\).
    • Finding the side of the square (a) in terms of ‘x’: \(2a^2 = 4x\).
      o \(a^2 = 2x\). Therefore \(a = \sqrt{(2x)}\).

Area of a square \(= a^2 = (\sqrt{(2x)})^2 = 2x\).

Correct Answer: Option E
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Re: If the length of a diagonal of a square is 2x^(1/2), what is the area [#permalink] New post   23 Aug 2023, 06:15
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