Measurement and Geometry

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Questions 1 - 10

The graphic below shows a blueprint for a swimming pool.

Swimming pool dimensions

If the pool is going to be 66 inches deep, how many cubic feet of water will it be able to hold? (1 ft = 12 in)

Explanation

Notice that the outer dimensions of the blueprint are the dimensions for the entire pool, including the concrete, while the inner dimensions are for the part of the pool that will be filled with water. Therefore, we want to focus on just the inner dimensions.

Notice that the depth is given in inches, while the dimensions are in feet. Convert 66 inches to feet by dividing 66 by 12, since 12 inches makes a foot:

$66;in\cdot\frac{1 ft}{12 in}=5.5 ft$

The inch units cancel out and leave us with just the feet units. 66 in is 5.5 ft.

Now we have all of the information we need to solve for the volume of the pool. The pool is a rectangular prism, and the formula for volume of a rectangular prism is

$length\cdot width\cdot height$

(In this case, the "height" of the swimming pool is its depth.)

The blueprint shows that the pool is 40 ft long and 30 ft wide. Plugging in the measurements from the problem, we get

$40,ft\cdot30,ft\cdot5.5,ft$

Multiplying this out, we get .

The graphic below shows a blueprint for a swimming pool.

Swimming pool dimensions

If the pool is going to be 66 inches deep, how many cubic feet of water will it be able to hold? (1 ft = 12 in)

Explanation

Notice that the depth is given in inches, while the dimensions are in feet. Convert 66 inches to feet by dividing 66 by 12, since 12 inches makes a foot:

$66;in\cdot\frac{1 ft}{12 in}=5.5 ft$

The inch units cancel out and leave us with just the feet units. 66 in is 5.5 ft.

Now we have all of the information we need to solve for the volume of the pool. The pool is a rectangular prism, and the formula for volume of a rectangular prism is

$length\cdot width\cdot height$

(In this case, the "height" of the swimming pool is its depth.)

The blueprint shows that the pool is 40 ft long and 30 ft wide. Plugging in the measurements from the problem, we get

$40,ft\cdot30,ft\cdot5.5,ft$

Multiplying this out, we get .

Give the coordinates of the midpoint of the line segment whose endpoints are the intercepts of the line of the equation

None of the other choices gives the correct response.

Explanation

The x-intercept and y-intercept of the line of an equation can be found by substituting 0 for and , respectively, as follows:

x-intercept:

Substitute and simplify:

Solve for by dividing by 2 on both sides:

$2x \div 2 = 40 \div 2$

The x-intercept of the line is located at

The y-intercept can be found by setting and solving for in a similar fashion:

$5y \div 5 = 40 \div 5$

The y-intercept is located at .

The midpoint of a line segment, given its endpoints $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ , is located at

$\left ( \frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2} \right )$ ;

substituting accordingly,

$\frac{x_{1}+x_{2}}{2} =\frac{20+8}{2} = \frac{28}{2} = 14$

$\frac{y_{1}+y_{2}}{2} = \frac{ 0+8}{2} = \frac{8}{2} = 4$

The midpoint is located at .

$\sin a^{\circ } = \frac{2}{5}$

$\cos (90-a)^{\circ } = ?$

None of the other choices gives the correct response.

$\frac{5}{\sqrt{21}}$

$\frac{2}{\sqrt{29}}$

$\frac{5}{\sqrt{29}}$

$\frac{2}{\sqrt{21}}$

Explanation

An identity of trigonometry is

$\sin a^{\circ} = \cos (90-a)^{\circ}$

for any value of .

Since $\sin a^{\circ } = \frac{2}{5}$ , it immediately follows that $\cos (90-a)^{\circ} =\frac{2}{5}$ .

This response is not among the given choices.

$\cos N = \frac{x}{3}$

Evaluate $\sin N$ in terms of .

$\frac{\sqrt{x^{2}-9}}{3}$

$\frac{\sqrt{x^{2}-9}}{x}$

$\frac{3}{x}$

$\frac{x}{\sqrt{x^{2}-9}}$

$\frac{3}{\sqrt{x^{2}-9}}$

Explanation

Suppose we allow be the lengths of the opposite leg and adjacent leg and the hypotenuse, respectively, of the right triangle with an acute angle measuring . The cosine is defined to be the ratio of the length of the adjacent side to that of the hypotenuse, so

$\cos N = \frac{A}{H} = \frac{x}{3}$

We can set the lengths of the adjacent leg and the hypotenuse to and 3, respectively. By the Pythagorean Theorem, the length of the opposite leg is

$O = \sqrt{H^{2}- A^{2}} = \sqrt{x^{2}-3^{2}} = \sqrt{x^{2}-9 }$

The sine of the angle is equal to the ratio of the length of the opposite leg to that of the hypotenuse, so

$\sin N = \frac{O}{H} = \frac{\sqrt{x^{2}-9 }}{3}$ .

$\sin a^{\circ } = \frac{4}{7}$

$\tan a^{\circ } = ?$

$\frac{4}{\sqrt{33}}$

$\frac{\sqrt{33}}{4}$

$\frac{7}{4}$

$\frac{7}{\sqrt{33}}$

$\frac{\sqrt{33}}{7}$

Explanation

The sine of an angle is defined to be the ratio of the length of the opposite leg of a right triangle to the length of its hypotenuse. Therefore, we can set . By the Pythagorean Theorem:
the adjacent leg of the triangle has measure

$A = \sqrt{H^{2}-O^{2}} = \sqrt{7^{2}-4^{2}} = \sqrt{49-16} = \sqrt{33}$

The tangent of the angle is the ratio of the length of the opposite leg to that of the adjacent leg, which is

$\tan a ^{\circ }= \frac{O}{A} = \frac{4}{\sqrt{33}}$ ,

the correct response.

$\bigtriangleup ABC \sim \bigtriangleup DEF$ with scale factor 5:4, with $\bigtriangleup ABC$ the larger triangle.

Complete the sentence: the area of $\bigtriangleup ABC$ is % greater than that of $\bigtriangleup DEF$ .

(Select the closest whole percent)

Explanation

The ratio of the areas of two similar triangles is equal to the square of their scale factor. The scale factor is equal to $\frac{5}{4}$ , so the ratio of the areas is the square of this, or $\left ( \frac{5}{4} \right ) ^{2} = \frac{25}{16}$ .

This makes the area of larger triangle $\bigtriangleup ABC$ equal to $\frac{25}{16} \times 100 % = 156 \frac{1}{4} %$ of that of smaller triangle $\bigtriangleup DEF$ —or, equivalently, $56 \frac{1}{4} %$ greater.

Two of the angles of a triangle are congruent; the third has measure ten degrees greater than either one of the first two. What is the measure of the third angle?

$66 \frac{2}{3} ^{\circ }$

$63 \frac{1}{3} ^{\circ }$

$53 \frac{1}{3} ^{\circ }$

$56 \frac{2}{3} ^{\circ }$

$73\frac{1}{3}^{\circ }$

Explanation

Let be the measure of the third angle. Since its measure is ten degrees greater than either of the others, then the common measure of the other two is . The sum of the measures of the angles of a triangle is 180 degrees, so

To solve for , first ungroup and collect like terms:

Isolate ; first add 20:

Divide by 3:

$3x \div 3 = 200 \div 3$

Since $200 \div 3 = 66 \textup{ R }2$ ,

$x = 66 \frac{2}{3}$ .

The third angle measures $66 \frac{2}{3} ^{\circ }$ .

$\cos N = \frac{x}{3}$

Evaluate $\sin N$ in terms of .

$\frac{\sqrt{x^{2}-9}}{3}$

$\frac{\sqrt{x^{2}-9}}{x}$

$\frac{3}{x}$

$\frac{x}{\sqrt{x^{2}-9}}$

$\frac{3}{\sqrt{x^{2}-9}}$

Explanation

$\cos N = \frac{A}{H} = \frac{x}{3}$

We can set the lengths of the adjacent leg and the hypotenuse to and 3, respectively. By the Pythagorean Theorem, the length of the opposite leg is

$O = \sqrt{H^{2}- A^{2}} = \sqrt{x^{2}-3^{2}} = \sqrt{x^{2}-9 }$

The sine of the angle is equal to the ratio of the length of the opposite leg to that of the hypotenuse, so

$\sin N = \frac{O}{H} = \frac{\sqrt{x^{2}-9 }}{3}$ .

Two of the angles of a triangle are congruent; the third has measure ten degrees greater than either one of the first two. What is the measure of the third angle?

$66 \frac{2}{3} ^{\circ }$

$63 \frac{1}{3} ^{\circ }$

$53 \frac{1}{3} ^{\circ }$

$56 \frac{2}{3} ^{\circ }$

$73\frac{1}{3}^{\circ }$