GRE Math : How to find the perimeter of a rectangle

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Example Question #1 : How To Find The Perimeter Of A Rectangle

Quantity A

The perimeter of a square with a diagonal of 12 inches.

Quantity B

The perimeter of a rectangle with a diagonal of 10 inches and longer sides that are three times the length of its shorter sides.

Possible Answers:

Quantity B is greater.

Quantity A is greater.

The quantities are equal.

The relationship cannot be determined from the information given.

Correct answer:

Quantity A is greater.

Explanation:

The diagonal of a square creates a 45-45-90 triangle; therefore, considering x as the value for the sides of the square in A, set up the ratio: 1/√2 = x/12 → x = 12/√2

Simplify the square root out of the denominator: x = (12^√2)/2 = 6^√2. The perimeter is equal to 4x; therefore, quantity A is 24^√2.

Use the Pythagorean Theorem for B. We know that one side of the triangle is x and the other must be 3x. Furthermore, we know the hypotenuse is 10; therefore:

x²+ (3x)² = 10²→ x²+ 9x² = 100 → 10x² = 100 → x² = 10 → x = √10.

Transform the √10 into √2 * √5 ; therefore, x = √2 * √5 . The perimeter of the rectangle will equal 3x + 3x + x + x = 8x = 8√2 * √5 .

Compare these two values. Quantity A has a coefficient of 24 in for its √2 . Quantity B has a coefficient of 8√5, which must be smaller than 24, because 8 * 3 = 24, but √5 is less than 3 (which would be √9 ); therefore, A is larger.

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Example Question #2 : How To Find The Perimeter Of A Rectangle

A rectangle has a length that is twice that of its height. If the perimeter of that rectangle is $20\:in$ , what is its area?

Possible Answers:

$400\:in^2$

$\frac{10}{3}\:in^2$

$\frac{250}{9}\:in^2$

$\frac{200}{9}\:in^2$

$\frac{150}{7}\:in^2$

Correct answer:

$\frac{200}{9}\:in^2$

Explanation:

Based on our first sentence, we know:

L=2H

The second sentence means:

$P=20\:in$ , which is the same as:

2L+2H=20

Now, we can replace with in the second equation:

2(2H)+2H=20

4H+2H=20

6H=20

3H=10

Therefore, $H=\frac{10}{3}\:in$

Now, this means that:

$L=2*\frac{10}{3}=\frac{20}{3}\:in$

If these are our values, then the area of the rectangle is:

$A=LH=\frac{10}{3}*\frac{20}{3}=\frac{200}{9}\:in^2$

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Example Question #42 : Quadrilaterals

A rectangle with an area of 64 square units is one-fourth as wide as it is long. What is the perimeter of the rectangle?

Possible Answers:

Correct answer:

Explanation:

Begin by setting up a ratio of width to length for the rectangle. We know that the width is one-fourth the length, therefore

$\frac{1}{4}L=W$ .

Now, we can substitute W in the rectangle area formula with our new variable to solve for the length.

$A=W\cdot L$

$A=\left(\frac{1}{4}L\right)\cdot L$

$A=\frac{1}{4}L^2$

Solve for length:

$64=\frac{1}{4}L^2$

256=L^2

L=16

Using the length, solve for Width:

$W=\frac{1}{4}(16)$

W=4

Now, plug the length and width into the formula for the perimeter of a rectangle to solve:

P=2W+2L

P=2(4)+2(16)

P=8+32

P=40

The perimeter of the rectangle is 40.

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GRE Math : How to find the perimeter of a rectangle

Study concepts, example questions & explanations for GRE Math

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Example Questions

Example Question #1 : How To Find The Perimeter Of A Rectangle

Example Question #2 : How To Find The Perimeter Of A Rectangle

Example Question #42 : Quadrilaterals

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