Areas and Perimeters of Polygons - SSAT Upper Level Quantitative

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Question

Which of the following shapes is NOT a quadrilateral?

Answer

A quadrilateral is any two-dimensional shape with sides. The only shape listed that does not have sides is a triangle.

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Question

Find the area of a regular hexagon that has side lengths of .

Answer

Use the following formula to find the area of a regular hexagon:

$\text{Area}=\frac{3\sqrt3}{2}\times \text{side}^2$ .

Now, substitute in the length of the side into this equation.

$\text{Area}=\frac{3\sqrt3}{2}\times6^2$

$\text{Area}=\frac{3\sqrt3}{2}\times 36$

$\text{Area}=\frac{108\sqrt3}{2}=54\sqrt3$

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Question

Find the area of a regular hexagon that has a side length of .

Answer

Use the following formula to find the area of a regular hexagon:

$\text{Area}=\frac{3\sqrt3}{2}\times \text{side}^2$ .

Now, substitute in the length of the side into this equation.

$\text{Area}=\frac{3\sqrt3}{2}\times 8^2$

$\text{Area}=\frac{3\sqrt3}{2}\times 64$

$\text{Area}=\frac{192\sqrt3}{2}=96\sqrt3$

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Question

Find the area of a regular hexagon that has a side length of .

Answer

Use the following formula to find the area of a regular hexagon:

$\text{Area}=\frac{3\sqrt3}{2}\times \text{side}^2$ .

Now, substitute in the length of the side into this equation.

$\text{Area}=\frac{3\sqrt3}{2}\times 12^2$

$\text{Area}=\frac{3\sqrt3}{2}\times 144$

$\text{Area}=\frac{432\sqrt3}{2}=216\sqrt3$

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Question

Find the area of a regular hexagon that has side lengths of .

Answer

Use the following formula to find the area of a regular hexagon:

$\text{Area}=\frac{3\sqrt3}{2}\times \text{side}^2$ .

Now, substitute in the length of the side into this equation.

$\text{Area}=\frac{3\sqrt3}{2}\times 7^2$

$\text{Area}=\frac{3\sqrt3}{2}\times 49$

$\text{Area}=\frac{147}{2}\sqrt3$

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Question

Find the area of a regular hexagon that has side lengths of .

Answer

Use the following formula to find the area of a regular hexagon:

$\text{Area}=\frac{3\sqrt3}{2}\times \text{side}^2$ .

Now, substitute in the length of the side into this equation.

$\text{Area}=\frac{3\sqrt3}{2}\times 10^2$

$\text{Area}=\frac{3\sqrt3}{2}\times 100$

$\text{Area}=\frac{300\sqrt3}{2}=150\sqrt3$

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Question

Which shape is NOT a quadrilateral?

Answer

A quadrilateral has to have sides, a circle does not have any sides.

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Question

Find the area of a regular hexagon with side lengths .

Answer

Use the following formula to find the area of a regular hexagon:

$\text{Area}=\frac{3\sqrt3}{2}\times \text{side}^2$ .

Now, substitute in the length of the side into this equation.

$\text{Area}=\frac{3\sqrt3}{2}\times 20^2$

$\text{Area}=\frac{3\sqrt3}{2}\times 400$

$\text{Area}=\frac{1200\sqrt3}{2}=600\sqrt3$

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Question

Find the area of a regular hexagon that has side lengths of .

Answer

Use the following formula to find the area of a regular hexagon:

$\text{Area}=\frac{3\sqrt3}{2}\times \text{side}^2$ .

Now, substitute in the length of the side into this equation.

$\text{Area}=\frac{3\sqrt3}{2}\times x^2$

$\text{Area}=\frac{3x^2}{2}\sqrt3$

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Question

Find the area of a regular hexagon with side lengths of .

Answer

Use the following formula to find the area of a regular hexagon:

$\text{Area}=\frac{3\sqrt3}{2}\times \text{side}^2$ .

Now, substitute in the length of the side into this equation.

$\text{Area}=\frac{3\sqrt3}{2}\times (2t)^2$

$\text{Area}=\frac{3\sqrt3}{2}\times 4t^2$

$\text{Area}=\frac{12t^2\sqrt3}{2}=6t^2\sqrt3$

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Question

Find the area of a regular hexagon with side lengths of .

Answer

Use the following formula to find the area of a regular hexagon:

$\text{Area}=\frac{3\sqrt3}{2}\times \text{side}^2$ .

Now, substitute in the length of the side into this equation.

$\text{Area}=\frac{3\sqrt3}{2}\times (3m)^2$

$\text{Area}=\frac{3\sqrt3}{2}\times 9m^2$

$\text{Area}=\frac{27m^2}{2}\sqrt3$

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Question

Rectangle A has length 40 inches and height 24 inches. Rectangle B has length 30 inches and height 28 inches. Rectangle C has length 72 inches, and its area is the mean of the areas of the other two rectangles. What is the height of Rectangle C?

Answer

The area of a rectangle is the product of the length and its height, Rectangle A has area $40 \times 24 = 960$ square inches; Rectangle B has area $30 \times 28 = 840$ square inches.

The area of Rectangle C is the mean of these areas, or

$\frac{960+840}{2} = 900$ square inches, so its height is this area divided by its length:

$900 \div 72 = 12 \frac{1}{2}$ inches.

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Question

Find the area of a regular hexagon with side lengths of $\frac{1}{2}$ .

Answer

Use the following formula to find the area of a regular hexagon:

$\text{Area}=\frac{3\sqrt3}{2}\times \text{side}^2$ .

Now, substitute in the length of the side into this equation.

$\text{Area}=\frac{3\sqrt3}{2}\times \left(\frac{1}{2}\right)^2$

$\text{Area}=\frac{3\sqrt3}{2}\times \frac{1}{4}$

$\text{Area}=\frac{3\sqrt3}{8}$

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Question

Find the area of a regular hexagon with side lengths of .

Answer

Use the following formula to find the area of a regular hexagon:

$\text{Area}=\frac{3\sqrt3}{2}\times \text{side}^2$ .

Now, substitute in the length of the side into this equation.

$\text{Area}=\frac{3\sqrt3}{2}\times (8n)^2$

$\text{Area}=\frac{3\sqrt3}{2}\times 64n^2$

$\text{Area}=\frac{192n^2\sqrt3}{2}=96n^2\sqrt3$

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Question

Find the area of a regular hexagon with side lengths .

Answer

Use the following formula to find the area of a regular hexagon:

$\text{Area}=\frac{3\sqrt3}{2}\times \text{side}^2$ .

Now, substitute in the length of the side into this equation.

$\text{Area}=\frac{3\sqrt3}{2}\times (12c)^2$

$\text{Area}=\frac{3\sqrt3}{2}\times 144c^2$

$\text{Area}=\frac{432c^2\sqrt3}{2}=216c^2\sqrt3$

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Question

A parallelogram has the base length of and the altitude of . Give the area of the parallelogram.

Answer

The area of a parallelogram is given by:

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

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Question

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 .

Answer

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

The area of a parallelogram is given by:

Where:

is the length of any base
is the corresponding altitude

So we can write:

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

$t\times t=12\times 3$

$t^{2}=36$

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Question

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 .

Answer

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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Question

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 .

Answer

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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Question

The longer diagonal of a rhombus is 20% longer than the shorter diagonal; the rhombus has area . Give the length of the shorter diagonal in terms of .

Answer

Let be the length of the shorter diagonal. If the longer diagonal is 20% longer, then it measures 120% of the length of the shorter diagonal; this is

$\frac{120}{100} = \frac{120 \div 20}{100 \div 20} = \frac{6}{5}$

of , or $\frac{6}{5}D$ .

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

$\frac{1}{2} \cdot \frac{6}{5}D \cdot D = N$

$\frac{3}{5}D^{2} = N$

$\frac{5}{3} \cdot \frac{3}{5}D^{2} = \frac{5}{3} \cdot N$

$D^{2} = \frac{5N}{3}$

$D =\sqrt{ \frac{5N}{3}}$

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