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India
Class VII

Calculating Side Lengths Using Pythagoras' Theorem

Lesson

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.  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. 

 

Remember!
$a^2+b^2$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.  

When you have to round to a required number of decimal places, look to the digit in the following place value column.  If that value is 5 or higher, we round up.  If that value is less than 5 we round down.  This lesson will help you with rounding decimals if you need a refresher. 

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 $=$= $a^2+b^2$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 $2$2 decimal places

 

Question 2

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

Round your answer to two decimal places.

A right-angled triangle with one leg measuring 3 cm, the other leg measuring 9 cm, and the hypotenuse measuring c cm. The right angle is indicated by a small square symbol.

 

 

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 1 extra step of rearranging to consider.  

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


Question 3

Find the length of unknown side $b$b of a right 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:

$c^2$c2 $=$= $a^2+b^2$a2+b2 start with the formula
$6^2$62 $=$= $4^2+b^2$42+b2 fill in the values we know
$b^2$b2 $=$= $6^2-4^2$6242 rearrange to get the $b^2$b2 on its own
$b^2$b2 $=$= $36-16$3616 evaluate the right-hand side
$b^2$b2 $=$= $20$20  
$b$b $=$= $\sqrt{20}$20 take the square root of both sides
$b$b $=$= $4.47$4.47 mm  round to the required number of decimal places

Reflect: This gives us a triangle with hypotenuse $6$6 mm and side lengths $4$4 and $4.47$4.47 mm.  It's good to check at the end that you haven't ended up with a side length longer than the hypotenuse, as the hypotenuse has to be the longest length. 

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

Consider the triangle below. One of the side lengths is unknown.

A right-angled triangle is shown with its sides labeled. The right angle is indicated at the vertex with a small square. The shorter side is labeled 6 $m$m. The longer side, whose dimension is unknown, is labeled $b$b $m$m. The hypotenuse is labeled 13 $m$m.
  1. Find the length of the unknown side in this triangle, rounding your answer to two decimal places (if necessary).

 

Outcomes

7.G.PT.4

Pythagoras Theorem (Verification only)

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