ACT Pythagorean Theorem and Coordinate Distance

For a right triangle, use $a^2+b^2=c^2$, with $c$ opposite the right angle. Add the squared legs to find the hypotenuse; subtract a known leg’s square from the hypotenuse’s square to find the other leg. Coordinate distance uses the same idea: horizontal and vertical changes form perpendicular legs.

ACT’s mathematics standards include the Pythagorean theorem and distance formula. This guide connects those skills through original Kibo examples. None of the problems below are official ACT questions.

Identify the right angle and the hypotenuse first

The legs meet at the $90^\circ$ angle. The hypotenuse sits across from that angle and is the longest side. Rotating a diagram does not change which side is the hypotenuse. Use a marked right angle or information that establishes perpendicular sides, such as adjacent sides of a rectangle.

OpenStax’s Pythagorean theorem lesson explains the side relationship and how to solve for either a leg or the hypotenuse. All lengths must use the same unit before you square them.

The square root gives a length, so choose the nonnegative result. For a triangle with three positive side lengths, the hypotenuse must be longer than either leg. A negative value under the square root is a reason to check your side labels or the given information.

Check a triangle before assuming it is right

Suppose a triangle has side lengths $7$, $9$, and $12$ centimeters. The longest side would have to be the hypotenuse if the triangle were right. But $7^2+9^2=49+81=130$, while $12^2=144$. The required equality fails, so this triangle is not right. A sketch that looks square at one corner cannot override those lengths.

Worked example: the diagonal of a rectangular panel

A rectangular display panel measures $7.5$ centimeters by $10$ centimeters. What is its diagonal length? The diagonal divides the rectangle into two right triangles, and it is opposite each right angle.

$$d^2=7.5^2+10^2=56.25+100=156.25$$

$$d=\sqrt{156.25}=12.5\text{ cm}$$

Check: $12.5^2=156.25$, and $12.5$ is longer than either side. Adding $7.5+10=17.5$ would measure a route along two edges. Multiplying $7.5\times10=75$ would give the area in square centimeters. Neither answers the diagonal-length question.

Worked example: finding a missing leg

A taut support cable is $26$ meters long. It runs from a point on a vertical pole to an anchor $10$ meters from the pole’s base on level ground. How high above the ground is the attachment? Assume the cable is a straight segment and the pole is perpendicular to the ground.

The cable is the hypotenuse. If $h$ is the attachment height in meters, then:

$$10^2+h^2=26^2$$

$$h^2=676-100=576$$

$$h=\sqrt{576}=24$$

The attachment is $24$ meters high. Substitution checks it: $10^2+24^2=100+576=676=26^2$. Adding the two known squares would incorrectly treat the cable as a leg. Subtracting the lengths directly, $26-10=16$, also misses the squared relationship.

Coordinate distance is a right-triangle problem

For points $(x_1,y_1)$ and $(x_2,y_2)$, the horizontal separation is $|x_2-x_1|$, and the vertical separation is $|y_2-y_1|$. Their squares give:

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

OpenStax derives the distance formula from horizontal and vertical legs. Pair the two horizontal coordinates and the two vertical coordinates. Put parentheses around negative coordinates before subtracting.

Worked example: exact distance and a two-leg route

Find the straight-line distance from $A=(-6,2)$ to $B=(3,-4)$, first exactly and then to the nearest tenth of a unit.

The coordinate changes are $3-(-6)=9$ and $-4-2=-6$. The downward change is negative, but its length is $6$. Squaring the signed difference gives the same result:

$$d=\sqrt{9^2+(-6)^2}=\sqrt{81+36}=\sqrt{117}=3\sqrt{13}$$

The exact distance is $3\sqrt{13}$ units. Its decimal value is about $10.8167$, so the nearest-tenth answer is $10.8$ units. Keep the radical until the requested rounding step. Check the exact answer by squaring: $(3\sqrt{13})^2=9\times13=117$.

A route that goes $9$ units right and then $6$ units down has length $15$ units. That is longer than the straight segment. If a question asks how far someone travels along those two segments, add the route lengths. If it asks how far apart the endpoints are, use the straight-line distance.

Changing the order of the points gives differences $-9$ and $6$, whose squares are unchanged. If the points share a horizontal or vertical coordinate, one difference is zero and the formula reduces to the absolute value of the other difference. Identical points have distance zero.

Practice before revealing the answers

For each problem, identify the requested length, write the equation, and check the result against the picture or coordinates.

1. A rectangular banner is $11$ inches wide and $13$ inches tall. Find its diagonal exactly and to the nearest tenth of an inch.
2. A right triangle has hypotenuse $29$ centimeters and one leg $20$ centimeters. Find the other leg.
3. Find the distance between $(-7,-3)$ and $(-2,9)$.
4. Find the distance between $(-4,8)$ and $(-4,-6)$. Explain why a full two-leg calculation is unnecessary.

A quick check before you move on

• Shape: What establishes the right angle?
• Side: Is the unknown a leg, a hypotenuse, or a route made of several segments?
• Arithmetic: Did you square each difference and take the final square root?
• Answer: Are the units consistent, is the length reasonable, and did you round only as requested?

If squaring negative differences caused trouble, review minus signs and parentheses in ACT math. For a wrong setup, use the practice error-log template to record which side you mislabeled, then retry with the explanation covered.

Ready for another question? Try Kibo’s free ACT practice.

Prepared with AI assistance. Original examples were checked with separate calculations; no human review is claimed. Sources checked October 27, 2026. ACT is a registered trademark of ACT, Inc., which is not affiliated with or endorsing this guide.