6.01 Pythagoras' theorem
In a right-angled triangle the largest angle in the triangle is $90^\circ$90°. The side across from the right angle will be the largest side. We call this side the hypotenuse.
All three sides of a right-angled triangle are related by the equation shown below:
The two smaller sides will be called $a$a and $b$b, and the hypotenuse (the longest side) will be $c$c. This relationship between sides in a right-angled triangle is called Pythagoras' theorem. We can use this theorem to find both hypotenuses and short sides.
Worked examples
Example 1
Is the following triangle a right-angled triangle?
Think: If its three side lengths satisfy $a^2+b^2=c^2$a2+b2=c2 then the triangle will be a right-angled triangle.
Do:
| $a^2+b^2$a2+b2 | $=$= | $15^2+20^2$152+202 |
Calculating left-hand side |
| $$ | $=$= | $225+400$225+400 |
|
| $$ | $=$= | $625$625 |
Evaluating |
| $c^2$c2 | $=$= | $25^2$252 |
Calculating right-hand side |
| $$ | $=$= | $625$625 |
Evaluating |
| $a^2+b^2$a2+b2 | $=$= | $c^2$c2 |
|
Reflect: We can also skip some of the working out if we realise that $\left(15,20,25\right)$(15,20,25) is a multiple of $\left(3,4,5\right)$(3,4,5), because each number has been multiplied by $5$5. This means it will also be a Pythagorean triple and the triangle will be a right-angled triangle.
EXAMPLE 2
Find the length of the hypotenuse of a right-angled triangle whose two other sides measure $10$10 cm and $12$12 cm. Round your answer to two decimal places.
Think: Here we want to find $c$c, and are given $a$a and $b$b. We can substitute the known values for $a$a and $b$b into Pythagoras' formula $a^2+b^2=c^2$a2+b2=c2 and then solve for $c$c.
Do:
| $c^2$c2 | $=$= | $a^2+b^2$a2+b2 |
Start with the formula |
| $c^2$c2 | $=$= | $10^2+12^2$102+122 |
Fill in the values for $a$a and $b$b |
| $c^2$c2 | $=$= | $100+144$100+144 |
Evaluate the squares |
| $c^2$c2 | $=$= | $244$244 |
Add the $100$100 and $144$144 together |
| $c$c | $=$= | $\sqrt{244}$√244 |
Take the square root of both sides |
| $c$c | $=$= | $15.62$15.62 cm |
Rounded to two decimal places |
Reflect: We can see that the formula gives us a hypotenuse length of $15.62$15.62 cm. This is larger than both our shorter sides, as we should expect. If we got a number smaller than one (or both) of the short sides we know we have made a mistake in our calculation.
EXAMPLE 3
A right-angled triangle has a hypotenuse of length $20$20 m and one short side that has a length of $11$11 m. Find the exact length of the other short side, and find its length rounded to two decimal places.
Think: Since we want to find the length of a short side, we will be solving for either $a$a or $b$b- let's choose $a$a. Since we want to find $a$a, our given values will be $b$b and $c$c which we can substitute into Pythagoras' formula $a^2+b^2=c^2$a2+b2=c2 and then solve for $a$a.
Do: We will substitute $b=11$b=11 and $c=20$c=20 into the formula for Pythagoras' theorem:
| $a^2+b^2$a2+b2 | $=$= | $c^2$c2 |
Start with the formula |
| $a^2+11^2$a2+112 | $=$= | $20^2$202 |
Fill in the values for $b$b and $c$c |
| $a^2$a2 | $=$= | $20^2-11^2$202−112 |
Subtract $11^2$112 from both sides to make $a^2$a2 the subject |
| $a^2$a2 | $=$= | $400-121$400−121 |
Evaluate the squares |
| $a^2$a2 | $=$= | $279$279 |
Subtract $121$121 from $400$400 |
| $a$a | $=$= | $\sqrt{279}$√279 m |
Take the square root of both sides |
| $a$a | $=$= | $16.70$16.70 m |
Rounded to two decimal places |
The exact length is $\sqrt{279}$√279 m, and the rounded length is $16.70$16.70 m.
Reflect: When finding a short side, our answer should always be shorter than the hypotenuse. If our answer is longer, we know we have made a mistake.
The most important thing to remember when finding a short side is that the two lengths need to go into different parts of the formula.
If you get the lengths around the wrong way, you will probably end up with the square root of a negative number (and a calculator error).
Pythagoras' theorem relates the three sides of a right-angled triangle, $a$a and $b$b are the two smaller sides, and the longest side, called the hypotenuse, is $c$c.
We can also test to see if a triangle is right-angled by checking to see if its three sides satisfy $a^2+b^2=c^2$a2+b2=c2.
To find the hypotenuse:
$c^2=a^2+b^2$c2=a2+b2
To find a shorter side:
$a^2=c^2-b^2$a2=c2−b2 or $b^2=c^2-a^2$b2=c2−a2
We can take the square root of both sides to give us the following formulas:
$c=\sqrt{a^2+b^2}$c=√a2+b2, $a=\sqrt{c^2-b^2}$a=√c2−b2, $b=\sqrt{c^2-a^2}$b=√c2−a2
Practice questions
Question 1
Find the length of the unknown side $c$c in the triangle below. Write each step of working as an equation.
Question 2
Find the length of the unknown side $x$x in the triangle below. Write each step of working as an equation.
Question 3
A movie director wants to shoot a scene where the hero of the film fires a grappling hook from the roof of one building to the roof of another. If the first building is $37$37 m tall, the other building is $54$54 m tall and the street between them is $10$10 m wide, what is the minimum length $l$l of rope needed for the grappling hook? Round your answer to two decimal places.