3.03 Apply the Pythagorean theorem
Using the Pythagorean theorem
The Pythagorean theorem states that in a right triangle the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. The theorem can be written algebraically.
$a^2+b^2=c^2$a2+b2=c2
where $c$c represents the length of the hypotenuse and $a$a, $b$b are the two shorter sides. To see why this is true you can check out the lesson here.
We can use the formula to find any side if we know the lengths of the two others.
If after we are done solving for the third side, we find that all three side lengths are whole numbers, the three side lengths may be referred to as a Pythagorean triple.
Worked examples
Question 1
Find the length of the hypotenuse of a right triangle whose two other sides measure $3$3 cm and $4$4 cm.
Think: Here we want to find $c$c, and are given $a$a and $b$b.
Do: We will substitute the values we know in the formula and then solve to find $c$c.
| $c^2$c2 | $=$= | $3^2+4^2$32+42 |
Fill in the values for $a$a and $b$b |
| $c^2$c2 | $=$= | $9+16$9+16 |
Evaluate the squares |
| $c^2$c2 | $=$= | $25$25 |
Add the numbers together |
| $c$c | $=$= | $\sqrt{25}$√25 |
Take the square root of both sides |
| $c$c | $=$= | $5$5 cm |
Reflect: Because all the lengths for the sides of this triangle are whole numbers and they satisfy Pythagorean theorem, the numbers $(3,4,5)$(3,4,5) form a Pythagorean triple.
Question 2
Find the length of the hypotenuse, $c$c in this triangle.
Applying this relationship
If we need to find one of the shorter side lengths ($a$a or $b$b), using the formula we will have one extra step of rearranging to consider.
Question 3
Find the length of unknown side $b$b of a right triangle whose hypotenuse is $10$10 mm and one other side is $6$6 mm.
Think: Here we want to find $b$b, the length of a shorter side.
Do:
| $c^2$c2 | $=$= | $a^2+b^2$a2+b2 | Start with the formula |
| $10^2$102 | $=$= | $6^2+b^2$62+b2 | Fill in the values we know |
| $b^2$b2 | $=$= | $10^2-6^2$102−62 | Rearrange to get the b**2 on its own |
| $b^2$b2 | $=$= | $100-36$100−36 | Evaluate the right-hand side |
| $b^2$b2 | $=$= | $64$64 | |
| $b$b | $=$= | $8$8 | Take the square root of both sides |
Reflect: Again, we can see that because all the lengths for the sides of the triangle are whole numbers and they satisfy the Pythagorean theorem they are said to form a Pythagorean triple.
| $a^2+b^2$a2+b2 | $=$= | $c^2$c2 |
| other side lengths | hypotenuse |
The value $c$c is used to represent the hypotenuse which is the longest side of the triangle. The other two lengths are $a$a, $b$b.
Use the letters provided to you in the questions, if no letters are provided you can use $a$a and $b$b for either of the sides.
Question 4
Calculate the value of $b$b in the triangle below. 
Question 5
Calculate the value of $a$a in the triangle below.
A right-angled triangle with the right angle at the bottom right corner, indicated by a small square symbol. The side opposite the right angle is the hypotenuse and measures 17 cm. The horizontal leg of the right angle, and also the base of the triangle, measures 15 cm. The vertical leg of the right angle, and also the height of the triangle, measures $a$a cm.
Pythagorean Triples
A Pythagorean triple (sometimes called a Pythagorean triple) is an ordered triple $\left(a,b,c\right)$(a,b,c) of three positive integers such that $a^2+b^2=c^2$a2+b2=c2.
If $\left(a,b,c\right)$(a,b,c) is a triple then $\left(b,a,c\right)$(b,a,c) is also a triple, since $b^2+a^2$b2+a2 is the same as $a^2+b^2$a2+b2. So the order of the first two numbers in the triple doesn't matter.
$\left(6,8,10\right)$(6,8,10) is also a Pythagorean triple, but it can be considered a multiple of another known Pythagorean triple, since $6$6, $8$8 and $10$10 have a common factor of $2$2. If we divide each number in the triple by this common factor, we get the known Pythagorean triple $\left(3,4,5\right)$(3,4,5).
A triangle whose sides form a Pythagorean triple will always be a right triangle.
Worked example
Question 6
The two smallest numbers in a Pythagorean triple are $20$20 and $21$21. What number, $c$c, will complete the triple?
Think: Here we have $a=20$a=20 and $b=21$b=21 We can use the Pythagorean theorem and solve for $c$c.
Do: Using the Pythagorean theorem,
| $c^2$c2 | $=$= | $a^2+b^2$a2+b2 | Start by writing the Pythagorean Theorem |
| $=$= | $20^2+21^2$202+212 | Substitute in the lengths of the legs | |
| $=$= | $400+441$400+441 | Square both | |
| $=$= | $841$841 | Find the sum | |
| so, $c$c | $=$= | $\sqrt{841}$√841 | Then, take the square root of both sides of the equation |
| $c$c | $=$= | $29$29 | Since 841 was a perfect square, $c$c is equal to $29$29 |
The missing value is $c=29$c=29, forming the triple $\left(20,21,29\right)$(20,21,29).
Practice questions
Question 7
Sean knows the two largest numbers in a Pythagorean Triple, which are $41$41 and $40$40. What number, $a$a, does Sean need to complete the triple?
Question 8
We would like to find the hypotenuse in a right triangle with shorter side lengths $6$6 and $8$8, using our knowledge of common Pythagorean triples. Below are some common Pythagorean triples. The two shorter sides $6$6, $8$8 and its hypotenuse will be multiples of the sides in which of the triples? They will be multiples of the Pythagorean triple: ($\editable{}$,$\editable{}$,$\editable{}$)
What number when multiplied by $3$3 and $4$4 gives $6$6 and $8$8 respectively?
$\editable{}$ Hence, what is the length of the hypotenuse in the triangle with two shorter sides $6$6 and $8$8? $\editable{}$
A $\left(3,4,5\right)$(3,4,5)
B $\left(5,12,13\right)$(5,12,13)
C $\left(8,15,17\right)$(8,15,17)
D $\left(7,24,25\right)$(7,24,25)
Applications using the Pythagorean theorem
Remember that for the Pythagorean theorem to be applicable to a practical problem, we must be considering the lengths of sides of a triangle that we are certain is a right triangle.
$a^2+b^2=c^2$a2+b2=c2, where
- $c$c is the length of the hypotenuse, and
- $a$a and $b$b are the lengths of the other two sides
There is a way to shorten our work. We can rearrange the Pythagorean theorem to find formulas for each side length.
To find the hypotenuse: $c=\sqrt{a^2+b^2}$c=√a2+b2
To find a shorter side: $a=\sqrt{c^2-b^2}$a=√c2−b2
To apply the Pythagorean theorem to real-life situations,
- Look for right triangles
- Choose which side, hypotenuse or a shorter side, you are trying to find
- Find the lengths of the other two sides
- Apply the relevant formula and substitute the lengths of the other two sides
Let's look at some examples so we can see this in action.
Practice questions
Question 9
Consider a cone with slant height $13$13m and perpendicular height $12$12m.
Find the length of the radius, $r$r, of the base of this cone.
Hence, find the length of the diameter of the cone's base.
Question 10
Find the length of the unknown side, $x$x, in the given trapezoid.
Give your answer correct to two decimal places.
A right trapezoid $ABDC$ABDC is depicted as suggested by the two adjacent right angles $\angle BAC$∠BAC or $\angle CAB$∠CAB on vertex $9$9 and $\angle DCA$∠DCA or $\angle ACD$∠ACD on vertex $7$7. Side $AB$AB or $BA$BA and Side $DC$DC or $CD$CD are the parallel sides of the trapezoid and Side $AB$AB or $BA$BA is longer than side $DC$DC or $CD$CD. Side $CA$CA or $AC$AC measures 9 units and is perpendicular to the two parallel sides. Side $CA$CA or $AC$AC is the base of the figure. Side $AB$AB or $BA$BA is measured as 13 units. Side $DC$DC or $CD$CD is measured as 7 units. Side $BD$BD or $DB$DB is labeled as x units.