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KS4 Mathematics S2 Pythagoras’ Theorem

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**S2.1 Introducing Pythagoras’ Theorem**

Contents S2 Pythagoras’ Theorem A S2.1 Introducing Pythagoras’ Theorem A S2.2 Identifying right-angled triangles A S2.3 Pythagorean triples A S2.4 Finding unknown lengths A S2.5 Applying Pythagoras’ Theorem in 2-D A S2.6 Applying Pythagoras’ Theorem in 3-D

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**Right-angled triangles**

A right-angled triangle contains a right angle. The longest side opposite the right angle is called the hypotenuse. Review the definition of a right-angled triangle. Tell pupils that in any triangle the angle opposite the longest side will always be the largest angle and vice-versa. Ask pupils to explain why no other angle in a right-angled triangle can be larger than or equal to the right angle. Ask pupils to tell you the sum of the two smaller angles in a right-angled triangle. Recall that the sum of the angles in a triangle is always equal to 180°. Recall, also, that two angles that add up to 90° are called complementary angles. Conclude that the two smaller angles in a right-angled triangle are complementary angles.

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**Identify the hypotenuse**

Use this exercise to identify the hypotenuse in right-angled triangles in various orientations.

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**The history of Pythagoras’ Theorem**

Pythagoras’ Theorem concerns the relationship between the sides of a right angled triangle. The theorem is named after the Greek mathematician and philosopher, Pythagoras of Samos. Explain that the theorem is named after Pythagoras because he, or a member of his society, was the first person known to have formally proven the result. Although the Theorem is named after Pythagoras the result was known to many ancient civilizations including the Babylonians, Egyptians and Chinese, at least 1000 years before Pythagoras was born.

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Pythagoras’ Theorem Pythagoras’ Theorem states that the square formed on the hypotenuse of a right-angled triangle … … has the same area as the sum of the areas of the squares formed on the other two sides. Pythagoras’ Theorem may be expressed as a relationship between areas, as shown here, or a relationship between side lengths. In fact, the area of any similar shapes may by drawn on the sides of a right-angled triangle. The area of the shape drawn on the hypotenuse, will be equal to the sum of the areas of the shapes drawn on the two shorter sides. See S9.4 Circles for a problem involving the areas of semi-circles drawn on the sides of a right-angled triangle.

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**Showing Pythagoras’ Theorem**

Drag the vertices of the triangle to change the lengths of the sides and rotate the right-angled triangle. Ask a volunteer to come to the board and use the pen tool to demonstrate how to find the area of each square. For tilted squares this can be done by using the grid to divide the squares into triangles and squares. Alternatively, a larger square can be drawn around the tilted square and the areas of the four surrounding triangles subtracted. Reveal the areas of the squares and verify that the area of the largest square is always equal to the sum of the areas of the squares on the shorter sides.

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**Pythagoras’ Theorem c2 a2 b2**

If we label the length of the sides of a right-angled triangle a, b and c as follows, then the area of the largest square is c × c or c2. c2 The areas of the smaller squares are a2 and b2. c a2 a We can write Pythagoras’ Theorem as This slide shows how Pythagoras’ Theorem can be written as a relationship between the side lengths of a triangle with sides a, b and c, where c is the hypotenuse. Ask pupils to tell you what a2 is equal to. (c2 – b2). Ask pupils to tell you what b2 is equal to. (c2 – a2). b b2 c2 = a2 + b2

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**A proof of Pythagoras’ Theorem**

There many proofs of Pythagoras’s Theorem. Pupils could be asked to research these on the internet and make posters to show them. In this proof we can see that the area of the two large squares is the same: (a + b)2 Both of these squares contain four identical triangles with side lengths a, b and c. In each of the large squares the four triangles are arranged differently to show visually that the area of the square with side length c is equal to the sum of the areas of the squares with side lengths a and b. This can be shown more formally considering the first arrangement. The area the red square with side length c is equal to the area of the large square with side length (a + b) minus the area of the four green right-angled triangles. The area of the four green right-angled triangles is 4 × ½ab = 2ab. c2 = (a + b)2 – 2ab Expanding, c2 = a2 + 2ab + b2 – 2ab c2 = a2 + b2

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**S2.2 Identifying right-angled triangles**

Contents S2 Pythagoras’ Theorem A S2.1 Introducing Pythagoras’ Theorem A S2.2 Identifying right-angled triangles A S2.3 Pythagorean triples A S2.4 Finding unknown lengths A S2.5 Applying Pythagoras’ Theorem in 2-D A S2.6 Applying Pythagoras’ Theorem in 3-D

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Pythagoras’ Theorem Pythagoras’ Theorem states that for a right-angled triangle with a hypotenuse of length c and the shorter sides of lengths a and b c a c2 = a2 + b2 b We can use Pythagoras’ Theorem to check whether a triangle is right-angled given the lengths of all the sides, Stress that Pythagoras’ Theorem is only true for right-angled triangles. If we are given the lengths of all three sides of a triangle, therefore, we can use Pythagoras’ Theorem to check whether of not the triangle is a right-angled triangle by squaring the length of the two shorter sides, adding the squares together and checking whether or not this is equal to the hypotenuse squared. If we are given the lengths of two of the sides in a right-angled triangle, we can also use Pythagoras’ Theorem to find the lengths of the unknown side. This is demonstrated in S2.4 Finding unknown lengths. to find the length of a missing side in a right-angled triangle given the lengths of the other two sides.

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**Identifying right-angled triangles**

If we are given the lengths of the three sides of a triangle, we can use Pythagoras’ Theorem to find out whether or not the triangle contains a right angle. For example, A triangle has sides of length 4 cm, 7 cm and 9 cm. Is this a right-angled triangle? If the sum of the squares on the two shorter sides is equal to the square on the longest side, the triangle has a right angle. = = 65 65 92 No, this is not a right-angled triangle.

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**Identifying right-angled triangles**

Use this activity to demonstrate how the relationship between the sum of the squares on the two shorter sides and the square on the hypotenuse changes as the right angle is modified. If the right angle becomes obtuse, c2 gets bigger. If the right-angle becomes acute, c2 gets smaller. We can use this to decide whether a non-right-angled triangle has three acute angles or whether it contains an obtuse angle. Ask pupils to explain why it is impossible for a triangle to contain more than one obtuse angle.

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**Identifying right-angled triangles**

If the sum of the squares on the two shortest sides of a triangle is less than the sum of the square on the longest side, all the angles in the triangle must be acute. In other words, if a triangle with sides of length a, b and c, where c is the longest side has, c2 < a2 + b2 then all the angles in the triangle must be acute.

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**Identifying right-angled triangles**

If the sum of the squares on the two shortest sides of a triangle is greater than the sum of the square on the longest side, one of the angles in the triangle must be obtuse. In other words, if a triangle with sides of length a, b and c, where c is the longest side has, c2 > a2 + b2 then one of the angles in the triangle must be obtuse.

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**Identifying right-angled triangles**

A triangle has sides of length 5 cm, 8 cm and 10 cm. Is the angle opposite the longest side acute, obtuse or right-angled? 10 cm 8 cm 5 cm ? = = 89 102 = 100 100 > 89 so the angle opposite the longest side is an obtuse angle.

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**Obtuse, acute or right-angled triangle**

In this activity pupils must find the sum of the squares of the two shorter sides. If this is less than the square of the longest side, the angle opposite the longest side must be acute. If this is more than the square of the longest side, the angle opposite the longest side must be obtuse. If this is the same as the square of the longest side, the triangle obeys Pythagoras’ Theorem and the angle opposite the longest side must therefore be a right angle. Pupils may use their calculators for lengths involving decimals.

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**Contents S2 Pythagoras’ Theorem A S2.1 Introducing Pythagoras’ Theorem**

S2.2 Identifying right-angled triangles A S2.3 Pythagorean triples A S2.4 Finding unknown lengths A S2.5 Applying Pythagoras’ Theorem in 2-D A S2.6 Applying Pythagoras’ Theorem in 3-D

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Pythagorean triples A triangle has sides of length 3 cm, 4 cm and 5 cm. Does this triangle have a right angle? Using Pythagoras’ Theorem, if the sum of the squares on the two shorter sides is equal to the square on the longest side, the triangle has a right angle. = = 25 Explain that a Pythagorean triple is three whole numbers that obey Pythagoras’ Theorem. Many Pythagorean triples where known to people in ancient times. = 52 Yes, the triangle has a right-angle. The numbers 3, 4 and 5 form a Pythagorean triple.

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Pythagorean triples Ancient Egyptians used the fact that a triangle with sides of length 3, 4 and 5 contained a right-angle to mark out field boundaries and for building.

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Pythagorean triples Three whole numbers a, b and c, where c is the largest, form a Pythagorean triple if, a2 + b2 = c2 3, 4, 5 is the simplest Pythagorean triple. Write down every square number from 12 = 1 to 202 = 400. Use these numbers to find as many Pythagorean triples as you can. Write down any patterns that you notice. Pupils should be able to find six Pythagorean triples using these numbers. These are shown on the next slide. They may notice that some of these are multiples of the 3, 4, 5 triangle.

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**Pythagorean triples How many of these did you find? 9 + 16 = 25**

= 52 3, 4, 5 = 100 = 102 6, 8, 10 = 169 = 132 5, 12, 13 = 225 = 152 9,12, 15 = 289 = 172 8, 15, 17 Pupils should notice that the Pythagorean triples that have not been circled are multiples of the the 3, 4, 5 Pythagorean triple. Tell pupils that a primitive Pythagorean triple is a Pythagorean triple that is not a multiple of another Pythagorean triple. Ask pupils to explain why any multiple of a Pythagorean triple must be another Pythagorean triple (using similar triangles). = 400 = 202 12, 16, 20 The Pythagorean triples 3, 4, 5; 5, 12, 13 and 8, 15 17 are called primitive Pythagorean triples.

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**Similar right-angled triangles**

The following right-angled triangles are similar. 10 ? 15 6 9 8 12 ? Remind pupils that the corresponding lengths in two mathematically similar shapes are always in the same ratio. That is, one shape is an enlargement of the other. Ask pupils if they can work out which Pythagorean triple both these triangles are based on. For the first triangle, 6 and 8 share a common factor of 2. Dividing these lengths by 2, gives us 3 and 4. For the second triangle, 9 and 15 share a common factor of 3. Dividing these lengths by 3, gives us 3 and 5. These triangles are both enlargements of a right-angled triangle with sides of length 3, 4 and 5. Use this to find the lengths of the missing sides. Alternatively, we could use a scale factor of 3/2 to scale from the smaller triangle to the larger triangle, or a scale factor of 2/3 to scale from the larger triangle to the smaller triangle. Find the lengths of the missing sides. Check that Pythagoras’ Theorem holds for both triangles.

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**Similar right-angled triangles**

The following right-angled triangles are similar. 10 ? 15 6 9 8 12 ? Check the lengths of the missing sides by showing that Pythagoras’ Theorem holds for both triangles. = = = 100 = 225 = 102 = 152

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**Similar right-angled triangles**

The following right-angled triangles are similar. 60 45 ? 32 ? 24 68 51 Ask pupils if they can work out which Pythagorean triple both these triangles are based on. For the first triangle, 60 and 68 share a common factor of 4. Dividing these lengths by 4, gives us 15 and 17. For the second triangle, 24 and 51 share a common factor of 3. Dividing these lengths by 3, gives us 8 and 17. These triangles are both enlargements of a right-angled triangle with sides of length 8, 15 and 17. Use this to find the lengths of the missing sides. Alternatively, we could use a scale factor of 3/4 to scale from the larger triangle to the smaller triangle, or a scale factor of 4/3 to scale from the smaller triangle to the larger triangle. Find the lengths of the missing sides. Check that Pythagoras’ Theorem holds for both triangles.

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**Similar right-angled triangles**

The following right-angled triangles are similar. 60 45 32 24 68 51 Pupils can verify these calculations using their calculators. Make sure that pupils can locate both the x2 key and the key on their calculators. = = = 4624 = 2601 = 682 = 512

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**Similar right-angled triangles**

The following right-angled triangles are similar. 2.4 1.8 ? 2 1.5 2.5 ? 3 Ask pupils if they can work out which Pythagorean triple both these triangles are based on. For the first triangle, 2 and 1.5 are half of 4 and 3 respectively. For the second triangle, 2.4 and 3 are 3/5 of 4 and 5 respectively. These triangles are both fractional enlargements of a right-angled triangle with sides of length 3, 4 and 5. Use this to find the lengths of the missing sides. Alternatively, we could use a scale factors. To scale from the smaller triangle to the larger triangle we scale from 2 to 2.4. In other words we would multiply by 2.4/2 or 6/5 or 1.2. To scale from the larger triangle to the smaller triangle we scale from 2.4 to 2. In other words we would multiply by 2/2.4 or 5/6 or divide by 1.2. Find the lengths of the missing sides. Check that Pythagoras’ Theorem holds for both triangles.

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**Similar right-angled triangles**

The following right-angled triangles are similar. 2.4 1.8 2 1.5 2.5 3 Conclude by explaining that most Pythagoras problems on the non-calculator paper will involve triangles based on Pythagorean triples. This is because these triangles have side lengths that can be written exactly (the side lengths are rational numbers). Right-angled-triangles that are not based on Pythagorean triples have at least one side length that cannot be written as a terminating or recurring decimal, that is a side length that is irrational. Such lengths can only by written exactly as surds (unresolved roots). See N4.2 Terminating and recurring decimals and N2.2 Surds = = = 6.25 = 9 = 2.52 = 32

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**S2.4 Finding unknown lengths**

Contents S2 Pythagoras’ Theorem A S2.1 Introducing Pythagoras’ Theorem A S2.2 Identifying right-angled triangles A S2.3 Pythagorean triples A S2.4 Finding unknown lengths A S2.5 Applying Pythagoras’ Theorem in 2-D A S2.6 Applying Pythagoras’ Theorem in 3-D

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Pythagoras’ Theorem Pythagoras’ Theorem states that for a right-angled triangle with a hypotenuse of length c and the shorter sides of lengths a and b c a c2 = a2 + b2 b We can use Pythagoras’ Theorem to check whether a triangle is right-angled given the lengths of all the sides, Stress that Pythagoras’ Theorem is only true for right-angled triangles. If we are given the lengths of all three sides of a triangle, therefore, we can use Pythagoras’ Theorem to check whether of not the triangle is a right-angled triangle by squaring the length of the two shorter sides, adding the squares together and checking whether or not this is equal to the hypotenuse squared. This is demonstrated in S2.2 Identifying right-angles. If we are given the lengths of two of the sides in a right-angled triangle, we can also use Pythagoras’ Theorem to find the lengths of the unknown sides. to find the length of a missing side in a right-angled triangle given the lengths of the other two sides.

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**Finding the length of the hypotenuse**

Use Pythagoras’ Theorem to calculate the length of side a. a 5 cm 12 cm Using Pythagoras’ Theorem, a2 = a2 = Talk through the calculation to find the length of the hypotenuse. Stress that if a2 = 169, to find a we need to square root 169. If pupils do not know the square root of 169, ensure that they are able to locate and use the key on their calculators. Encourage pupils to learn at least the first 15 square numbers. Pupils should also be encouraged to look at the triangle to make sure that their answer seems about right compared to the other two given sides. a2 = 169 a = 169 a = 13 cm

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**Finding the length of the hypotenuse**

Use Pythagoras’ Theorem to calculate the length of side PR. 0.7 m Q 2.4 m R Using Pythagoras’ Theorem PR2 = PQ2 + QR2 Substituting the values we have been given, PR2 = This example uses the letters at each vertex to label the sides. Stress to pupils that they must start by stating Pythagoras’ Theorem for the given triangle and show every step in their calculation. They should also draw a diagram if one has not been given. Again encourage pupils to think about whether the answer seems sensible given the lengths of the other two sides. PR2 = PR2 = 6.25 PR = 6.25 PR = 2.5 m

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**Finding the length of the hypotenuse**

6 cm Use Pythagoras’ Theorem to calculate the length of side x to 2 decimal places. x 3 cm Using Pythagoras’ Theorem x2 = x2 = x2 = 45 x = 45 x = 6.71 cm

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**Finding the length of the shorter sides**

Use Pythagoras’ Theorem to calculate the length of side a. 26 cm 10 cm a Using Pythagoras’ Theorem, a = 262 a2 = 262 – 102 a2 = 676 – 100 Explain that when we use Pythagoras’ Theorem to find the length of one of the shorter sides, we have to subtract the square of the given shorter side from the square of the hypotenuse. a2 = 576 a = 576 a = 24 cm

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**Finding the length of the shorter sides**

A Use Pythagoras’ Theorem to calculate the length of side AC to 2 decimal places. 5 cm B C 8 cm Using Pythagoras’ Theorem AB2 + AC2 = BC2 Substituting the values we have been given, 52 + AC2 = 82 AC2 = 82 – 52 PR2 = 39 PR = 39 PR = 6.24 cm

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**Finding the length of the shorter sides**

Use Pythagoras’ Theorem to calculate the length of side x to 2 decimal places. 15 cm 7 cm x Using Pythagoras’ Theorem, x = 152 x2 = 152 – 72 x2 = 225 – 49 x2 = 176 x = 176 x = cm

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**Find the correct equation**

One of the most common errors when solving problems involving Pythagoras’ Theorem is failing to identify which side is the hypotenuse before writing the equation of the relationship between the sides. For each of the examples in the activity ask pupils to identify the hypotenuse before selecting the correct equation. Encourage pupils to go through the process of identifying the hypotenuse and writing down Pythagoras’ Theorem before attempting to solve any problem involving the lengths of the sides of right-angled triangles.

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Complete this table

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**S2.5 Applying Pythagoras’ Theorem in 2-D**

Contents S2 Pythagoras’ Theorem A S2.1 Introducing Pythagoras’ Theorem A S2.2 Identifying right-angled triangles A S2.3 Pythagorean triples A S2.4 Finding unknown lengths A S2.5 Applying Pythagoras’ Theorem in 2-D A S2.6 Applying Pythagoras’ Theorem in 3-D

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**Finding the lengths of diagonals**

Pythagoras’ Theorem has many applications. For example, we can use it to find the length of the diagonal of a rectangle given the lengths of the sides. Use Pythagoras’ Theorem to calculate the length of the diagonal, d. 10.2 cm 13.6 cm d d2 = d2 = d2 = 289 d = 289 d = 17 cm

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**Finding the lengths of diagonals**

Use Pythagoras’ Theorem to calculate the length d of the diagonal in a square of side length 7 cm. d 7 cm Using Pythagoras’ Theorem d2 = d2 = d2 = 98 d = 98 d = 9.90 cm (to 2 d.p.)

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**Finding the lengths of diagonals**

Show that the length of the diagonal d in a square of side length a can be found by the formula d = √2 a d a Using Pythagoras’ Theorem d2 = a2 + a2 d2 = 2a2 d = 2a2 Explain that we can use this formula to work out the length of the diagonal in a square given the length of its side. Ask pupils to explain why it is better to leave 2 as it is in the formula, rather than trying to write it as a decimal. For example, in a square of side length 12 cm, the length of the diagonal is 2 × 12 = cm (to 2 d.p.). Ask pupils how we could use the formula to work out the side length of a square given the length of the diagonal. d = 2 a Use this formula to find the length of the diagonal in a square with side length 12 cm to 2 decimal places.

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**Finding the height of an isosceles triangle**

Use Pythagoras’ Theorem to calculate the height h of this isosceles triangle. 5.8 cm h 8 cm Using Pythagoras’ Theorem in half of the isosceles triangle, we have h = 5.82 h2 = 5.82 – 42 5.8 cm 4 cm h Explain that the height divides the isosceles triangle into two equal parts and so the length of the base in the right-angled triangle must be half of the length of the base of the original isosceles triangle. Ask pupils when it might be important to know the perpendicular height of a triangle, for example, when trying to work out its area. h2 = – 16 h2 = 17.64 h = 17.64 h = 4.2 cm

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**Finding the height of an equilateral triangle**

Use Pythagoras’ Theorem to calculate the height h of an equilateral triangle with side length 4 cm. 4 cm h Using Pythagoras’ Theorem in half of the equilateral triangle, we have h = 42 h 4 cm 2 cm h2 = 42 – 22 Explain that like the isosceles triangle, the height divides the equilateral triangle into two equal parts and so the length of the base in the right-angled triangle must be half of the length of the base of the original equilateral triangle. h2 = 16 – 4 h2 = 12 h = 12 h = 3.46 cm (to 2 d.p.)

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**Finding the height of an equilateral triangle**

Show that the height h in an equilateral triangle of side length a can be found by the formula h = √3 a a h 2 Using Pythagoras’ Theorem in half of the equilateral triangle, we have We can think of this as 1a2 – a2 1 4 h = a2 4 a2 h a 2 h2 = a2 – 4 a2 Explain that we can think of a2 – a2/4 as 1a2 – ¼a2. 1 – ¼ = ¾ so we have ¾a2. The square root of 3a2/4 is 3a/2. Conclude that we can find the height of an equilateral triangle by multiplying the side length by 3/2 (= to 3 significant figures), or we can find the side length of an equilateral triangle by multiplying the height by 2/3 (= 1.15 to 3 significant figures). h2 = 4 3a2 h = a 2 3

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Ladder problem Change the length and the position of the ladder to generate various problems.

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Flight path problem Use Pythagoras’ Theorem to calculate the distance of the aeroplane from the starting point.

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**Applying Pythagoras’ Theorem twice**

Sometimes we have to apply Pythagoras’ Theorem twice to find a required length. For example, Find the length of side a. To solve this problem we need to find the length of the other missing side which we can call b. 4 cm 9 cm 18 cm a b2 = 182 – (4 + 9)2 b2 = 324 – 169 b b2 = 155 Explain that it is not necessary to square root 155 to find the value of b because we need the value as b2 to find a. Now, a2 = b2 + 42 a2 = a2 = 171 a = cm (to 2 d.p.)

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Find the given lengths 1 2 3 4 5 6 –1 –2 –3 –4 –5 –6 y x A(–4, 5) The points A(–4, 5), B(–4, –2), and C (6,–2) form a right angled triangle. 7 units What is length of the line AB? B(–4, –2) C (6, –2) AB = 5 – –2 The length of AB is found by subtracting the values of the y-coordinates of points A and B. = 5 + 2 = 7 units

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Find the given lengths 1 2 3 4 5 6 –1 –2 –3 –4 –5 –6 y x A(–4, 5) The points A(–4, 5), B(–4, –2), and C (6,–2) form a right angled triangle. 7 units What is length of the line BC? B(–4, –2) C (6, –2) 10 units BC = 6 – –4 The length of BC is found by subtracting the values of the x-coordinates of points B and C. = 6 + 4 = 10 units

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Find the given lengths 1 2 3 4 5 6 –1 –2 –3 –4 –5 –6 y x A(–4, 5) The points A(–4, 5), B(–4, –2), and C (6,–2) form a right angled triangle. 12.21 units (to 2 d.p.) 7 units What is length of the line AC? B(–4, –2) C (6, –2) 10 units Using Pythagoras, AC2 = AB2 + AC2 = = = 149 AC = 149 = units (to 2 d.p.)

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**Finding the distance between two points**

We can use Pythagoras’ Theorem to find the distance between any two points on a co-ordinate grid. Find the distance between the points A(7, 6) and B(3, –2). 1 2 3 4 5 6 7 –1 –2 –3 x y A(7, 6) We can add a third point at C to form a right-angled triangle. AB2 = AC2 + BC2 Stress that AC is the vertical distance between the points A and B and BC is the horizontal distance. The distance between any two points can be found by square rooting the sum of the the horizontal distance squared and the vertical distance squared. = = B(3, –2) C (7, –2) = 80 AC = 80 = 8.94 units (to 2 d.p.)

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**Finding the distance between two points**

What is the distance between two points with coordinates A(xA, yA) and B(xB, yB)? The horizontal distance between the points is xB – xA The vertical distance between the points is yB – yA Using Pythagoras’ Theorem the square of the distance between the points A(xA, yA) and B(xB, yB) is This slide shows the generalization for finding the distance between two points. (xB – xA)2 + (yB – yA)2 The distance between the points A(xA, yA) and B(xB, yB) is (xB – xA)2 + (yB – yA)2

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**Solving problems using Pythagoras’ Theorem**

As we have seen Pythagoras’ Theorem can be used to solve a great many problems involving the lengths in right-angled triangles. When solving a problem using Pythagoras’ Theorem always do the following: Draw a clearly labeled diagram of the situation, making sure you identify which side is the hypotenuse and which are the shorter sides. State Pythagoras’ Theorem for the given side lengths before substituting any values. Make sure that your answer is sensible given the lengths of the other two sides.

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**S2.6 Applying Pythagoras’ Theorem in 3-D**

Contents S2 Pythagoras’ Theorem A S2.1 Introducing Pythagoras’ Theorem A S2.2 Identifying right-angled triangles A S2.3 Pythagorean triples A S2.4 Finding unknown lengths A S2.5 Applying Pythagoras’ Theorem in 2-D A S2.6 Applying Pythagoras’ Theorem in 3-D

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**The longest diagonal in a cuboid**

Pythagoras’ Theorem can also be applied to three-dimensional problems. For example, what is the length of the longest diagonal in a cuboid measuring 5 cm by 7 cm by 8 cm? H Start by labeling the vertices of the cuboid. G 7 cm 5 cm 8 cm E F The longest diagonal is CE. Justify the fact that CE, AG, BH and DF are of equal length in a cuboid. Ask pupil to explain why ACE must be a right-angled triangle. We could also use AG, BH or DF. D C CE is the hypotenuse of right-angled triangle ACE. A B

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**The longest diagonal in a cuboid**

Pythagoras’ Theorem can also be applied to three-dimensional problems. For example, what is the length of the longest diagonal in a cuboid measuring 5 cm by 7 cm by 8 cm? 7 cm 5 cm 8 cm A B C D E F G H Before we can find the length of CE, we need to find the length of AC. We can do this by applying Pythagoras’ Theorem to triangle ABC. Ask pupils to explain how we know that ABC is a right-angled triangle. AC2 = AC2 = AC2 = 74

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**The longest diagonal in a cuboid**

Pythagoras’ Theorem can also be applied to three-dimensional problems. For example, what is the length of the longest diagonal in a cuboid measuring 5 cm by 7 cm by 8 cm? H Now we know that AC2 = 74, we can use Pythagoras’ Theorem to work out the length of CE. G E F CE2 = AE2 + AC2 8 cm Explain that we do not need to square root 74 to find the length of AC. This is because the value of AC2 for our formula. We would be square rooting only to square again. Errors caused by rounding could result. CE2 = D C CE2 = 5 cm A B 7 cm CE2 = 138 CE = cm (to 2 d.p.)

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**The longest diagonal in a cuboid**

Show that the longest diagonal d in a cuboid measuring a by b by c is given by the formula d = (a2 + b2 + c2) Let’s call the diagonal across the base e. Using Pythagoras’ Theorem, a b c e2 = a2 + b2 d and d2 = e2 + c2 So, e d2 = a2 + b2 + c2 d = (a2 + b2 + c2)

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**The longest diagonal in a cuboid**

A cuboid has sides of length 8 cm, 10 cm and 11 cm. Use the formula d = (a2 + b2 + c2) to find the length of the longest diagonal d to 2 d.p. Substituting the values into the formula, we have d = ( ) = ( ) = 285 = cm (to 2 d.p.)

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