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The area of square ABCD is x^ 2 + 18 √ 3 x + 243 . What is

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by BTGmoderatorAT » Sun Feb 11, 2018 9:31 am
The area of square ABCD is x^ 2 + 18 √ 3 x + 243 . What is the length of the square's diagonal?

A) x+9√3
B) 4x+36√3
C) (x+9√3)^2
D)√2 (x+9√3)
E) (x+9√3)/√2

Is there a strategic approach to this question? Can any experts show?
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by [email protected] » Sun Feb 11, 2018 11:03 am
Hi ardz24,

We're told that the area of a SQAURE is x^ 2 + 18 √ 3 x + 243 . We're asked for the length of the square's diagonal. While this question might look a bit 'scary', this question has a great built-in 'pattern shortcut' that you can use - along with the answer choices - to answer this question without too much trouble.

To start, since we're dealing with a SQUARE, we have to consider how the diagonal relates to the length of a side. If we call the side "X", then the diagonal is "√2(X)." Given that relationship, there would be a pretty good chance that the diagonal would have a "√2" in it somewhere.

The obvious choices is Answer D: (√2)(x+9√3)
If that is the diagonal, then the sides of the square would be (x+9√3)
Thus, the area would be....
(x+9√3)(x+9√3) =
X^2 + 2(9X√3) + (9√3)^2 =
X^2 + 18X√3 + 243
This is an exact match for what we were told, so this MUST be the answer.

Final Answer: D

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by EconomistGMATTutor » Sun Feb 11, 2018 11:12 am
The area of square ABCD is x^ 2 + 18 √ 3 x + 243 . What is the length of the square's diagonal?

A) x+9√3
B) 4x+36√3
C) (x+9√3)^2
D)√2 (x+9√3)
E) (x+9√3)/√2

Is there a strategic approach to this question? Can any experts show?
Hi ardz24,
Let's take a look at your question.

We know that the area of a square is always the square of the length of side.
So we can find the length of side using the area.
$$Area=x^2+18\sqrt{3}x+243$$
If s represents the length of side of square, then,
$$s^2=x^2+18\sqrt{3}x+243$$
Use the formula a^2+2ab+b^2=(a+b)^2 to write the RHS as a square:
$$s^2=x^2+2\left(x\right)\left(9\sqrt{3}\right)+\left(9\sqrt{3}\right)^2$$
$$s^2=\left(x+9\sqrt{3}\right)^2$$
$$s=\left(x+9\sqrt{3}\right)$$
Now we can find the length of the diagonal of the square using Pythagorean theorem,
$$\left(Diagonal\right)^2=\left(x+9\sqrt{3}\right)^2+\left(x+9\sqrt{3}\right)^2$$
$$Diagonal=\sqrt{2\left(x+9\sqrt{3}\right)^2}$$
$$Diagonal=\sqrt{2}\left(x+9\sqrt{3}\right)$$
Therefore, option D is correct.

Hope it helps.
I am available if you'd like any follow up.
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by Jeff@TargetTestPrep » Tue Feb 13, 2018 5:38 pm
ardz24 wrote:The area of square ABCD is x^ 2 + 18 √ 3 x + 243 . What is the length of the square's diagonal?

A) x+9√3
B) 4x+36√3
C) (x+9√3)^2
D)√2 (x+9√3)
E) (x+9√3)/√2
If we let s = the length of a side of the square, we must have s^2 = x^2 + 18√3x + 243. Furthermore, we can factor x^2 + 18√3x + 243 as (x + c)(x + c) where c is a constant. To determine c, we can just take the square root of 243, which is √243 = √81 * √3 = 9√3. So:

x^2 + 18√3x + 243 = (x + 9√3)(x + 9√3) (Notice that 2 * x * 9√3 = 18√3x)

So we see that s = x + 9√3 and since the diagonal = s√2, the diagonal = (x + 9√3)√2.

Answer: D

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