The Pythagorean Theorem on the SAT
The Pythagorean theorem is on the reference sheet, but the questions rarely tell you to use it. You will find it inside ladders leaning on walls, diagonals of rectangles, distances between points, and the heights of isosceles triangles.
In a right triangle with legs $a$ and $b$ and hypotenuse $c$, $a^2 + b^2 = c^2$. The hypotenuse is the side across from the right angle, and it is always the longest side.
Before you calculate, look for a triple you already know: $3$-$4$-$5$, $5$-$12$-$13$, $8$-$15$-$17$, $7$-$24$-$25$, or a multiple of one, like $6$-$8$-$10$ or $9$-$12$-$15$. When you spot one, you don't need to calculate at all.
$a^2 + b^2 = c^2$. The hypotenuse $c$ is always across from the right angle, and it is always the longest side.
Three examples
Example 1: a ladder
A $25$-foot ladder leans against a wall with its base $7$ feet from the wall. How high up the wall does it reach?
- The ladder is the hypotenuse: $7^2 + h^2 = 25^2$.
- $h^2 = 625 - 49 = 576$.
- $h = 24$ feet. This is the $7$-$24$-$25$ triple.
Example 2: the height of an isosceles triangle
An isosceles triangle has sides $13$, $13$ and $10$. Find its area.
- The height cuts the base of $10$ into two halves of $5$.
- Each half is a right triangle with hypotenuse $13$ and leg $5$: $h = \sqrt{169 - 25} = 12$.
- Area $= \dfrac{1}{2}(10)(12) = 60$.
Example 3: distance between points
Find the distance between $(1, 2)$ and $(7, 10)$.
- The run is $7 - 1 = 6$ and the rise is $10 - 2 = 8$.
- $d = \sqrt{6^2 + 8^2} = \sqrt{100} = 10$.
- The $6$-$8$-$10$ triple again.
What goes wrong
- Adding when you should subtract. If you know the hypotenuse, the missing leg is $\sqrt{c^2 - a^2}$.
- Putting the wrong side as $c$. The hypotenuse is across from the right angle, never one of the legs.
- Forgetting the square root. $6^2 + 8^2 = 100$ is $c^2$. The side is $10$.
- Adding the sides before squaring. $(a + b)^2$ is not $a^2 + b^2$.
- Using the full base in an isosceles triangle. The height splits the base in half.
Where the theorem comes up
| You see | Do this |
|---|---|
| Two legs, find the hypotenuse | $c = \sqrt{a^2 + b^2}$. |
| Hypotenuse and one leg | Subtract: $b = \sqrt{c^2 - a^2}$. |
| Diagonal of a rectangle | The length and width are the legs. |
| Distance between two points | The run and the rise are the legs. |
| Height of an isosceles triangle | The height splits the base in half and makes two right triangles. |
| "Is it a right triangle?" | Check whether the two shorter sides squared add up to the longest side squared. |
Practice questions
Look for a triple before you calculate. Then open the solution to check.
- A$14$
- B$10$
- C$48$
- D$100$
Show solution
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- A$4, 5, 6$
- B$5, 6, 8$
- C$9, 12, 15$
- D$6, 7, 9$
Show solution
- A$21$
- B$\sqrt{63}$
- C$108$
- D$15$
Show solution
- A$4\sqrt{2}$
- B$8$
- C$16$
- D$2\sqrt{2}$
Show solution
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- A$6$
- B$2\sqrt{11}$
- C$8$
- D$2\sqrt{6}$
Show solution
Show solution
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