Pythagoras' theorem - Part 2 - KS3 Maths

Key points

The image shows a right angled triangle. The right angled triangle has its right angled vertex in the bottom left corner. The vertical side to the left is labelled a, the horizontal side is labelled b, and the diagonal side is labelled c. Written right: the formula, a, squared plus b squared equals c squared. The c and the c squared are coloured orange. The triangle is coloured purple. — Image caption,
𝒄 = the length of the hypotenuse.

An understanding of how to use Pythagoras’ theorem to find missing sides in a right-angled triangle is essential for applying the theorem in different contexts.
Pythagoras’ theorem can be used to find the distance between two points. This is done by joining the points together to form the of a right-angled triangle and using the theorem \(a\)² + \(b\)² = \(c\)² to find the length of the hypotenuse.
Pythagoras’ theorem can also be used to find missing lengths in shapes, such as rectangles and once the shape has been split into right-angled triangles.

How to find the distance between two points

Follow these steps to calculate the distance between two on a set of . The same steps can be taken to calculate the length of a where the two end points are the two coordinates.

Join the two coordinates with a straight line. This will become the of a right-angled triangle and is the length that needs to be found.
Draw a and line from the two coordinates to form a right-angled triangle.
Calculate the horizontal length of the triangle by finding the difference between the x-coordinates. This is side a of the right-angled triangle.
Calculate the vertical length of the triangle by finding the difference between the y-coordinates. This is side b of the right-angled triangle.
Substitute the values of \(a\) and \(b\) into Pythagoras’ theorem: \(a\)² + \(b\)² = \(c\)².
Add the squares together, then find the square root to calculate the c, the hypotenuse. The hypotenuse is the distance between the two points.

Example

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Find the distance between the points P and Q.
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(3,1) is the coordinate that is 3 along the x-axis, and 1 up the y-axis. (6,5) is the coordinate that is 6 along the x-axis, and 5 up the y-axis. The line segment PQ is the line between the two points P and Q. Join the two coordinates to make a line segment.
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Draw a horizontal line (going across) from P, and a vertical line (going down) from point Q, to form a right-angled triangle.
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The width of the triangle is the difference between the x-coordinates 3 and 6. The width is 6 – 3 = 3
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The height of the triangle is the difference between the y-coordinates 1 and 5. The height is 5 – 1 = 4
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Pythagoras’ theorem can now be used on the right-angled triangle to calculate the length PQ. PQ is the hypotenuse so should be labelled 𝒄, and the other two sides should be labelled 𝒂 and 𝒃. It does not matter which way round 𝒂 and 𝒃 are labelled. Substitute the values of 𝒂 and 𝒃 into the formula 𝒂² + 𝒃² = 𝒄² to give the equation 3² + 4² = 𝒄².
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Add 9 + 16 together to give the value of 𝒄² = 25. To find the value of 𝒄, calculate the square root of 25. This gives the final answer of 𝒄 = 5, therefore PQ = 5

Question

Find the distance between points R and S.

The image shows a set of axes. The horizontal axis is labelled x. The values go up in ones from zero to six. The vertical axis is labelled y. The values go up in ones from zero to six. Two points have been marked on the axes. Point R has coordinate, two comma five. Point S has coordinate, four comma one. A dashed line has been drawn between points R and S. A horizontal line, has been drawn from S to coordinate two comma one. A vertical line, has been drawn from coordinate, two comma one to R. The horizontal distance between point S and coordinate two comma one has been marked with a horizontal arrow and labelled as two. The vertical distance between coordinate, two comma one and point R has been marked with a vertical arrow and labelled as four. The side of length two has been labelled as a, the side of length four has been labelled as b, and R S has been labelled c. The a, b, and c are coloured blue. The two, four, arrows and solid lines are coloured pink. The dashed line is coloured orange.

Applying Pythagoras’ theorem to other shapes

Pythagoras’ theorem can be used to find the diagonal of a rectangle. The width and height of the rectangle become \(a\) and \(b\) in the formula and \(c\) is the diagonal length.

\(a\)² + \(b\)² = \(c\)²

can be split into two right-angled triangles. This means Pythagoras’ theorem can be used to find the lengths of missing sides, such as the height. It is important to remember that when an isosceles triangle is cut in half this way, the base length is also cut in half.

Examples

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Find the length of the diagonal EG to 1 decimal place (1 dp).
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The diagonal EG is the line that goes from E to G. This creates two right-angled triangles.
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Taking the triangle DEG, Pythagoras’ theorem can be used to calculate the length of EG. EG is the hypotenuse so should be labelled as 𝒄. ED can be labelled as 𝒂. DG can be labelled as 𝒃. Substitute the values of 𝒂 and 𝒃 into the formula 𝒂² + 𝒃² = 𝒄² to give the equation 4² + 6² = 𝒄², 16 + 36 = 52. To find the value of 𝒄, calculate the square root of 52. This gives the final answer of 𝒄 = 7.211... When rounded to 1 decimal place, the length of the diagonal EG is 7.2 cm.
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Find the perpendicular height of the isosceles triangle.
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An isosceles triangle has two sides of the same length (25 cm). The perpendicular height is the height at a right angle to the base of the triangle. The isosceles triangle can be split into two right-angled triangles. Pythagoras’ theorem can then be used on either of the two right-angled triangles. The 25 cm length becomes the hypotenuse so should be labelled 𝒄. The base of the isosceles triangle has been cut in half, so is 7 cm. This can be labelled as 𝒂. The height can be labelled as 𝒃.
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Substitute the values of 𝒂, 𝒃 and 𝒄 into the equation 𝒂² + 𝒃² = 𝒄². 𝒂² = 7² and 𝒄² = 25². 𝒃² is not known. This leads to the equation 7² + 𝒃² = 25². Calculate the value of the squares. 7² = 49, and 25² = 625. This leads to the equation 49 + 𝒃² = 625. To work out the value of 𝒃², subtract 49 from both sides of the equation. This leads to the equation 𝒃² = 576. The inverse of squaring is square rooting, so to find 𝒃, calculate the square root of 576. This gives the answer of 𝒃 = 24. Therefore, the height of the triangle is 24 cm.
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Calculate the length of 𝒙, the base of the isosceles triangle.
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The isosceles triangle can be split into two right-angled triangles. The base of the isosceles triangle has been cut in half, so the base of the right-angled triangle is given a different letter, 𝒚. 𝒚 is half of 𝒙. Pythagoras’ theorem can then be used on either of the two right-angled triangles. The 3.4 m length becomes the hypotenuse so should be labelled 𝒄. The height is 3 m, and can be labelled as 𝒂. The base of the right-angled triangle can be labelled as 𝒃.
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Substitute the values into the equation 𝒂² + 𝒃² = 𝒄². 𝒂² = 3² and 𝒄² = 3.4². 𝒃² can be substituted with 𝒚² . This leads to the equation 3² + 𝒚² = 3.4² . Calculate the value of the squares. 3² = 9, and 3.4² = 11.56. This leads to the equation 9 + 𝒚² = 11.56. To work out the value of 𝒚² , subtract 9 from both sides of the equation. This leads to the equation 𝒚² = 2.56. The inverse (opposite) of squaring is square rooting, so to find 𝒚, calculate the square root of 2.56. This gives the answer of 𝒚 = 1.6cm. However, this is not the final answer.
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The aim is to find the base of the isosceles triangle, 𝒙. The base of each right-angled triangle is 1.6 cm. 𝒙 is double this length. 𝒙 = 2 x 1.6 = 3.2 cm.

Question

A question asks to calculate the height of the isosceles triangle, \(b\). The start of the working is shown on the right.

What mistake has been made?

An image of an isosceles triangle. The isosceles triangle has a base of twenty four centimetres. The two equal sloping sides measure thirty seven centimetres. The vertical height of the triangle is labelled b. Written right: a, squared plus b squared equals c squared. Written below: twenty four squared plus b squared equals thirty seven squared. The b is coloured blue, and the triangle is coloured pink.

Visual proof of Pythagoras' theorem

Explore a visual way of showing Pythagoras’ theorem below.

Example

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This triangle will be used to show Pythagoras’ theorem that 𝒂² + 𝒃² = 𝒄². The hypotenuse of the right-angled triangle is labelled as 𝒄. The other two sides are labelled 𝒂 and 𝒃.
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The whole square is made up of 4 right-angled triangles with lengths 𝒂 and 𝒃, and 2 squares. The area of the top square is 𝒂 x 𝒂 = 𝒂². The area of the bottom square is 𝒃 x 𝒃 = 𝒃².
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The four triangles can be rearranged to make the diagram on the right. The length of the hypotenuse of each triangle is 𝒄. The area of the orange square is 𝒄 x 𝒄 = 𝒄².
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The diagram on the left contains four right-angled triangles, and two squares (with areas 𝒂² and 𝒃²). The diagram on the right contains the same four right-angled triangles, and one square (with area 𝒄²).The two squares on the left (with areas 𝒂² and 𝒃²) must take up the same amount of space as the square on the right (with area 𝒄²). Therefore, 𝒂² + 𝒃² = 𝒄².
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The diagram on the left contains 4 right - angled triangles, and two squares ( with areas 𝒂² and 𝒃².) The diagram on the right contains the same 4 right - angled triangles, and one square ( with area 𝒄²). The two squares on the left ( with areas 𝒂² and 𝒃²) must take up the same amount of space as the square on the right ( with area 𝒄²). Therefore, 𝒂² + 𝒃² = 𝒄².