Rectangles

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Geometry › Rectangles

Questions 1 - 10

If the perimeter of a rectangle is , and the width of the rectangle is , what is the area of a rectangle?

Explanation

Recall how to find the perimeter of a rectangle:

$\text{Perimeter}=2(\text{length}+\text{width})$

Since we are given the width and the perimeter, we can solve for the length.

$\text{length}+\text{width}=\frac{\text{Perimeter}}{2}$

$\text{length}=\frac{\text{Perimeter}}{2}-\text{width}$

Substitute in the given values for the width and perimeter to find the length.

$\text{length}=\frac{36}{2}-17$

Simplify.

$\text{length}=18-17$

Solve.

$\text{length}=1$

Now, recall how to find the area of a rectangle.

$\text{Area}=\text{length}\times\text{width}$

Substitute in the values of the length and width to find the area.

$\text{Area}=17 \times 1$

Solve.

$\text{Area}=17$

If a rectangle has a width of and a length that is double the width, what would be the area of the rectangle? Round to the nearest tenth.

Explanation

To calculate the area of a triangle, we want to multiply the length by the width. Since the length is twice that of the width, which is , we can determine length as such:

$2.3;cm\cdot2=4.6;cm$

Now that we know the values for length and width, we can calculate the area of the triangle:

$Area=2.3;cm\cdot4.6;cm=10.6;cm^2$

If a rectangle has a width of and a length that is double the width, what would be the area of the rectangle? Round to the nearest tenth.

Explanation

To calculate the area of a triangle, we want to multiply the length by the width. Since the length is twice that of the width, which is , we can determine length as such:

$2.3;cm\cdot2=4.6;cm$

Now that we know the values for length and width, we can calculate the area of the triangle:

$Area=2.3;cm\cdot4.6;cm=10.6;cm^2$

If the perimeter of a rectangle is , and the width of the rectangle is , what is the area of a rectangle?

Explanation

Recall how to find the perimeter of a rectangle:

$\text{Perimeter}=2(\text{length}+\text{width})$

Since we are given the width and the perimeter, we can solve for the length.

$\text{length}+\text{width}=\frac{\text{Perimeter}}{2}$

$\text{length}=\frac{\text{Perimeter}}{2}-\text{width}$

Substitute in the given values for the width and perimeter to find the length.

$\text{length}=\frac{36}{2}-17$

Simplify.

$\text{length}=18-17$

Solve.

$\text{length}=1$

Now, recall how to find the area of a rectangle.

$\text{Area}=\text{length}\times\text{width}$

Substitute in the values of the length and width to find the area.

$\text{Area}=17 \times 1$

Solve.

$\text{Area}=17$

A rectangle has an area of $96 ; cm^{2}$ . The width is four less than the length. What is the perimeter?

Explanation

For a rectangle, area is and perimeter is , where is the length and is the width.

Let = length and = width.

The area equation to solve becomes , or $x^{2}-4x-96=0$ .

To factor, find two numbers the sum to -4 and multiply to -96. -12 and 8 will work:

x2 + 8x - 12x - 96 = 0

x(x + 8) - 12(x + 8) = 0

(x - 12)(x + 8) = 0

Set each factor equal to zero and solve:

or .

Therefore the length is and the width is , giving a perimeter of .

A rectangle has an area of $96 ; cm^{2}$ . The width is four less than the length. What is the perimeter?

Explanation

For a rectangle, area is and perimeter is , where is the length and is the width.

Let = length and = width.

The area equation to solve becomes , or $x^{2}-4x-96=0$ .

To factor, find two numbers the sum to -4 and multiply to -96. -12 and 8 will work:

x2 + 8x - 12x - 96 = 0

x(x + 8) - 12(x + 8) = 0

(x - 12)(x + 8) = 0

Set each factor equal to zero and solve:

or .

Therefore the length is and the width is , giving a perimeter of .

What is the area of a rectangle that has a length of and a width of ?

Explanation

Recall how to find the area of a rectangle:

$\text{Area}=\text{length}\times\text{width}$

Now, plug in the given length and width to find the area.

When multiplying decimals together first move the decimal over so that the number is a whole integer.

$0.75\rightarrow 75$

Now we multiple the integers together.

$0.75\cdot 85\rightarrow 75\cdot 85=6375$

From here, we need to move the decimal place back. In this particular problem we moved the decimal over a total of two decimal places.

Therefore our answer becomes,

$6375\rightarrow 63.75$

$\text{Area}=85 \times 0.75= 63.75$

What is the area of a rectangle that has a length of and a width of ?

Explanation

Recall how to find the area of a rectangle:

$\text{Area}=\text{length}\times\text{width}$

Now, plug in the given length and width to find the area.

$\text{Area}=12 \times 10= 120$

If the perimeter of a rectangle is , and the length of the rectangle is , what is the area of the rectangle?

Explanation

First, use the information given about the perimeter to find the width of the rectangle.

Recall how to find the perimeter of a rectangle:

$\text{Perimeter}=2(\text{length}+\text{width})$

From this equation, we can solve for the width.

$\text{length}+\text{width}=\frac{\text{Perimeter}}{2}$

$\text{width}=\frac{\text{Perimeter}}{2}-\text{length}$

Substitute in the information from the question to find the width of the rectangle.

$\text{Width}=\frac{48}{2}-10$

Simplify.

$\text{Width}=14$

Now, recall how to find the area of a rectangle:

$\text{Area}=\text{length}\times\text{width}$

Substitute in the information about the length and width to find the area for the rectangle in question.

$\text{Area}=10 \times 14$

Solve.

$\text{Area}=140$

If the perimeter of a rectangle is , and the width of the rectangle is , what is the area of the rectangle?

Explanation

Recall how to find the perimeter of a rectangle:

$\text{Perimeter}=2(\text{length}+\text{width})$

Since we are given the width and the perimeter, we can solve for the length.

$\text{length}+\text{width}=\frac{\text{Perimeter}}{2}$

$\text{length}=\frac{\text{Perimeter}}{2}-\text{width}$

Substitute in the given values for the width and perimeter to find the length.

$\text{length}=\frac{36}{2}-12=6$

Simplify.

$\text{length}=\frac{36}{2}-12=6$

Solve.

$\text{length}=\frac{36}{2}-12=6$

Now, recall how to find the area of a rectangle.

$\text{Area}=\text{length}\times\text{width}$

Substitute in the values of the length and width to find the area.

$\text{Area}=12 \times 6$

Solve.

$\text{Area}= 72$

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