# Video: Pack 2 • Paper 3 • Question 17

Pack 2 • Paper 3 • Question 17

03:07

### Video Transcript

The diagram shows a rectangular field. Lowercase 𝑙 is 3.25 metres to the nearest centimetre. Capital 𝐿 is 6.4 metres to the nearest centimetre. Calculate the lower bound for the length of the diagonal of the field. Show all of your working.

Our first step is to work out the maximum and minimum or upper and lower bounds of the length and width of the rectangle. The width of the rectangle was 3.25 metres to the nearest centimetre. This means that our lower bound will be 3.245 and our upper bound will be 3.255.

It’s important to note that when we write our inequality that 𝑙 can be greater than or equal to 3.245, but must be less than 3.255. This is because a length of 3.255 would be rounded up to 3.26 metres. The length of the rectangular field was 6.4 metres to the nearest centimetre. This means that our lower bound would be 6.395 and our upper bound would be 6.405.

Once again, when writing the inequality, it is important to note that 𝐿 can be greater than or equal to 6.395, but must be less than 6.405. We now have the maximum and minimum length and width of the rectangular field. As our question asked for the lower bound, we need to use the values 3.245 and 6.395.

In order to calculate the diagonal of the field, we need to consider the right-angled triangle drawn. Pythagoras’s theorem states that 𝑎 squared plus 𝑏 squared is equal to 𝑐 squared, where 𝑐 is the longest side of our right-angled triangle, known as the hypotenuse. Substituting in the values from our triangle gives us 𝑑 squared is equal to 3.245 squared plus 6.395 squared. Square rooting both sides of this equation gives us 𝑑 is equal to the square root of 3.245 squared plus 6.395 squared. Typing this into the calculator gives us a value of 𝑑 of 7.171.

This means that the lower bound for the length of the diagonal of the field is 7.171 metres.