Measurement

New Zealand

Level 4

Lesson

We know the area of a rectangle can be worked out by multiplying its length by its width, but did you know that we can also use that rule to help us work out the area of a triangle? It's a pretty neat thing to know, so let's see how this can be true.

Now that we have seen that the rule works, we can use it to work out the area of triangles.

Area of a triangle = $\frac{1}{2}$12 base × perpendicular height

Did you notice we've only looked at right-angled triangles, where we can make out the perpendicular height using the existing sides of the triangle? Well, the rule definitely works for all triangles, so let's just check, so we can trust the rule for all triangles.

We can use the rule:

Area of a triangle = $\frac{1}{2}$12 base × perpendicular height

for any triangle. The important thing is that you choose one of the triangle's sides as the base, and then use the perpendicular height. For a right-angled triangle, it's where the right angle is formed with the base. For others, it is the line from the base to the opposite corner (vertex), where a right angle is formed. It may even be outside the triangle, like the last image below.

This video helps you identify the height of triangles, so don't worry if you aren't sure. It also shows you how to calculate the area of any triangle.

you could change the base and height of a triangle, and see how it changes the area. You can, in this applet! Play around and watch what happens as you change the area. You can modify the height and base, as well as move the top point.

Remember!

The rule for the area of a triangle can be written in different ways, but they all mean the same thing:

Area = $\frac{1}{2}$12 × base × height

Area = $\frac{1}{2}bh$12`b``h`

Area = $\frac{1}{2}\left(b\right)\left(h\right)$12(`b`)(`h`)

When you know the area of a triangle, you can use the area to work out the missing base or height of the triangle. Let's see how we can find the missing values when we know the area and one of the dimensions of our triangle.

In this applet, you get to change the dimensions of the triangle, to change the area. The area is shown, so you can choose to reveal the base or the height. Can you work out the missing dimension? Reveal the answer to check your answer.

Now that we know how to calculate the area of triangles, we can use it to work out actual problems we may have. That can include working out the base or perpendicular height, if we know the area of our triangle. You may be surprised at how many triangles are around you. In the video you'll see some common triangles, and even see how triangles are a part of chatterboxes, or fortune-tellers. Do you remember making these?

Did you see the chatterbox in the video and wonder how you could make one? Here's a link to find out more:

https://www.cleverpatch.com.au/ideas/by-product-type/paper-and-card/chatterbox

While it is a little bit of fun, it's a great opportunity to see how the dimensions of the triangle and square (a type special rectangle) are related. Maybe you could make one and include questions about the area of triangles!

In this applet, you have a rectangle and $6$6 triangles. The area of the rectangle is known, but you have to work out the area of the triangles. You can reveal up to $3$3 clues if you need them. Grab the triangles by the dots and move them and rotate them to fit in the rectangle.

Remember!

Sometimes you can use the features of one shape to help solve problems with another.

Use the picture to answer the following questions about the area of the rectangle and the triangle.

Firstly, find the area of the entire rectangle.

Now find the area of the triangle.

Area of triangle $=$= $\frac{1}{2}$12 $\times$× area of rectangle m ^{2}Area of triangle $=$= $\frac{1}{2}\times\editable{}$12× m ^{2}(Fill in the value for the area of the rectangle.) Area of triangle $=$= $\editable{}$ m ^{2}(Complete the multiplication.)

Find the area of the triangle with base length $10$10 m and perpendicular height $8$8 m shown below.

Find the value of $h$`h` in the triangle with base length $6$6 cm if its area is $54$54 cm^{2}.

A gutter running along the roof of a house has a cross-section in the shape of a triangle. If the area of the cross-section is $50$50 cm^{2}, and the length of the base of the gutter is $10$10 cm, find the perpendicular height $h$`h` of the gutter.

Use side or edge lengths to find the perimeters and areas of rectangles, parallelograms, and triangles and the volumes of cuboids