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SAT Mathematics : SAT Math

Study concepts, example questions & explanations for SAT Mathematics

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All SAT Mathematics Resources

137 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #8 : Using Radians

$\frac{37\pi}{18}$ radians is equivalent to how many degrees?

Possible Answers:

$350^\circ$

$10^\circ$

$185^\circ$

$370^\circ$

Correct answer:

$370^\circ$

Explanation:

1 radian is equal to $\frac{180}{\pi}$ degrees. Using this conversion factor,

$\frac{37\pi}{18}\times\frac{180}{\pi}=37\times10=370$ .

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Example Question #9 : Using Radians

Simplify your answer.

Convert $45^{\circ}$ to radians:

Possible Answers:

$\frac{\pi }{2}$

$\frac{1 }{4}$

$\frac{\pi }{8}$

$\frac{\pi }{4}$

Correct answer:

$\frac{\pi }{4}$

Explanation:

We know that:

$360^{\circ}=2\pi$ Radians

since the giving angle was in degrees then we multiply

$45^{\circ}*\frac{\pi}{180^{\circ}}= \frac{\pi}{4}$

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Example Question #10 : Using Radians

Give your answer in terms of $\pi$ .

Convert $165^{\circ}$ to radians:

Possible Answers:

$\frac{12\pi}{11}$

$\frac{11\pi}{24}$

$\frac{11\pi}{6}$

$\frac{11\pi}{12}$

Correct answer:

$\frac{11\pi}{12}$

Explanation:

To convert degrees to radians, we need to multiply the given degree by $\frac{\pi}{180^{\circ}}$ .

$165*^{\circ}\frac{\pi}{180^{\circ}}=\frac{165\pi}{180}$

To simplify, we get:

$\frac{11\pi}{12}$

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Example Question #1 : Applying The Equation Of A Circle

What is the equation for a circle of radius 12, centered at the intersection of the two lines:

$y_{1}=4x+3$

and

$y_{2}=5x+44$ ?

Possible Answers:

$(x-22)^{2}+(y-3)^{2}=12$

$(x+41)^{2}+(y+161)^{2}=141$

None of the other answers

$(x-41)^{2}+(y-161)^{2}=144$

Correct answer:

$(x+41)^{2}+(y+161)^{2}=141$

Explanation:

To begin, let us determine the point of intersection of these two lines by setting the equations equal to each other:

$4x+3=5x+44;\, 3=x+44;\, -41=x$

To find the y-coordinate, substitute into one of the equations. Let's use $y_{1}$ :

y=4*-41+3=-164+3=-161

The center of our circle is therefore: (–41, –161).

Now, recall that the general form for a circle with center at $(x_{0},y_{0})$ is:

$(x-x_{0})^{2}+(y-y_{0})^{2}=r^{2}$

For our data, this means that our equation is:

$(x+41)^{2}+(y+161)^{2}=12^{2}$ or $(x+41)^{2}+(y+161)^{2}=144$

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Example Question #2 : Applying The Equation Of A Circle

What is the radius of a circle with the equation x(x-8)+y(y-6)=24 ?

Possible Answers:

Correct answer:

Explanation:

We need to expand this equation to $x^{2}-8x+y^{2}-6y=24$ and then complete the square.

This brings us to $x^{2}-8x+16+y^{2}-6y+9=24+16+9$ .

We simplify this to $(x-4)^{2}+(y-3)^{2}=49$ .

Thus the radius is 7.

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Example Question #1 : Applying The Equation Of A Circle

A circle has its origin at (0,0) . The point (5,7) is on the edge of the circle. What is the radius of the circle?

Possible Answers:

$r=6\sqrt{2}$

$r=\sqrt{74}$

There is not enough information to answer this question.

$r=2\sqrt{18}$

Correct answer:

$r=\sqrt{74}$

Explanation:

The radius of the circle is equal to the hypotenuse of a right triangle with sides of lengths 5 and 7.

r^2=5^2+7^2

r^2=25+49= 74

$r=\sqrt{74}$

This radical cannot be reduced further.

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Example Question #3 : Applying The Equation Of A Circle

A circle with a radius of five is centered at the origin. A point on the circumference of the circle has an x-coordinate of two and a positive y-coordinate. What is the value of the y-coordinate?

Possible Answers:

$\sqrt{(21)}$

$\sqrt{(23)}$

Correct answer:

$\sqrt{(21)}$

Explanation:

Recall that the general form of the equation of a circle centered at the origin is:

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

We know that the radius of our circle is five. Therefore, we know that the equation for our circle is:

$x^{2}+y^{2}=5^{2}$

$x^{2}+y^{2}=25$

Now, the question asks for the positive y-coordinate when . To solve this, simply plugin for :

$2^{2}+y^{2}=25$

$4+y^{2}=25$

$y^{2}=21$

$y=\pm \sqrt{(21)}$

Since our answer will be positive, it must be $\sqrt{(21)}$ .

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Example Question #2 : Applying The Equation Of A Circle

The following circle is moved $\frac{\pi}{2}$ spaces to the left. Where is its new center located?

$(x-\frac{\pi}{2})^2 + (y - \frac{3\pi}{2})^2 = \pi$

Possible Answers:

$(\pi, 0)$

$(0, \pi)$

(1, 2)

$(2\pi, 1)$

Correct answer:

$(0, \pi)$

Explanation:

Remember that the general equation for a circle with center (h, k) and radius is (x-h)^2 + (y-k)^2 = r^2 .

With that in mind, our original center is at $(\frac{\pi}{2}, \frac{3\pi}{2})$ .

If we move the center $\frac{\pi}{2}$ units to the left, that means that we are subtracting $\frac{\pi}{2}$ from our given coordinates.

$\frac{\pi}{2} - \frac{\pi}{2} = 0$

$\frac{3\pi}{2} - \frac{\pi}{2} = \frac{2\pi}{2} = \pi$

Therefore, our new center is $(0, \pi)$ .

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Example Question #4 : Applying The Equation Of A Circle

A square on the coordinate plane has vertices at the points with coordinates (9,0) , (9,9) , (0, 9) , and (0,0) . Give the equation of the circle that circumscribes the square.

Possible Answers:

$\left ( x -9 \right )^{2} + \left ( y- 9\right )^{2} = \frac{81}{2}$

$\left ( x - \frac{9}{2}\right )^{2} + \left ( y- \frac{9}{2}\right )^{2} = \frac{81}{4}$

$\left ( x -9 \right )^{2} + \left ( y- 9\right )^{2} = 81$

$\left ( x - \frac{9}{2}\right )^{2} + \left ( y- \frac{9}{2}\right )^{2} = \frac{81}{2}$

Correct answer:

$\left ( x - \frac{9}{2}\right )^{2} + \left ( y- \frac{9}{2}\right )^{2} = \frac{81}{2}$

Explanation:

The equation of the circle on the coordinate plane with radius and center (h,k ) is

$(x-h)^{2} + (y-k)^{2} = r ^{2}$

The figure referenced is below:

Screen shot 2020 09 29 at 11.32.47 am

The center of the circle is at the point of intersection of the diagonals, which, as is the case with any rectangle, bisect each other. Therefore, looking at the diagonal with endpoints (0,0) and (9,9) , we can set $x_{1} = y_{1} = 0 , x_{2} = y_{2} = 9$ in the midpoint formula:

$h =\frac{ x_{1}+ x_{2} }{2} = \frac{0+ 9}{2} = \frac{9}{2}$

and

$k=\frac{ y_{1}+ y_{2} }{2} = \frac{0+ 9}{2} = \frac{9}{2}$

The center of the circumscribing circle is therefore $\left ( \frac{9}{2}, \frac{9}{2} \right )$ .

The radius of the circumscribing circle is the distance from this point to any point on the circle. The distance formula can be used here:

$r= \sqrt{(x_{2} - x_{1})^{2} +(y_{2} - y_{1})^{2} }$

Since we are actually trying to find $r ^{2}$ , we will use the form

$r^{2}= (x_{2} - x_{1})^{2} +(y_{2} - y_{1})^{2}$

Choosing the radius with endpoints (0,0) and $\left ( \frac{9}{2}, \frac{9}{2} \right )$ , we set $x_{1} = y_{1} = 0 , x_{2} = y_{2} = \frac{9}{2}$ and substitute:

$r^{2}= \left ( \frac{9}{2} - 0 \right ) ^{2} + \left ( \frac{9}{2} - 0 \right ) ^{2}$

$= \left ( \frac{9}{2} \right )^{2} + \left ( \frac{9}{2} \right )^{2}$

$= \frac{81}{4 } + \frac{81}{4 }$

$= \frac{81}{2}$

Setting $h =k= \frac{9}{2}$ and $r^{2}= \frac{81}{2}$ and substituting in the circle equation:

$(x-h)^{2} + (y-k)^{2} = r ^{2}$

$\left ( x - \frac{9}{2}\right )^{2} + \left ( y- \frac{9}{2}\right )^{2} = \frac{81}{2}$ , the correct response.

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Example Question #5 : Applying The Equation Of A Circle

Screen shot 2020 09 29 at 11.34.42 am

The above figure shows a circle on the coordinate axes with its center at the origin. $\overarc {AB}$ has length $6 \pi$ .

Give the equation of the circle.

Possible Answers:

$x^{2}+ y^2 =4$

$x^{2}+ y^2 =16$

$x^{2}+ y^2 =64$

$x^{2}+ y^2 =36$

Correct answer:

$x^{2}+ y^2 =64$

Explanation:

A $135^{\circ }$ arc of a circle represents $\frac{135}{360 } = \frac{3}{8}$ of the circle, so the length of the arc is three-eighths its circumference. Set up the equation and solve for :

$\frac{3}{8}C = 6 \pi$

$\frac{8} {3} \cdot \frac{3}{8}C =\frac{8} {3} \cdot 6 \pi$

$C = 16 \pi$

The equation of a circle on the coordinate plane is

$(x- h)^{2}+ (y - k)^2 = r^{2}$ ,

where (h, k) are the coordinates of the center and is the radius.

The radius of a circle can be determined by dividing its circumference by $2 \pi$ , so

$r = \frac{C}{2 \pi } = \frac{16 \pi }{2 \pi } = 8$

The center of the circle is (0,0) , so h = k = 0 . Substituting 0, 0, and 8 for , , and , respectively, the equation of the circle becomes

$(x- 0)^{2}+ (y - 0)^2 =8^{2}$ ,

$x^{2}+ y^2 =64$ .

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SAT Mathematics : SAT Math

Study concepts, example questions & explanations for SAT Mathematics

All SAT Mathematics Resources

Example Questions

Example Question #8 : Using Radians

Example Question #9 : Using Radians

Example Question #10 : Using Radians

Example Question #1 : Applying The Equation Of A Circle

Example Question #2 : Applying The Equation Of A Circle

Example Question #1 : Applying The Equation Of A Circle

Example Question #3 : Applying The Equation Of A Circle

Example Question #2 : Applying The Equation Of A Circle

Example Question #4 : Applying The Equation Of A Circle

Example Question #5 : Applying The Equation Of A Circle

All SAT Mathematics Resources

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