Geometry - SSAT Upper Level: Quantitative

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Question

Give the equation of the line through and .

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Answer

First, find the slope:

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} =\frac{9-1}{2-8}=\frac{8}{-6}=-\frac{4}{3}$

Apply the point-slope formula:

$y-y_{1}= m (x-x_{1})$

$y-1= -\frac{4}{3} (x-8)$

$y-1= -\frac{4}{3} x- \left (-\frac{4}{3} \right )8$

$y-1= -\frac{4}{3} x+ \frac{32}{3}$

$y= -\frac{4}{3} x+ \frac{35}{3}$

Rewriting in standard form:

$\frac{4}{3} x + y= \frac{35}{3}$

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Question

A line can be represented by . What is the slope of the line that is perpendicular to it?

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Answer

You will first solve for Y, to get the equation in form.

represents the slope of the line, which would be $-\frac{2}{3}$ .

A perpendicular line's slope would be the negative reciprocal of that value, which is $\frac{3}{2}$ .

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Question

Examine the above diagram. What is ?

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Answer

Use the properties of angle addition:

$\left ( x - 13\right )+ (y + 25) +43= 180$

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Question

Give the equation of a line that passes through the point and has an undefined slope.

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Answer

A line with an undefined slope has equation for some number ; since this line passes through a point with -coordinate 4, then this line must have equation

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Question

Give the equation of a line that passes through the point and has slope 1.

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Answer

We can use the point slope form of a line, substituting $m = 1, x_{1} = 4, y _{1} = 7$ .

$y- y_{1} = m (x-x_{1})$

or

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Question

Find the equation the line goes through the points and .

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Answer

First, find the slope of the line.

$\text{Slope}=\frac{y_2-y_1}{x_2-x_1}=\frac{2-4}{0-3}=\frac{-2}{-3}=\frac{2}{3}$

Now, because the problem tells us that the line goes through , our y-intercept must be .

Putting the pieces together, we get the following equation:

$y=\frac{2}{3}x+2$

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Question

A line passes through the points and . Find the equation of this line.

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Answer

To find the equation of a line, we need to first find the slope.

$\text{Slope}=\frac{y_2-y_1}{x_2-x_1}=\frac{-2-6}{-2-(-4)}=\frac{-8}{2}=-4$

Now, our equation for the line looks like the following:

To find the y-intercept, plug in one of the given points and solve for . Using , we get the following equation:

Solve for .

Now, plug the value for into the equation.

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Question

What is the equation of a line that passes through the points and ?

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Answer

First, we need to find the slope of the line.

$\text{Slope}=\frac{y_2-y_1}{x_2-x_1}=\frac{-7-1}{-2-8}=\frac{-8}{-10}=\frac{4}{5}$

Next, find the -intercept. To find the -intercept, plug in the values of one point into the equation , where is the slope that we just found and is the -intercept.

$1=8(\frac{4}{5})+b$

Solve for .

$1=\frac{32}{5}+b$

$b=-\frac{27}{5}$

Now, put the slope and -intercept together to get $y=\frac{4}{5}x-\frac{27}{5}$

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Question

Are the following two equations parallel?

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Answer

When two lines are parallal, they must have the same slope.

Look at the equations when they are in slope-intercept form, where b represents the slope.

We must first reduce the second equation since all of the constants are divisible by .

This leaves us with . Since both equations have a slope of , they are parallel.

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Question

Reduce the following expression:

$\frac{10a^{3}b^{5}c^{-2}}{2a^{2}}$

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Answer

For this expression, you must take each variable and deal with them separately.

First divide you two constants .

Then you move onto and when you divide like exponents you must subtract the exponents leaving you with .

is left by itself since it is already in a natural position.

Whenever you have a negative exponential term, you must it in the denominator.

This leaves the expression of $\frac{5ab^{5}}{c^{2}}$ .

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Question

What is the area of a circle with a diameter of , rounded to the nearest whole number?

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Answer

The formula for the area of a circle is

$\dpi{100} \pi r^{2}$

Find the radius by dividing 9 by 2:

$\dpi{100} \frac{9}{2}=4.5$

So the formula for area would now be:

$\dpi{100} \pi r^{2}=\pi (4.5)^{2}=20.25\pi \approx 63.6= 64$

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Question

What is the area of a circle that has a diameter of inches?

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Answer

The formula for finding the area of a circle is $\pi r^{2}$ . In this formula, represents the radius of the circle. Since the question only gives us the measurement of the diameter of the circle, we must calculate the radius. In order to do this, we divide the diameter by .

$\frac{15}{2}=7.5$

Now we use for in our equation.

$\pi (7.5)^{2}=176.7146 : in^{2}$

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Question

What is the area of a circle with a diameter equal to 6?

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Answer

First, solve for radius:

$r=\frac{d}{2}=\frac{6}{2}=3$

Then, solve for area:

$A=r^2\pi=3^2\pi=9\pi$

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Question

The diameter of a circle is $4\ cm$ . Give the area of the circle.

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Answer

The area of a circle can be calculated using the formula:

$Area=\frac{\pi d^2}{4}$ ,

where is the diameter of the circle, and $\pi$ is approximately .

$Area=\frac{\pi d^2}{4}=\frac{\pi\times 4^2}{4}=4\pi \Rightarrow Area\approx 4\times 3.14\Rightarrow Area\approx 12.56 \ cm^2$

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Question

The diameter of a circle is . Give the area of the circle in terms of .

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Answer

The area of a circle can be calculated using the formula:

$Area=\frac{\pi d^2}{4}$ ,

where is the diameter of the circle and $\pi$ is approximately .

$Area=\frac{\pi (4t)^2}{4}=\frac{16\pi t^2}{4}=4\pi t^2 \Rightarrow Area\approx 4\times 3.14\times t^2$

$\Rightarrow Area\approx 12.56t^2$

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Question

The radius of a circle is $\frac{2}{\sqrt{\pi }}$ . Give the area of the circle.

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Answer

The area of a circle can be calculated as $Area=\pi r^2$ , where is the radius of the circle, and $\pi$ is approximately .

$Area=\pi r^2=\pi\times (\frac{2}{\sqrt{\pi}})^2=\pi\times \frac{4}{\pi}\Rightarrow Area=4$

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Question

The circumference of a circle is inches. Find the area of the circle.

Let $\pi = 3.14$ .

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Answer

First we need to find the radius of the circle. The circumference of a circle is $Circumference =2\pi r$ , where is the radius of the circle.

$12.56=2\times 3.14\times r\Rightarrow r=2\ in$

The area of a circle is $Area=\pi r^2$ where is the radius of the circle.

$Area=\pi r^2=3.14\times 2^2=12.56\ in^2$

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Question

The perpendicular distance from the chord to the center of a circle is $\sqrt{2}t$ , and the chord length is $2\sqrt{2}t$ . Give the area of the circle in terms of .

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Answer

Chord length = $2\sqrt{r^2-d^2}$ , where is the radius of the circle and is the perpendicular distance from the chord to the circle center.

Chord length = $2\sqrt{r^2-d^2}\Rightarrow (2\sqrt{2})t=2\sqrt{r^2-(\sqrt{2}t)^2}$

$\Rightarrow 8t^2=4(r^2-2t^2)\Rightarrow 4r^2=16t^2\Rightarrow r^2=4t^2\Rightarrow r=2t$

$Area=\pi r^2$ , where is the radius of the circle and $\pi$ is approximately .

$Area=\pi r^2=3.14\times (2t)^2=12.56t^2$

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Question

A circle on the coordinate plane has equation

$x^{2} + y^{2} = 66$ .

Which of the following gives the area of the circle?

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Answer

The equation of a circle on the coordinate plane is

$x^{2} + y^{2} = r^{2}$ ,

where is the radius. Therefore, in this equation,

$r^{2} = 66$ .

The area of a circle is found using the formula

$A = \pi r^{2}$ ,

so we substitute 66 for $r^{2}$ , yielding

$A = 66 \pi$ .

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Question

Give the area of the above figure.

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Answer

The figure is a semicircle - one-half of a circle - with radius 5.5, or $\frac{11}{2}$ . Its area is one-half of the square of the radius multiplied by $\pi$ - that is,

$A = \frac{1}{2} \pi r^{2}$

$A = \frac{1}{2} \pi \left (\frac{11}{2} \right )^{2}$

$A = \frac{1}{2} \pi \cdot \frac{121}{4}$

$A = \frac{121\pi}{8 }$

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