 # Area Under a Graph Worksheets, Questions and Revision

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## What you need to know

The area underneath the line on a speed-time graph represents the distance travelled. When the graph is curved, it’s more difficult to find the area but we may still want an idea of the distance travelled. So, we estimate it.

Example: Below is a speed-time graph. Using 4 strips of equal width, find an approximation for the distance travelled in the first 4 seconds.

The question asks us to use “4 strips of equal width”. Given that we’re looking for the area under the graph in the first 4 seconds, each one of these strips must have a width of 1 second on the x-axis.

To form these strips then, we should start by drawing vertical lines at 1, 2, 3, and 4 seconds on the x-axis.  The result should look like the graph seen on the left.

Now, to find the area of these strips, we have to turn them into trapeziums.

To do this, we draw straight lines connecting the tops of each of the red lines we just drew, starting from the origin.

So, we draw a line that goes from (0, 0) to the top of the red line at 1, and then one that goes from there to the top of the red line at 2, and so on. We will then have 3 trapeziums and one triangle which we’ll label A, B, C, and D.

The aim is to find the areas of A, B, C and D and add them up. Make sure you’re familiar with the formula for triangles and trapeziums by clicking here.

To find the areas of these shapes we need their dimensions. They’re all 1 wide, but we don’t know how tall they are, so we have to try our best to accurately read the values off the y-axis.

Reading off the values in order, we get

$\text{when }t=1,\,s=4.5,$
$\text{when }t=2,\,s=6.4,$
$\text{when }t=3,\,s=7.8,$
$\text{when }t=4,\,s=9$

Now that we have all the values, we can calculate the areas. Recall:

$\text{area of triangle }=\dfrac{1}{2}bh,\,\,\,\text{ and }\,\,\,\text{area of trapezium }=\dfrac{1}{2}(a+b)h$

So, firstly we get

$\text{area of A }=\dfrac{1}{2}\times 1 \times 4.5=2.25$

Then, see that we’re looking at the trapeziums propped up on their side. So, for all the trapeziums, $h$ is not the “height” from where we’re looking, instead, $\mathbf{h}$ is the width, which in this case is just 1. Then, $a$ and $b$ for each trapezium are the appropriate values we just read off the y-axis – the ones at the corners of the shape we want.

For example, looking at the “heights” of the corners of shape B, we see $a=4.5$ and $b=6.4$. Therefore, we get

$\text{area of B }=\dfrac{1}{2}\times (4.5+6.4) \times 1=5.45$

$\text{area of C }=\dfrac{1}{2}\times (6.4+7.8) \times 1=7.1$

$\text{area of D }=\dfrac{1}{2}\times (7.8+9) \times 1=8.4$

Therefore, our final estimate for the area under the graph, and so the distance travelled, is

$2.25+5.45+7.1+8.4=23.2\text{ m}$

NOTE: because of the way we calculate this estimate, people’s answers will naturally vary a little, but this doesn’t mean they’re wrong. In an exam, there will be a range of answers that they will give you full marks for, you just have to try to be as accurate as you can with your estimate.

### Example Questions

We’ll need 4 strips of equal width over the course of 20 seconds. $20\div 4=5$, so each strip must be 5 seconds wide on the x-axis. We start by drawing vertical lines every 5 seconds going from the x-axis up to the graph. Then, connecting each of the points where those vertical lines meet the graph, we get our 4 strips: 3 trapeziums and 1 triangle. Lastly, drawing some horizontal lines from the ‘corners’ of our trapeziums to ensure we can read off the y-values, our picture should look like the graph below.

For ease, we’ve labelled the four shapes that we going to find the areas of with the letters A, B, C, and D. Shape A is a triangle, so reading the y-value from the graph we get

$\text{area of A }=\dfrac{1}{2}\times 5 \times 3.5=8.75$

The other 3 shapes are trapeziums. Reading the remaining y-values from the graph, we get

$\text{area of B }=\dfrac{1}{2}\times (3.5+14)\times 5=43.75$ $\text{area of C }=\dfrac{1}{2}\times (14+32)\times 5=115$ $\text{area of D }=\dfrac{1}{2}\times (32+57)\times 5=222.5$

Now, adding up the results, we get the estimate of the distance travelled to be

$8.75+43.75+115+222.5=390\text{ m}$.

Note: Any answer between 385m and 395m is acceptable in this case.

#### Is this a topic you struggle with? Get help now.

We’ll need 3 strips of equal width over the course of 3 seconds, so each strip must be 1 second wide on the x-axis. We start by drawing vertical lines every second, between 7s and 10s, going from the x-axis up to the graph.

Then, connecting each of the points where those vertical lines meet the graph, we get our 3 strips: all trapeziums. Lastly, drawing some horizontal lines from the ‘corners’ of our trapeziums to ensure we can read off the y-values, our picture should look like the graph below. For ease, we’ve labelled the three shapes that we going to find the areas of with the letters A, B, and C. All 3 are trapeziums of “height” 1, so, reading the y-values from the graph we get

$\text{area of A }=\dfrac{1}{2}\times (26+28)\times 1=27$ $\text{area of B }=\dfrac{1}{2}\times (28+25)\times 1=26.5$ $\text{area of C }=\dfrac{1}{2}\times (25+15)\times 1=20$

Now, adding up the results, we get the estimate of the distance travelled to be

$27+26.5+20=73.5\text{ m}$.

Note: Any answer between 72m and 75m is acceptable in this case.