Area of a Quadrilateral - GED Math

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Question

Find the area of the trapezoid:

Answer

The area of a trapezoid is calculated using the following equation:

$A=\frac{b_1+b_2}{2}h=\frac{7+5}{2}(3)=\frac{12}{2}(3)=(6)(3)=18$

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Question

A rectangle has length 10 inches and width 5 inches. Each dimension is increased by 3 inches. By what percent has the area of the rectangle increased?

Answer

The area of a rectangle is its length times its width.

Its original area is $10 \times 5 = 50$ square inches; its new area is $13 \times 8 = 104$ square inches. The area has increased by

$\frac{104 - 50}{50 } \times 100 = \frac{54}{50 } \times 100 = 108 %$ .

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Question

A rectangle has length 10 inches and width 8 inches. Its length is increased by 2 inches, and its width is decreased by 2 inches. By what percent has the area of the rectangle decreased?

Answer

The area of a rectangle is its length times its width.

Its original area is $10 \times 8 = 80$ square inches; its new area is $12 \times 6 = 72$ square inches. The area has decreased by

$\frac{80 -72}{80} \times 100 % = \frac{8}{80} \times 100 % = 10 %$ .

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Question

The length of each side of a square is increased by 10%. By what percent has its area increased?

Answer

Let be the original sidelength of the square. Increasing this by 10% is the same as adding 0.1 times that sidelength to the original sidelength. The new sidelength is therefore

The area of a square is the square of its sidelength.

The area of the square was originally $s^{2}$ ; it is now

$\left ( 1.1s\right )^{2}= 1.21s^{2}$

That is, the area has increased by

$\frac{1.21s^{2}-s^{2}}{s^{2}} \times 100 % =\frac{0.21s^{2}}{s^{2}} \times 100 % = 0.21 \times 100 % = 21 %$ .

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Question

The above figure depicts the rectangular swimming pool at an apartment. The apartment manager needs to purchase a tarp that will cover this pool completely. However, because of the cutting device the pool store uses, the length and the width of the tarp must each be a multiple of four yards. Also, the tarp must be at least one yard longer and one yard wider than the pool.

What will be the minimum area of the tarp the manager purchases?

Answer

Three feet make a yard, so the length and width of the pool are $\frac{50}{3} = 16 \frac{2}{3}$ yards and $\frac{35}{3} = 11 \frac{2}{3}$ yards, respectively. Since the dimensions of the tarp must exceed those of the pool by at least one yard, the tarp must be at least $17 \frac{2}{3}$ yards by $12 \frac{2}{3}$ yards; but since both dimensions must be multiples of four yards, we take the next multiple of four for each.

The tarp must be 20 yards by 16 yards, so the area of the tarp is the product of these dimensions, or

$20 \times 16 = 320$ square yards.

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Question

Note: Figure NOT drawn to scale

Refer to the above figure, which shows a rectangular garden (in green) surrounded by a dirt path five feet wide throughout. What is the area of that dirt path?

Answer

The dirt path can be seen as the region between two rectangles. The outer rectangle has length and width 100 feet and 60 feet, respectively, so its area is

$A = 100 \cdot 60 = 6,000$ square feet.

The inner rectangle has length and width $(100 - 2\times 5) = 90$ feet and $(60 - 2\times 5) = 50$ feet, respectively, so its area is

$A = 90 \times 50 = 4,500$ square feet.

The area of the path is the difference of the two:

square feet.

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Question

Note: Figure NOT drawn to scale

Refer to the above figure, which shows a rectangular garden (in green) surrounded by a dirt path (in brown). The dirt path is feet wide throughout. Which of the following polynomials gives the area of the garden?

Answer

The length of the garden is feet less than that of the entire lot, or

.

The width of the garden is feet less than that of the entire lot, or

.

The area of the garden is their product:

$= 100 \cdot 60 - 100 \cdot 2y - 2y \cdot 60+2y \cdot 2y$

$= 6,000-200y-120y+4y^{2}$

$=4y^{2} - 320 y + 6,000$

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Question

Note: Figure NOT drawn to scale.

Refer to the above figure, which shows a rectangular garden (in green) surrounded by a dirt path (in brown). The dirt path is five feet wide throughout. Which of the following polynomials gives the area of the garden in square feet?

Answer

The length of the garden is $2 \times 5 = 10$ feet less than that of the entire lot, or

.

The width of the garden is $2 \times 5 = 10$ less than that of the entire lot, or

.

The area of the garden is their product:

$=2x \cdot x - 2x \cdot 10 - 10 \cdot x + 10 \cdot 10$

$= 2x ^{2} - 20x - 10 x + 100$

$= 2x ^{2} - 30 x + 100$

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Question

Note: Figure NOT drawn to scale

Refer to the above diagram.

Half of Rectangle is pink. .

Evaluate .

Answer

Rectangle has length and width , so it has area

$LW = 20 \times 15 = 300$ .

The area of Rectangle is twice that of Rectangle , or 600. Its length is

.

Its width is

.

Plug in what we know and solve for :

$600 =30 \times \left (15 + DF \right )$

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Question

Note: Figure NOT drawn to scale

What percent of Rectangle is pink?

Answer

The pink region is Rectangle . Its length and width are

so its area is the product of these, or

$14 \times 8 = 112$ .

The length and width of Rectangle are

so its area is the product of these, or

$20 \times 12 = 240$ .

So we want to know what percent 112 is of 240, which can be answered as follows:

$\frac{112}{240} \times 100 % = 46\frac{2}{3} %$

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Question

Note: Figure NOT drawn to scale

What percent of Rectangle is white?

Answer

The pink region is Rectangle . Its length and width are

so its area is the product of these, or

$14 \times 7 = 98$ .

The length and width of Rectangle are

so its area is the product of these, or

$20 \times 10 = 200$ .

The white region is Rectangle cut from Rectangle , so its area is the difference of the two:

.

So we want to know what percent 102 is of 200, which can be answered as follows:

$\frac{102}{200} \times 100 % = 51 %$

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Question

Note: Figure NOT drawn to scale.

Calculate the area of Rhombus in the above diagram if:

Answer

The area of a rhombus is half the product of the lengths of diagonals $\overline{AC}$ and $\overline{BD}$ . This is

$\frac{1}{2} \times 10 \times 24 = 120$ .

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Question

Find the area of a square with a side of .

Answer

Write the formula for the area of a square.

Substitute the side into the equation.

Simplify the equation.

The answer is:

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Question

If a rectangle has a length of 18cm and a width that is half the length, what is the area of the rectangle?

Answer

To find the area of a rectangle, we will use the following formula:

$A = l \cdot w$

where l is the length and w is the width of the rectangle.

Now, we know the length of the rectangle is 18cm. We also know the width is half the length. Therefore, the width is 9cm. So, we can substitute. We get

$A = 18\text{cm} \cdot 9\text{cm}$

$A = 162\text{cm}^2$

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Question

If a square has a length of 10in, find the area.

Answer

To find the area of a square, we will use the following formula:

$A = l \cdot w$

where l is the length and w is the width of the square.

Now, we know the length of the square is 10in. Because it is a square, all sides are equal. Therefore, the length is also 10in. So, we can substitute. We get

$A = 10\text{in} \cdot 10\text{in}$

$A = 100\text{in}^2$

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Question

Find the area of a rectangle with a width of 8in and a length that is two times the width.

Answer

To find the area of a rectangle, we will use the following formula:

$A = l \cdot w$

where l is the length and w is the width of the rectangle.

Now, we know the width of the rectangle is 8in. We also know the length of the rectangle is two times the width. Therefore, the length is 16in. So, we can substitute. We get

$A = 16\text{in} \cdot 8\text{in}$

$A = 128\text{in}^2$

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Question

Find the area of a square with a length of 11cm.

Answer

To find the area of a square, we will use the following formula:

$A = l \cdot w$

where l is the length and w is the width of the square.

Now, we know the length of the square is 11cm. Because it is a square, all sides are equal. Therefore, the width is also 11cm. So, we can substitute. We get

$A = 11\text{cm} \cdot 11\text{cm}$

$A = 121\text{cm}^2$

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Question

Find the area of a rectangle with a width of 7in and a length that is three times the width.

Answer

To find the area of a rectangle, we will use the following formula:

$A = l \cdot w$

where l is the length and w is the width of the rectangle.

Now, we know the width of the rectangle is 7in. We also know the length is three times the width. Therefore, the length is 21in. So, we can substitute. We get

$A =21\text{in} \cdot 7\text{in}$

$A = 147\text{in}^2$

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Question

A square has an area of $196\text{cm}^2$ . Find the length of one side.

Answer

A square has 4 equal sides. The formula to find the area of a square is

where b is the length of one side of the square. To find the length of one side of the square, we will solve for b.

Now, we know the area of the square is $196\text{cm}^2$ . So, we will substitute and solve for b. So,

$196\text{cm}^2 = b^2$

$\sqrt{196\text{cm}^2} = \sqrt{b^2}$

$14\text{cm} = b$

$b = 14\text{cm}$

Therefore, the length of one side of the square is 14cm.

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Question

What of the following is NOT a property of a quadrilateral?

Answer

A quadrilateral is a shape with four corners, or vertices, whose interior angles angles must sum up to 360 degrees.

The shape also has 4 edges.

The quadrilateral does not necessarily require two pairs of parallel lines.

Quadrilaterals may also be irregular.

The answer is:

$\textup{The quadrilateral must contain at least a set of parallel edges.}$

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