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4.01 Pythagoras' theorem

Lesson

Pythagoras' theorem states that the three sides of any right-angled triangle (a triangle that has a $90^\circ$90° angle), are related in a simple way.

Let's look at a right-angled triangle with sides measuring $a$a, $b$b and $c$c:

The side $c$c is the hypotenuse. The hypotenuse of a right-angled triangle can be readily identified because it is always:

  1. the longest side in the triangle, and
  2. opposite the right-angle.

 

Sides $a$a, $b$b and $c$c are related in the following way:

Pythagoras' theorem

For any right-angled triangle: 

"the square of the hypotenuse is equal to the sum of the squares of the two shorter sides"

 

$c^2=a^2+b^2$c2=a2+b2

 

where $c$c represents the length of the triangle's hypotenuse and $a$a and $b$b are the lengths of the two shorter sides. 

Pythagoras' theorem can be represented by the following diagram, where the combined areas of the squares produced by the two shorter sides is equal to the area of the square produced by the hypotenuse.

If two sides of any right-angled triangle are known, and one side is unknown, Pythagoras' theorem can be used to find the length of the unknown side. 

We can rearrange the equation to make the unknown side the subject:

If we want to solve for the hypotenuse: $c=\sqrt{a^2+b^2}$c=a2+b2
   
If we want to solve for either of the two shorter sides:

$b=\sqrt{c^2-a^2}$b=c2a2

$a=\sqrt{c^2-b^2}$a=c2b2

 

Worked example

A right-angled triangle has two shorter sides that are $10$10 cm and $12$12 cm in length. Find the length of the hypotenuse.

Think: We want to find the length of the longest side. The two shorter sides, $a$a and $b$b are given, so we can substitute directly into Pythagoras's theorem.

Do:

$c^2$c2 $=$= $a^2+b^2$a2+b2

 

  $=$= $10^2+12^2$102+122

(Substitute $a=10$a=10 and $b=12$b=12)

  $=$= $100+144$100+144

 

  $=$= $244$244

 

$c$c $=$= $\sqrt{244}$244

(Take the square root of both sides)

  $=$= $15.6204$15.6204...

 

  $=$= $15.62$15.62 cm ($2$2 d.p.)

(Round to 2 decimal places)

 

So, the hypotenuse is approximately $15.62$15.62 cm long.

 

Determine if a triangle is right-angled

If all we know about a triangle is the lengths of its sides, Pythagoras' theorem can be used to determine whether or not the triangle is right-angled.

For example, if a triangle has side lengths measuring $6$6 cm, $8$8 cm and $10$10 cm, we can substitute these values into Pythagoras' theorem. If the values satisfy the theorem, then the triangle must be right-angled.

The hypotenuse, $c$c, is clearly $10$10 cm, because it is the longest side. It doesn't matter how we assign $a$a and $b$b, but let's make $a=6$a=6 cm and $b=8$b=8 cm.

$c^2$c2 $=$= $10^2$102
  $=$= $100$100

 

$a^2+b^2$a2+b2 $=$= $6^2+8^2$62+82
  $=$= $36+64$36+64
  $=$= $100$100

 

Since $c^2=a^2+b^2$c2=a2+b2, a triangle with side-lengths $6$6, $8$8 and $10$10 units, must be a right-angled triangle.

 

 

Practice Questions

Question 1

Calculate the value of $c$c in the triangle below.

A right-angled triangle with a right angle shown at the bottom right corner. The base is labeled as 14 cm, the height on the right side is labeled as 48 cm, and the hypotenuse is labeled with c cm. There is a small square at the right angle indicating the 90-degree angle.

Question 2

Calculate the value of $b$b in the triangle shown. Give your answer correct to $2$2 decimal places.

Question 3

Find the length of the unknown side $b$b in a right-angled triangle whose hypotenuse measures $6$6 mm and one other side measures $4$4 mm.

Give your answer correct to two decimal places.

Question 4

Find the length of the unknown side, $x$x, in the given trapezium.

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.

 

 

Outcomes

MS11-4

performs calculations in relation to two-dimensional and three-dimensional figures

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