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.