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SAT Math : Geometry

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Example Questions

Example Question #1 : Distance Formula

What is the distance of the line drawn between points (–1,–2) and (–9,4)?

Possible Answers:

√5

Correct answer:

Explanation:

The answer is 10. Use the distance formula between 2 points, or draw a right triangle with legs length 6 and 8 and use the Pythagorean Theorem.

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Example Question #2 : Distance Formula

What is the distance between two points $\dpi{100} \small (6,14)$ and $\dpi{100} \small (-6,9)$ ?

Possible Answers:

$\dpi{100} \small 13$

$\dpi{100} \small 17$

$\dpi{100} \small -12$

$\dpi{100} \small 5$

$\dpi{100} \small 10\sqrt{3}$

Correct answer:

$\dpi{100} \small 13$

Explanation:

To find the distance between two points such as these, plot them on a graph.

Then, find the distance between the $\dpi{100} \small x$ units of the points, which is 12, and the distance between the $\dpi{100} \small y$ points, which is 5. The $\dpi{100} \small x$ represents the horizontal leg of a right triangle and the $\dpi{100} \small y$ represents the vertial leg of a right triangle. In this case, we have a 5,12,13 right triangle, but the Pythagorean Theorem can be used as well.

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Example Question #3 : Distance Formula

What is the distance between $(1,3)\ and\ (5,6)$ ?

Possible Answers:

$7$

$5$

$6$

$4$

$8$

Correct answer:

$5$

Explanation:

Let $P_{1}(1,3)$ and $P_{2}(5,6)$ and use the distance formula:

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

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Example Question #4 : Distance Formula

What is the distance between the point (-12,35) and the origin?

Possible Answers:

Correct answer:

Explanation:

The distance between 2 points is found using the distance or Pythagorean Theorem. Because values are squared in the formula, distance can never be a negative value.

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Example Question #2 : Distance Formula

Bill gets in his car and drives north for 30 miles at 40 mph. He then turns west and drives 40 mph for 40 minutes. Finally, he goes directly northeast 40 miles in 25 minutes.

Using the total distance traveled as a straight line ("as the crow flies") and the time spent traveling, which of the following is closest to Bill's average speed?

Possible Answers:

$35\ mph$

$32\ mph$

$27\ mph$

$30\ mph$

$40\ mph$

Correct answer:

$32\ mph$

Explanation:

Each part of the problem gives you 2 out of the 3 pieces of the rate/time/distance relationship, thus allowing you to find the third (if needed) by using the equation:

$Distance = rate\cdot time$

The problem is otherwise an application of geometry and the distance formula. We need to find the distance between 2 points, but we need to go step-by-step to find out where the final point is. The first two steps are relatively easy to follow. He travels 30 miles north. Now he turns west and travels for 40 minutes at 40 mph. This is $\frac{2}{3}$ of an hour, so we have $40\cdot \frac{2}{3}=\frac{80}{3}\ miles$ . Thus if we are looking at standard Cartesian coordinates (starting at the origin), we are now at the point $\left ( \frac{80}{3},30 \right )$ .

We are now on the last step: 40 miles northeast. We need to decipher this into - and - coordinate changes. To do this, we think of a triangle. Because we are moving directly northeast, this is a 45 degree angle to the horizontal. We can thus imagine a 45-45-90 triangle with a hypotenuse of 40. Now using the relationships on triangles we have:

$40^{2} = x^{2} + x^{2} = 2x^{2}\Rightarrow 40 = \sqrt{2x^{2}} = x\sqrt{2} \Rightarrow x = \frac{40}{\sqrt{2}} = 20\sqrt{2}$

So the final step moves us $20\sqrt{2}$ up and to the right. Moving this way from the point $\left ( \frac{80}{3},30 \right )$ leaves us at:

$\left ( \frac{80}{3}-20\sqrt{2}, 30+20\sqrt{2} \right ) \approx \left ( -1.6, 58.3\right )$

Using the distance formula for this point from the origin gives us a distance of ~58 miles.

Now for the time. We traveled 30 miles at 40 mph. This means we traveled for $\frac{30}{40}$ hours or 45 minutes.

We then traveled 40 mph for 40 minutes. Increasing our total time traveled to 85 minutes.

Finally, we traveled 40 miles in 25 minutes, leaving our total time traveled at 110 minutes. Returning to hours, we have $\frac{110}{60}$ hours or 1.83 hours.

Our final average speed traveled is then:

$\frac{58 miles}{1.83 hours} \approx 31.7 mph$

The closest answer to this value is 32 mph.

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Example Question #3 : Distance Formula

Steven draws a line that is 13 units long. If $(-4,1)$ is one endpoint of the line, which of the following might be the other endpoint?

Possible Answers:

$(9,14)$

$(13,13)$

$(3,7)$

$(5,12)$

$(1,13)$

Correct answer:

$(1,13)$

Explanation:

The distance formula is $\sqrt{((x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2})}$ .

Plug in $(-4,1)$ with each of the answer choices and solve.

Plug in $(1,13)$ :

$d=\sqrt{(1-(-4))^2+ (13-1)^2} = \sqrt{5^2+12^2}=\sqrt{25+144}=\sqrt{169}=13$

This is therefore the correct answer choice.

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Example Question #1 : How To Find The Length Of A Line With Distance Formula

What is the distance between the points (3,12) and (7,2) ?

Possible Answers:

$2\sqrt{29}$

$3\sqrt{5}$

$4\sqrt{2}$

Correct answer:

$2\sqrt{29}$

Explanation:

Plug the points into the distance formula and simplify:

distance² = (x₂ – x₁)² + (y₂ – y₁)² = (7 – 3)² + (2 – 12)² = 4² + 10² = 116

distance = √116 = √(4 * 29) = 2√29

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Example Question #2 : How To Find The Length Of A Line With Distance Formula

What is the distance between $(3,4)$ and $(8,16)$ ?

Possible Answers:

$12$

$20$

$13$

$23$

$17$

Correct answer:

$13$

Explanation:

The formula for the distance between two points is $d = \sqrt{(x_{2} - x_{1})^2+(y_{2} - y_{1})^2}$ .

Plug in the points:

$d=\sqrt{(8-3)^2+(16-4)^2}$

$d = \sqrt{5^2+12^2} = 13$

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Example Question #9 : Distance Formula

Suppose a line is connected from two points. What is the distance of the line if the line is connected from (3,-9) to (-1,6) ?

Possible Answers:

$\sqrt{31}$

$\sqrt{241}$

$\sqrt{11}$

$\sqrt{19}$

$3\sqrt{17}$

Correct answer:

$\sqrt{241}$

Explanation:

Write the distance formula and substitute the values into the formula.

$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Let (x_1,y_1)=(3,-9) and (x_2,y_2)=(-1,6) .

Plugging in our coordinates into the distance formula we get the following.

$D=\sqrt{((-1)-3)^2+(6-(-9))^2}= \sqrt{(-4)^2+(15)^2}$

$D= \sqrt{16+225}=\sqrt{241}$

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Example Question #11 : Distance Formula

Find the Distance of the line shown below:

Screen shot 2015 10 27 at 3.11.25 pm

Possible Answers:

$2\sqrt{5}$

$\sqrt{53}$

$\sqrt{13}$

Correct answer:

$\sqrt{53}$

Explanation:

The distance formula is $^{\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}}$ . In the graph shown above the coordinates are (-5,8) and (2,10) . When you plug the coordinates into the equation you get:

$\sqrt{(-5-2)^{2}+(8-10)^{2}}$ , which then simplifies to $\sqrt{(-7)^{2}+(2)^{2}}$

$\sqrt{49+4}=\sqrt{53}$ , because is a prime number there is no need to simplify.

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SAT Math : Geometry

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #1 : Distance Formula

Example Question #2 : Distance Formula

Example Question #3 : Distance Formula

Example Question #4 : Distance Formula

Example Question #2 : Distance Formula

Example Question #3 : Distance Formula

Example Question #1 : How To Find The Length Of A Line With Distance Formula

Example Question #2 : How To Find The Length Of A Line With Distance Formula

Example Question #9 : Distance Formula

Example Question #11 : Distance Formula

All SAT Math Resources

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