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Pythagoras' Theorem
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Key Stage 3

Review of Pythagoras' Theorem

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Pythagorean Theorem

Pythagorean Theorem states that 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. 

So if two sides of a triangle are known and one side is unknown this relationship can be used to find the length of the unknown side.

We can rearrange the equation any which way to make the unknown side the subject.

$c=\sqrt{a^2+b^2}$c=√a2+b2

$b=\sqrt{c^2-a^2}$b=√c2−a2

$a=\sqrt{c^2-b^2}$a=√c2−b2

Examples

Question 1

Find the length of the hypotenuse of a right-angled triangle whose two other sides measure $10$10 cm and $12$12 cm.

Think: Here we want to find $c$c, and are given $a$a and $b$b.

Do:

$c^2$c2 $=$= $10^2+12^2$102+122
$c$c $=$= $\sqrt{10^2+12^2}$√102+122
$c$c $=$= $15.62$15.62 cm (rounded to $2$2 decimal places)

 

Question 2

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

Think: Here we want to find $b$b, the length of a shorter side.

Do:

$6^2$62 $=$= $b^2+4^2$b2+42
$b^2$b2 $=$= $6^2-4^2$62−42
$b$b $=$= $\sqrt{6^2-4^2}$√62−42
$b$b $=$= $4.47$4.47 mm (rounded to $2$2 decimal places)

Here are some worked examples.

Question 3

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 4

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

Give your answer correct to two decimal places.

A right-angled triangle is presented with the right angle at the top left. The side adjacent the right angle is labeled with the variable "b." The triangle's height, on the left side, is marked with the number 10, and the hypotenuse, slanting down from the left to the right, is labeled with the number 24. 

 

Question 5

Find the length of the unknown side in this right-angled triangle, expressing your answer as a decimal approximation to two decimal places.

 

We can combine this knowledge of Pythagoras, with our existing work on trigonometric ratios. 

Remember the Trigonometric Ratios

$\sin\theta=\frac{opposite}{hypotenuse}$sinθ=oppositehypotenuse​

$\cos\theta=\frac{adjacent}{hypotenuse}$cosθ=adjacenthypotenuse​

$\tan\theta=\frac{opposite}{adjacent}$tanθ=oppositeadjacent​

 

We can use these ratios to find unknown side lengths and angles in 2D shapes other than triangles, 3D shapes, as well as real-life situations.

The process is exactly the same. We just need to locate the right-angled triangle, the use to applicable trig ratio to work out the unknown value.

Let's look at some examples so we can see this in action.

 

Worked Examples

Question 6

Consider a cone with slant height $13$13m and perpendicular height $12$12m.

A cone with a circular base. The cone altitude, illustrated by a vertical dashed line, measures 12 meters, highlighted by a scale line on the left. The radius of the base circle is represented by a horizontal dashed line and is labeled r. These two lines are perpendicular, forming a right angle, which is denoted by a small square symbol. The slant height of the cone, which stretches from the apex to a point on the circumference of the base and opposite to the right angle, measures 13 meters, as indicated by the slanted scale line placed on the right. Together, the radius of the base (base), the altitude of the cone (height), and the slant height (hypotenuse) compose a right-angled triangle.

  1. Find the length of the radius, $r$r, of the base of this cone.

  2. Hence, find the length of the diameter of the cone's base.

Question 7

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$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$13 units. Side $DC$DC or $CD$CD is measured as $7$7 units. Side $BD$BD or $DB$DB is labeled as $x$x units.

 

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