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SSAT Upper Level Math : Areas and Perimeters of Polygons

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Example Questions

Example Question #81 : Areas And Perimeters Of Polygons

Rectangle A has length 40 inches and height inches; Rectangle B has length 30 inches and height inches; Rectangle C has height inches, and its area is the sum of those of the other two rectangles. What is its length?

Possible Answers:

$23\frac{1}{3} \textrm{ in}$

$22\frac{1}{2} \textrm{ in}$

$\frac{45H}{2} \textrm{ in}$

The correct answer is not among the other choices.

$\frac{70H}{3} \textrm{ in}$

Correct answer:

The correct answer is not among the other choices.

Explanation:

The area of a rectangle is the product of the length and its height.

Rectangle A has area 40H square inches, and Rectangle B has an area of $30 \cdot 2H = 60 H$ square inches, so the sum of their areas is 40H + 60 H = 100 H square inches. This is the area of Rectangle C; divide it by height inches to get a length of

$\frac{100 H}{4H} = \frac{100}{4} = 25$ inches. This answer is not among the given choices.

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Example Question #81 : Areas And Perimeters Of Polygons

The perimeter of a rectangle is 490 centimeters. The width of the rectangle is three-fourths of its length. What is the area of the rectangle?

Possible Answers:

$29,400 \textrm{ cm}^{2}$

$14,700 \textrm{ cm}^{2}$

$58,800 \textrm{ cm}^{2}$

$11,025 \textrm{ cm}^{2}$

$19,600\textrm{ cm}^{2}$

Correct answer:

$14,700 \textrm{ cm}^{2}$

Explanation:

Let be the length of the rectangle. Then its width is three-fourths of this, or $\frac{3}{4} L$ . The perimeter is the sum of the lengths of its sides, or

$L + \frac{3}{4} L + L + \frac{3}{4} L = \frac{7}{2} L$ .

Set this equal to 490 centimeters and solve for :

$\frac{7}{2} L = 490$

$\frac{2} {7} \cdot \frac{7}{2} L =\frac{2} {7} \cdot 490$

L = 140

The length of the rectangle is 140 centimeters; the width is three-fourths of this, or

$\frac{3}{4} \times 140 = 105$ centimeters.

The area is the product of the length and the width:

$140 \times 105 = 14,700$ square centimeters.

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Example Question #13 : How To Find The Area Of A Rectangle

A basketball team wants to paint a 4-foot wide border around its court to make sure fans don't get too close to the action. If the court is 94 by 50 feet, and one can of paint can cover 300 square feet, how many cans of paint does the team need to ensure that the entire border is painted?

(Assume that you cannot buy partial cans of paint.)

Possible Answers:

Correct answer:

Explanation:

We begin this problem by finding the difference of two areas: the larger rectangle bounded by the outer edge of the border and the smaller rectangle that is the court itself.

The larger rectangle is 5916 square feet $\left(94+4+4\right\times\left(50+4+4 \right )$ , and the court is 4700 square feet $(94\times 50)$ .

The difference, 1296 , repesents the area of the border.

Now we divide this by 300 , which is just a bit over . But since we can't leave square feet unpainted, we have to round up to cans of paint.

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Example Question #81 : Areas And Perimeters Of Polygons

Box

The above diagram shows a rectangular solid. The shaded side is a square. Give the total surface area of the solid.

Possible Answers:

500

1,000

800

2,000

Correct answer:

1,000

Explanation:

A square has four sides of equal length, as seen in the diagram below.

Box 2

All six sides are rectangles, so their areas are equal to the products of their dimensions:

Top, bottom, front, back (four surfaces): $20 \times 10 = 200$

Left, right (two surfaces): $10 \times 10 = 100$

The total area: $4 \times 200 + 2 \times 100 = 800 + 200 = 1,000$

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Example Question #82 : Areas And Perimeters Of Polygons

Box

The above diagram shows a rectangular solid. The shaded side is a square. In terms of , give the surface area of the solid.

Possible Answers:

400x

$2x^{2} + 80x$

80x + 800

$20 x^{2}$

Correct answer:

$2x^{2} + 80x$

Explanation:

Since a square has four sides of equal length, the solid looks like this:

Box

The areas of each of the individual surfaces, each of which is a rectangle, are the product of their dimensions:

Front, back, top, bottom (four surfaces): 20 x

Left, right (two surfaces): $x \cdot x = x^{2}$

The total surface area is therefore

$2 \cdot x^{2} + 20x \cdot 4 = 2 x^{2} + 80x$

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Example Question #821 : Geometry

Parallelogram

Figure NOT drawn to scale

The above figure shows Rhombus ABCD ; and are midpoints of their respective sides. Rhombus ABCD has area 900.

Give the area of Rectangle AXCY .

Possible Answers:

450

675

300

225

Correct answer:

450

Explanation:

A rhombus, by definition, has four sides of equal length. Therefore, AB= CD , and, by the Multiplication Property, $\frac{1}{2 }AB=\frac{1}{2 } CD$ . Also, since and are the midpoints of their respective sides, $AX = XB = \frac{1}{2 }AB$ and $\frac{1}{2 } CD = CY = YD$ . Combining these statements, and letting n = AX :

$n = AX = XB = \frac{1}{2 }AB=\frac{1}{2 } CD = CY = YD$

Also, both $\overline{AY}$ and $\overline{XC}$ are altitudes of the rhombus; they are congruent, and we will call their common length (height).

The figure, with the lengths, is below.

Rhombus

The area of the entire Rhombus ABCD is the product of its height and the length of a base , so

$A_{1} = h \cdot 2n = 2hn =900$ .

Rectangle AXCY has as its length and width and , so its area is their product , Since

2 hn =900 ,

From the Division Property, it follows that

$2 hn \div 2 = 900 \div 2$ ,

and

hn = 450 .

This makes 450 the area of Rectangle AXCY .

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Example Question #81 : Areas And Perimeters Of Polygons

A parallelogram has the base length of and the altitude of 2t-1 . Give the area of the parallelogram.

Possible Answers:

2t^2+t

t^2

2t^2-t

t^2+t

2t^2

Correct answer:

2t^2-t

Explanation:

The area of a parallelogram is given by:

Area=ba

Where is the base length and is the corresponding altitude. So we can write:

Area=t(2t-1)=2t^2-t

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Example Question #2 : Area Of A Parallelogram

A parallelogram has a base length of which is 3 times longer than its corresponding altitude. The area of the parallelogram is 12 square inches. Give the .

Possible Answers:

$6\sqrt{2}$

$6\sqrt{3}$

Correct answer:

Explanation:

Base length is so the corresponding altitude is $\frac{t}{3}$ .

The area of a parallelogram is given by:

Area=ba

Where:

is the length of any base
is the corresponding altitude

So we can write:

$12=t\times \frac{t}{3}\Rightarrow t\times t=12\times 3\Rightarrow t^2=36\Rightarrow t=6$

$t\times t=12\times 3$

$t^{2}=36$

t=6

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Example Question #2 : Area Of A Parallelogram

The length of the shorter diagonal of a rhombus is 40% that of the longer diagonal. The area of the rhombus is . Give the length of the longer diagonal in terms of .

Possible Answers:

$\sqrt{\frac{5K}{2}}$

$\sqrt{5K}$

$K\sqrt{\frac{5}{2}}$

$\frac{1}{5}\sqrt{K}$

$K\sqrt{\frac{2}{5}}$

Correct answer:

$\sqrt{5K}$

Explanation:

Let be the length of the longer diagonal. Then the shorter diagonal has length 40% of this. Since 40% is equal to $\frac{40}{100 } = \frac{40 \div 20 }{100 \div 20 } = \frac{2}{5}$ , 40% of is equal to $\frac{2}{5}D$ .

The area of a rhombus is half the product of the lengths of its diagonals, so we can set up, and solve for , in the equation:

$\frac{1}{2} \cdot \frac{2}{5}D \cdot D = K$

$\frac{1}{5}D^{2}= K$

$5\cdot \frac{1}{5}D^{2}=5\cdot K$

$D^{2}=5K$

$D =\sqrt{5K}$

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Example Question #1 : Area Of A Parallelogram

The length of the shorter diagonal of a rhombus is two-thirds that of the longer diagonal. The area of the rhombus is square yards. Give the length of the longer diagonal, in inches, in terms of .

Possible Answers:

$12\sqrt{ 3Q}$

$3\sqrt{ 3Q}$

$4\sqrt{ 3Q}$

$\sqrt{ 3Q}$

$36\sqrt{ 3Q }$

Correct answer:

$36\sqrt{ 3Q }$

Explanation:

Let be the length of the longer diagonal in yards. Then the shorter diagonal has length two-thirds of this, or $\frac{2}{3}D$ .

The area of a rhombus is half the product of the lengths of its diagonals, so we can set up the following equation and solve for :

$\frac{1}{2} \cdot \frac{2}{3}D \cdot D = Q$

$\frac{1}{3}D ^{2} = Q$

$3\cdot \frac{1}{3}D ^{2} = 3 \cdot Q$

$D ^{2} = 3Q$

$D = \sqrt{3Q}$

To convert yards to inches, multiply by 36:

$\sqrt{ 3Q } \times 36 = 36\sqrt{ 3Q }$

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SSAT Upper Level Math : Areas and Perimeters of Polygons

Study concepts, example questions & explanations for SSAT Upper Level Math

All SSAT Upper Level Math Resources

Example Questions

Example Question #81 : Areas And Perimeters Of Polygons

Example Question #81 : Areas And Perimeters Of Polygons

Example Question #13 : How To Find The Area Of A Rectangle

Example Question #81 : Areas And Perimeters Of Polygons

Example Question #82 : Areas And Perimeters Of Polygons

Example Question #821 : Geometry

Example Question #81 : Areas And Perimeters Of Polygons

Example Question #2 : Area Of A Parallelogram

Example Question #2 : Area Of A Parallelogram

Example Question #1 : Area Of A Parallelogram

All SSAT Upper Level Math Resources

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