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## The area of one square is x^2 + 10x + 25 and the area

tagged by: M7MBA

This topic has 3 expert replies and 1 member reply

### Top Member

M7MBA Moderator
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#### The area of one square is x^2 + 10x + 25 and the area

Sun Feb 18, 2018 4:51 am
The area of one square is x^2 + 10x + 25 and the area of another square is 4x^2 - 12x + 9. If the sum of the perimeters of both squares is 32, what is the value of x?

A. 0
B. 2
C. 2.5
D. 4.67
E. 10

The OA is the option B.

What are the equations that I should use here? Experts, may you tell me how to use the quadratic equations? I need your help here.

### GMAT/MBA Expert

EconomistGMATTutor GMAT Instructor
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Top Reply
Mon Feb 19, 2018 12:08 pm
Quote:
The area of one square is x^2 + 10x + 25 and the area of another square is 4x^2 - 12x + 9. If the sum of the perimeters of both squares is 32, what is the value of x?

A. 0
B. 2
C. 2.5
D. 4.67
E. 10

The OA is the option B.

What are the equations that I should use here? Experts, may you tell me how to use the quadratic equations? I need your help here.
Hi M7MBA,
Let's take a look at your question.

We will first find the length of side for both the squares and then find the perimeters of both.
$$Area\ of\ first\ square\ =\ x^2+10x+25$$
Let length of side of the square is 'a', then,
$$a\times a=\ x^2+10x+25$$
$$a^2=\ x^2+10x+25$$
$$a^2=\ x^2+2\left(x\right)\left(5\right)+\left(5\right)^2$$
$$a^2=\ \left(x+5\right)^2$$
$$a=x+5$$
We know that perimeter of a square is 4 times the length of a side, therefore,
$$Perimeter\ of\ first\ square\ =\ 4\left(x+5\right)$$

$$Area\ of\ second\ square=4x^2-12x+9$$
Let length of side of the square is 'b', then,
$$b\times b=4x^2-12x+9$$
$$b^2=4x^2-12x+9$$
$$b^2=\left(2x\right)^2-2\left(2x\right)\left(3\right)+\left(3\right)^2$$
$$b^2=\left(2x-3\right)^2$$
$$b=2x-3$$
$$Perimeter\ of\ second\ square\ =\ 4\left(2x-3\right)$$

We know that the sum of the perimeters of both squares is 32, therefore,
$$4\left(x+5\right)+\ 4\left(2x-3\right)=32$$
$$4x+20+\ 8x-12=32$$
$$12x+8=32$$
$$12x=32-8$$
$$12x=24$$
$$x=\frac{24}{12}=2$$

Therefore, Option B is correct.

Hope it helps.
I am available if you'd like any help.

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Rich.C@EMPOWERgmat.com Elite Legendary Member
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Sun Feb 18, 2018 10:40 am
Hi M7MBA,

We're told that the area of one square is X^2 + 10X + 25, the area of another square is 4X^2 - 12X + 9 and the sum of the perimeters of both squares is 32. We're asked for the value of X. This question is based around standard Quadratic rules and a bit of Geometry.

To start, since we're dealing with two SQUARES, we know that the the reverse-FOIL of each area will be the same 'term' twice (since those terms represent the lengths of the sides of each square).

X^2 + 10X + 25 =
(X+5)(X+5)

4X^2 - 12X + 9 =
(2X - 3)(2X - 3)

The PERIMETER of a square = 4(side), so the two perimeters are....4(X+5) and 4(2X - 3). We're told that the SUM of those two perimeters is 32, so we can set up an equation and solve for X:
4(X+5) + 4(2X - 3) = 32
4X + 20 + 8X - 12 = 32
12X + 8 = 32
12X = 24
X = 2

Final Answer: B

GMAT assassins aren't born, they're made,
Rich

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### GMAT/MBA Expert

Scott@TargetTestPrep GMAT Instructor
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Wed Feb 21, 2018 1:28 pm
M7MBA wrote:
The area of one square is x^2 + 10x + 25 and the area of another square is 4x^2 - 12x + 9. If the sum of the perimeters of both squares is 32, what is the value of x?

A. 0
B. 2
C. 2.5
D. 4.67
E. 10
We have to determine the side length of each square. Since the area of the first square is x^2 + 10x + 25 = (x + 5)^2, its side is x + 5. Similarly, since the area of the second square is 4x^2 - 12x + 9 = (2x - 3)^2, its side is 2x - 3. Furthermore, the perimeter of the first square is 4(x + 5), and that of the second square is 4(2x - 3). Since the sum of the perimeters of the two squares is 32, we can create the following equation:

4(x + 5) + 4(2x - 3) = 32

4x + 20 + 8x - 12 = 32

12x + 8 = 32

12x = 24

x = 2

Answer: B

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### Top Member

deloitte247 Master | Next Rank: 500 Posts
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Sun Mar 11, 2018 11:54 am
$$A1=x^2+10x+25$$
$$A2=4x^2-12x+9$$
$$let\ the\ perimeter\ be\ represented\ by\ p1\ \&\ p2\ for\ A1\ and\ A2\ respectively$$
$$Therefore,\ p1+p2=32$$
$$A1=\left(x^2+5x\right)+\left(5x+25\right)$$
$$A1=x\left(x+5\right)+5\left(x+5\right)$$
$$A1=\left(x+5\right)\left(x+5\right)$$

$$A2=4x^2-12x+9$$
$$A2=\left(4x^2-6x\right)-\left(6x+9\right)$$
$$A2=2x\left(2x^{ }-3\right)-3\left(2x-3\right)$$
$$A2=\left(2x^{ }-3\right)\left(2x-3\right)$$
$$\sin ce\ perimeter\ =4l,\ \ \ therefore\ 4l1+4l2=32$$
$$4\left(2x-3\right)+4\left(2x+5\right)=32$$
$$8x-12+4x+20=32$$
$$12x+8=32$$
$$12x=24$$
$$x=\frac{24}{12}=2$$
Hence the correct answer is option B

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