Measurement

Ontario 09 Applied (MFM1P)

Solve contextual problems involving perimeters

Lesson

Once we know how to calculate the perimeter of a shape, or polygon, we can use this to solve real problems that we may come across.

The outside measurement can be used for lots of things, including measuring how much a new fence might cost.

A rectangular athletics field is $140$140 metres long and $40$40 metres wide. How far, in km will an athlete run by completing $6$6 laps of along the edge of the field?

We can also use the perimeter to compare things, as we do in this video. We need to hang a large television on the wall, but need to know if the wall is big enough. We then want to go one step further, and centre the television. How can we do that? Have a look and see how we can do both of these things.

The length of a rectangle is twice its width and its perimeter is $30$30 cm.

If the width of the rectangle is $x$

`x`cm, write an expression for the perimeter of the rectangle in terms of $x$`x`.Use the expression obtained above to find the width ($x$

`x`) of the rectangle.Find the length of the rectangle.

A present is contained in a cube shaped box. Ribbon is wrapped around the present as shown.

If the side length of the box is $10$10 cm, what is the shortest length of the ribbon needed to neatly go around the box without overlap?

The bow requires $8$8 cm of ribbon. How much ribbon is needed altogether to wrap the box in this way?

What is the total length of ribbon needed if we wrap the present with two lengths of ribbons, as shown below? (Assume a single bow is tied.)

Remember!

Once we know the perimeter of our shape, or even how to calculate the perimeter, we can work out other things. No matter the shape, perimeter is always the total of the outside lengths.

Solve problems involving the areas and perimeters of composite two-dimensional shapes (i.e., combinations of rectangles, triangles, parallelograms, trapezoids, and circles)