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Questions 1 - 10

The perimeter of a square is 48. What is the length of its diagonal?

$12\sqrt{2}$

$\sqrt{12}$

$24\sqrt{2}$

$48\sqrt{2}$

$2\sqrt{12}$

Explanation

Perimeter = side * 4

48 = side * 4

Side = 12

We can break up the square into two equal right triangles. The diagonal of the sqaure is then the hypotenuse of these two triangles.

Therefore, we can use the Pythagorean Theorem to solve for the diagonal:

$\sqrt{144+144}=\sqrt{hypotenuse^2}$

$hypotenuse=\sqrt{2\times 144}=\sqrt{2\times 12\times 12}=12\sqrt{2}$

Find the volume of the following sphere.

Sphere

$288 \pi m^3$

$270 \pi m^3$

$300 \pi m^3$

$312 \pi m^3$

$360 \pi m^3$

Explanation

The formula for the volume of a sphere is:

$V=\frac{4}{3} \pi r^3$

where is the radius of the sphere.

Plugging in our values, we get:

$V=\frac{4}{3} \pi (6m)^3$

$V=\frac{4}{3} \pi 216m^3 = 288 \pi m^3$

Find the area of the following rhombus:

Screen_shot_2014-03-01_at_8.57.04_pm

The perimeter of the rhombus is .

Explanation

The formula for the perimeter of a rhombus is:

Where is the length of the side

Plugging in our values, we get:

The formula for the area of a rhombus is:

$A = \frac{1}{2} (d_1)(d_2)$

Where is the length of one diagonal and is the length of another diagonal

By drawing the diagonals, we create a right triangle with the hypotenuse as and the side as .

Since we know that is a phythagorean triple, we can infer that the third side is .

Plugging in our values, we get:

$A = \frac{1}{2}(24m)(10m) = 120m^2$

The base angle of an isosceles triangle is 15 less than three times the vertex angle. What is the vertex angle?

Explanation

Every triangle contains 180 degrees. An isosceles triangle has one vertex angle and two congruent base angles.

Let = vertex angle and = base angle

So the equation to solve becomes .

What is the hypotenuse of a right triangle with side lengths and ?

Explanation

The Pythagorean Theorem states that . This question gives us the values of and , and asks us to solve for .

Take and and plug them into the equation as and :

$12^{2}+16^{2}=c^{2}$

Now we can start solving for :

$144+256=c^{2}$

$400=c^{2}$

$\sqrt{400}=\sqrt{c^{2}}$

The length of the hypotenuse is .

To the nearest tenth, give the area of a circle with diameter 17 inches.

$227.0 \textrm{ in}^{2}$

$907.9\textrm{ in}^{2}$

$289.0 \textrm{ in}^{2}$

$454.0\textrm{ in}^{2}$

$578.0\textrm{ in}^{2}$

Explanation

The radius of a circle with diameter 17 inches is half that, or 8.5 inches. The area of the circle is

$A = \pi r^{2} = \pi \cdot 8.5^{2} \approx 227.0$

What is the surface area of a sphere with a radius of ?

$576\pi$

$144\pi$

$288\pi$

$24\pi$

Explanation

To solve for the surface area of a sphere you must remember the formula:

First, plug the radius into the equation for :

$SA=(4)(12^{2})(\pi)$

Since $(12^{2})=144$ , the surface area is $(4)(144)(\pi)=576\pi$ .

The answer is therefore $576\pi$ .

Find the volume of the following sphere.

Sphere

$288 \pi m^3$

$270 \pi m^3$

$300 \pi m^3$

$312 \pi m^3$

$360 \pi m^3$

Explanation

The formula for the volume of a sphere is:

$V=\frac{4}{3} \pi r^3$

where is the radius of the sphere.

Plugging in our values, we get:

$V=\frac{4}{3} \pi (6m)^3$

$V=\frac{4}{3} \pi 216m^3 = 288 \pi m^3$

Triangle A: A right triangle with sides length , , and .

Triangle B: An equilateral triangle with side lengths .

Which triangle has a greater area?

Triangle B

Triangle A

The areas of the two triangles are the same.

There is not enough information given to determine which triangle has a greater area.

Explanation

The formula for the area of a right triangle is $A = \frac{1}{2}bh$ , where is the length of the triangle's base and is its height. Since the longest side is the hypotenuse, use the two smaller numbers given as sides for the base and height in the equation to calculate the area of Triangle A:

$A = \frac{1}{2}(6)(8)$

The formula for the area of an equilateral triangle is $A = \frac{s^2\sqrt{3}}{4}$ , where is the length of each side. (Alternatively, you can divide the equilateral triangle into two right triangles and find the area of each). Triangle B's area is thus calculated as:

$A = \frac{(8)^2\sqrt{3}}{4}$

$A = \frac{64\sqrt{3}}{4}$

$A = 16\sqrt{3}$

To determine which of the two areas is greater without using a calculator, rewrite the areas of the two triangles with comparable factors. Triangle A's area can be expressed as $24= 8\sqrt{3}\sqrt{3}$ , and Triangle B's area can be expressed as $16\sqrt{3}= (8)(2)\sqrt{3}$ . Since is greater than $\sqrt{3}$ , the product of the factors of Triangle B's area will be greater than the product of the factors of Triangle A's, so Triangle B has the greater area.