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If the length of a certain rectangle is decreased

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If the length of a certain rectangle is decreased

by sachin_yadav » Tue Apr 23, 2013 9:58 am
The following is an example question of a geometry from a prep book (building concepts). Didn't understand how to solve it.

If the length of a certain rectangle is decreased by 4 cm and the width is increased by 3 cm, a square with the same area as the original rectangle would result. Find the perimeter of the original rectangle.

No answer choices

Answer is 50 cm

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by Anju@Gurome » Tue Apr 23, 2013 10:10 am
sachin_yadav wrote:If the length of a certain rectangle is decreased by 4 cm and the width is increased by 3 cm, a square with the same area as the original rectangle would result. Find the perimeter of the original rectangle.
Let us assume the length and width of the rectangle are L and W, respectively, in cm.

Now, length of the square is (L - 4) cm and width of the square is (W + 3)
And, as it is a square, length = width ---> (L - 4) = (w + 3) ---> L = W + 7
So, area of the square = (L - 4)(W + 3) = (W + 3)²

And, area of the rectangle = LW = (W + 7)W

So, (W + 3)² = (W + 7)W
--> W² + 6W + 9 = W² + 7W
--> W = 9
--> L = (9 + 7) = 16

So, the perimeter of the rectangle = 2*(L + W) = 2*(16 + 9) = 2*25 = 50
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by sachin_yadav » Tue Apr 23, 2013 10:42 am
Thank you :D
Anju@Gurome wrote:
Let us assume the length and width of the rectangle are L and W, respectively, in cm.

Now, length of the square is (L - 4) cm and width of the square is (W + 3)
And, as it is a square, length = width ---> (L - 4) = (w + 3) ---> L = W + 7
So, area of the square = (L - 4)(W + 3) = (W + 3)²

And, area of the rectangle = LW = (W + 7)W

So, (W + 3)² = (W + 7)W
--> W² + 6W + 9 = W² + 7W
--> W = 9
--> L = (9 + 7) = 16

So, the perimeter of the rectangle = 2*(L + W) = 2*(16 + 9) = 2*25 = 50
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by rintoo22 » Tue Apr 23, 2013 12:14 pm
If the length of a certain rectangle is decreased by 4 cm and the width is increased by 3 cm, a square with the same area as the original rectangle would result. Find the perimeter of the original rectangle.
I would start with second half of the question which states
a square with the same area
. Therefore let x be the side of the square. Therefore the area fo the square is x^2....(1)

Now lets go to the first part
length of a certain rectangle is decreased by 4 cm and the width is increased by 3 cm
. So before the it was a square with area x, the sides fo the rectangle would have been (x+4) and (x-3). So the area is (x+4)(x-3)...(2)

And the question states that the area is the same
(x+4)(x-3)=x^2
x^2-3x+4x-12=x^2
x=12

Put the value of x => 2*(16) + 2*(9) = 50

Hope this helps.
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by GMATGuruNY » Wed Apr 24, 2013 2:42 am
sachin_yadav wrote:The following is an example question of a geometry from a prep book (building concepts). Didn't understand how to solve it.

If the length of a certain rectangle is decreased by 4 cm and the width is increased by 3 cm, a square with the same area as the original rectangle would result. Find the perimeter of the original rectangle.

40
50
60
70
80
I've added answer choices, which would be provided on the GMAT.
The perimeter of the square must be close in value to one of the answer choices, which represent the perimeter of the rectangle.
Since the length of the rectangle DECREASES by 4cm to form the square, the LENGTH of the rectangle is 4cm GREATER than each side of the square.
Since the width of the rectangle INCREASES by 3cm to form the square, the WIDTH of the rectangle is 3cm LESS than each side of the square.

Case 1: Each side of the square = 10
Area of the square = 10*10 = 100.
Area of the rectangle = (10+4)(10-3) = 14*7 = 98.
The areas are close, but not equal.

Case 2: Each side of the square = 11
Area of the square = 11*11 = 121.
Area of the rectangle = (11+4)(11-3) = 15*8 = 120.
Almost there.

Case 3: Each side of the square = 12
Area of the square = 12*12 = 144.
Area of the rectangle = (12+4)(12-3) = 16*9 = 144.
Success!

Thus, the perimeter of the rectangle = 2(16) + 2(9) = 50.

The correct answer is B.
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