Area Under A Graph Worksheets | Revision and Questions | MME

# Area Under a Graph Worksheets, Questions and Revision

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## What you need to know

Determining the area under a is useful to calculate the distance travelled.

Area Under Velocity Time Graph = Distance Travelled

If the graph is made up of straight lines it can be split into regular shapes and the area calculated. If the graph is curved then the area must be approximated this method will be shown below.

Make sure you are happy with the following topics before continuing:

## 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. This results in the graph shown.

Now

WE have our $4$ shapes, $A, B, C$ and $D$

Now we must find the area of all $4$ shapes.

Shape $\bf{A}$: Is a triangle

Area = $\dfrac{1}{2} \text{base} \times \text{Height}$

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

Shape $\bf{B}$: is a trapezium.

Area of a trapezium = $\dfrac{1}{2}(a+b)h$

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

Shape $\bf{C}$: is a trapezium.

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

Shape $\bf{D}$: is a trapezium.

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

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.

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.

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