15/11/2018 Starter L.O. To be able to Find the missing side of a right-angled triangle, given the other two sides Solve problems using Pythagoras’ theorem, such as finding the diagonal of a rectangle Solve problems using Pythagoras’ theorem in 3 dimensions Key Words hypotenuse, right angle, square, square root, theorem Starter

FINDING THE HYPOTENUSE 15/11/2018 CONTENTS – Click to go to… INTRODUCTION FINDING THE HYPOTENUSE FINDING ANY SIDE PROBLEM SOLVING PYTHAGORAS IN 3D REVISION

15/11/2018 INTRODUCTION

SQUARES AND SQUARE ROOTS 15/11/2018 SQUARES AND SQUARE ROOTS On your mini whiteboards… Quick Questions!

15/11/2018 Evaluate : 72

15/11/2018 x Square Area = 49 m2

15/11/2018 5 m Square Area = ?

15/11/2018 Evaluate : 81

15/11/2018 x Square Area = 144 m2

15/11/2018 10 m Square Area = ?

15/11/2018 Evaluate : 12

15/11/2018 Evaluate : 400

15/11/2018 Evaluate : 251/2

15/11/2018 Evaluate : 1

Identifying the hypotenuse 15/11/2018 Identifying the hypotenuse A right-angled triangle contains a right angle The side opposite the right angle is always the LONGEST It has a special name - HYPOTENUSE hypotenuse

15/11/2018 Use this exercise to identify the hypotenuse in right-angled triangles in various orientations.

Which side is the hypotenuse? 15/11/2018 Which side is the hypotenuse? f e a b d c g h i

15/11/2018 Discovery task! Make accurate constructions of the three triangles below MEASURE accurately the missing length on each one. Copy and fill in this table… Can you spot the pattern that links the last three columns? 8cm 6cm 12cm 5cm 4cm 3cm 𝒂 𝒃 𝒄 𝒂 𝟐 𝒃 𝟐 𝒄 𝟐 3 4 9 5 12 6 8

15/11/2018 Pythagoras proof activity From your worksheet… Colour in square 5 one colour, then cut it off Colour in the square with pieces 1-4 a different colour, hen cut it off, also cutting along the dotted lines to get four shapes from that square Try to arrange the five pieces you have so that they fit exactly into the larger bold square. Stick them down. Write a sentence that describes what you have discovered from this activity “I have shown that…_____________________ _____________________________________ _____________________________________”

15/11/2018 Pythagoras’ Theorem a + b = h 2 a b h

15/11/2018 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.

c b a X X X a2 X c2 X X b2 X X + = a2 b2 c2

15/11/2018 There are 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

15/11/2018 Pythagorean trees! – Do you think you could draw one?!

FINDING THE HYPOTENUSE 15/11/2018 FINDING THE HYPOTENUSE

a + b = h h a b 11/15/2018 Pythagoras’ Theorem – ALGEBRAIC METHOD Learn the formula! Substitute the two values that you know into the formula Solve the equation to find the unknown value a b h a + b = h 2

Example – Algebraic method 15/11/2018 Example – Algebraic method h 2 = a 2 + b 2 5 cm 12 cm x 1 h 2 = a 2 + b 2 4.6 cm 9.8 cm x 2

Pythagoras – QUICK METHOD 15/11/2018 Pythagoras – QUICK METHOD Decide if the side you are LOOKING FOR is the HYPOTENUSE, or one of the two SHORTER SIDES Use one of these quick and easy methods to work out what you need Looking for the LONGEST SIDE? (the hypotenuse!) Square it Add it Square root it Looking for a SHORTER SIDE? (not the hypotenuse!) Square it Add it Square root it

15/11/2018 Example – Quick method Square it… 4 2 = 16 1 + 3 cm 4 cm x 1 Square it… 3 2 = 9 + Add it… = 25 Square root it… 𝑥= 25 = 𝟓cm Square it… 9.8 2 = 96.04 4.6 cm 9.8 cm x 2 Square it… 4.6 2 = 21.16 + Add it… =117.2 Square root it… 𝑥= 117.2 =10.825… =𝟏𝟎.𝟖 cm (1dp)

15/11/2018 b = 15.62cm a = 7.21cm d = 9.68cm c = 7.21cm f = 15.43cm Task – Find the missing side, to 2d.p. 10 cm a b 1) 2) b = 15.62cm 4 cm 6 cm 12 cm a = 7.21cm 5.3 cm 5 cm c 8.1 cm 3) 4) d d = 9.68cm c = 7.21cm 7cm 10.1 cm 12 cm 8.5 cm 9.7 cm 5) 6) f = 15.43cm e e = 8.60cm f

15/11/2018 Task – Find the missing side, to 2d.p. 13.04 m 16.12 cm 1) 2) 3) x x 9.85 m 8 cm 9 m 7 m x 11 m 4 m 14 cm 4) 5) 6) 6 m x x 9.76 m 16.46 mm 7.81 m 5 m x 8.2 m 5.3 m 14.7 mm 7.4 mm

15/11/2018 FINDING ANY SIDE

a + b = h h a b 11/15/2018 Pythagoras’ Theorem – ALGEBRAIC METHOD Learn the formula! Substitute the two values that you know into the formula Solve the equation to find the unknown value a b h a + b = h 2

15/11/2018 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.

Example – Algebraic method 15/11/2018 Example – Algebraic method a 2 = h 2 − b 2 25 m 7 m x m

Pythagoras – QUICK METHOD 11/15/2018 Pythagoras – QUICK METHOD Decide if the side you are LOOKING FOR is the HYPOTENUSE, or one of the two SHORTER SIDES Use one of these quick and easy methods to work out what you need Looking for the LONGEST SIDE? (the hypotenuse!) Square it Add it Square root it Looking for a SHORTER SIDE? (not the hypotenuse!) Square it Add it Square root it

15/11/2018 Example – Quick method Square it… 15 2 =225 1 − 9cm 15cm 1 Square it… 9 2 = 81 − Subtract it… = 144 Square root it… 𝑥= 144 = 𝟏𝟐cm Square it… 7 2 = 49 x 5cm 7cm 2 Square it… 5 2 = 25 − Subtract it… = 24 Square root it… 𝑥= 24 =4.89897… =4.9 cm (1dp)

1) 2) 4) 3) 6) 5) 15/11/2018 h = 11.26cm g = 9.17cm i = 4.45cm Task – Find the missing side, to 2d.p. 5.2 cm 10 cm 12.4 cm 1) 2) h = 11.26cm 4 cm g h g = 9.17cm 4) j 3) 14.5 cm i 8.4 cm i = 4.45cm j = 23.86cm 25.3 cm 13.8cm 2.9 cm 6) 23 cm 5) l k k = 9.15cm l = 19.26cm 9.6 cm 30 cm

15/11/2018 Task – Find the missing sides, to 2.d.p 12.69 m 1) 2) 3) x 8 mm 9.75 m 24 mm 22.63 mm x 7 m 4) 5) 6) 14 m 52 mm 18 m 28 mm x x 20 m 17.29 m 5 m 14.28 m 43.82 mm x

15/11/2018 Task – Find the missing sides, to 2.d.p 5.3 1) 2) 3) 10.44 34 x x 14.32 27.50 9 6 x 13 20 4) 5) 6) x 1.8 18.25 4.48 14 9.80 x 23 x 14 10 4.1

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Exam question – Spot the errors! 15/11/2018 Exam question – Spot the errors! a) FIND PQ 19 2 + 15 2 = 361 + 225 =586𝑐𝑚 So PQ is 586cm b) FIND THE AREA OF PQR 15 × 19 2 = 142.5𝑐𝑚

PYTHAGORAS CATCH PHRASE! 11/15/2018 PYTHAGORAS CATCH PHRASE! On your mini whiteboards… Choose a question to answer. Get it right to reveal part of the picture and guess the catchphrase!

The diagram shows a trapezium ABCD . Calculate the length of AB. 6.71cm The diagram shows a trapezium ABCD . Calculate the length of AB. 10cm 7cm 8cm A D C B 5cm This rectangle has a diagonal of 13cm and a width of 12cm. Calculate it’s height. j 12cm 13cm 9m c 12.73m 151.99 miles Cardiff is 130 miles west of London. Hull is 200 miles north east of Cardiff. How far is Hull from London? 130 miles 200 miles Cardiff London Hull 3m 8m d 8.54m 23cm 16cm f 16.52cm 13.23cm 15cm a 20cm 12cm 8cm b 8.94cm 6.93m k 4m 8m

15/11/2018 BAD HAIR DAY!

15/11/2018 PROBLEM SOLVING

15/11/2018 Problem solving with Pythagoras Draw yourself a diagram which includes a right-angled triangle Identify which side is the hypotenuse Use Pythagoras in the normal way to find the missing side

15/11/2018 Change the length and the position of the ladder to generate various problems.

15/11/2018 Use Pythagoras’ Theorem to calculate the distance of the aeroplane from the starting point.

15/11/2018 Examples 1 2 d 6 cm 9.3 cm 7.8 cm 4.3 cm x cm Find the diagonal of the rectangle 6 cm 9.3 cm 1 d A rectangle has a width of 4.3 cm and a diagonal of 7.8 cm. Find its perimeter. 2 7.8 cm 4.3 cm x cm Perimeter = 2(6.5+4.3) = 21.6 cm

15/11/2018 Example A boat sails due East from a Harbour (H), to a marker buoy (B), 15 miles away. At B the boat turns due South and sails for 6.4 miles to a Lighthouse (L). It then returns to harbour. Make a sketch of the journey. What is the total distance travelled by the boat? H B L 15 miles 6.4 miles Total distance travelled = 21.4 + 16.3 = 37.7 miles

15/11/2018 Example A 12 ft ladder rests against the side of a house. The top of the ladder is 9.5 ft from the floor. How far is the base of the ladder from the house? 12 ft 9.5 ft L

PROBLEM SOLVING WITH PYTHAGORAS 11/15/2018 PROBLEM SOLVING WITH PYTHAGORAS On your mini whiteboards… Quick Questions!

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Finding the distance between two points 15/11/2018 Finding the distance between two points 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) 10 units C (6, –2) AB = 5 – –2 BC = 6 – –4 = 5 + 2 = 6 + 4 = 7 units = 10 units

Finding the distance between two points 15/11/2018 Finding the distance between two points 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 AB? B(–4, –2) 10 units C (6, –2) Using Pythagoras, AC2 = AB2 + AC2 = 72 + 102 = 49 + 100 = 149 AC = 149 = 12.21 units (to 2 d.p.)

Finding the distance between two points 15/11/2018 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 = 82 + 42 = 64 + 16 B(3, –2) C (7, –2) = 80 AC = 80 = 8.94 units (to 2 d.p.)

15/11/2018 Task – Coordinate cards Choose any TWO CARDS Sketch them and join with a line. This is your hypotenuse. Sketch in the horizontal and vertical sides to make a right angled triangle. Use Pythagoras to find the length of the hypotenuse Do as many as you can Difficulty Levels Green ( , ) coordinates in the first quadrant POSITIVE ONLY   Orange ( , )  coordinates in the other three quadrants POSITIVE AND NEGATIVE   Red ( , )   coordinates with decimal numbers + AND - INCLUDING DECIMALS   Blank   ( , ) make your own! AS DIFFICULT AS YOU LIKE

Finding the distance between two points - Algebraically 15/11/2018 Finding the distance between two points - Algebraically 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 (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

15/11/2018 Challenge problem! Use Pythagoras’ theorem to find the length and width of the rectangle and hence find its perimeter and area in surd form. Perimeter = √10 + 2√10 + √10 + 2√10 1 3 √10 6 2 2√10 = 6√10 units Area = √10 × 2√10 = 2 × √10 × √10 = 2 × 10 = 20 units2

Find the diagonals of the kite 15/11/2018 Find the diagonals of the kite 5 cm 12 cm 6 cm 5 cm x cm y cm

15/11/2018 Real life problem B A W 130 miles 170 miles B An aircraft leaves RAF Waddington (W) and flies on a bearing of NW for 130 miles and lands at a another airfield (A). It then takes off and flies 170 miles on a bearing of NE to a Navigation Beacon (B). From (B) it returns directly to Waddington. Make a sketch of the flight. How far has the aircraft flown?

15/11/2018 PYTHAGORAS IN 3D

Example – Find the distance BG 15/11/2018 Example – Find the distance BG Find FG first FG2 = 52 + 122 FG = (52 + 122) FG = 13 cm A B E F C D G H 3 cm 5 cm 12 cm Use FG to find BG BG2 = 32 + 132 FG = (32 + 132) FG = 13.3 cm 13 cm

Example – Find the distance BG 15/11/2018 Example – Find the distance BG Find EC first EC2 = 5.42 + 9.22 EC = (5.42 + 9.22) EC = 10.67cm A B D C E F 3.1 cm 9.2 cm 5.4 cm Use EC to find BG BE2 = 3.12 + 10.672 BE = (3.12 + 10.672) BE = 11.1 cm (1 dp) 10.67cm

15/11/2018 Problem solving My pencil case is in the shape of a cuboid. It measures 8cm, by 7cm by 5cm. I want to put my new pencil into the pencil case. The pencil measures 12cm. Does it fit?

15/11/2018 REVISION

15/11/2018 Right Angled Triangle Topics (printable version) Area of the triangle Questions involve a BASE and PERPENDICULAR HEIGHT It will ask you for the area of the triangle Use AREA = Pythagoras Involves __________ sides, ____ given and ____ missing If you want the LONGEST side (hypotenuse): square it, square it, ADD it, then square-root it If you want one of the SHORTER sides: square it, square it, SUBTRACT it, then square-root it Trigonometry Involves ____ sides and ____ angle ____ of these things will be given and ____ will be missing Use SOH___TOA HYPOTENUSE! O S H × 𝜃 A C H × 𝜃 O T A × 𝜃

15/11/2018 Task – Complete the review worksheet of Pythagoras questions Click to reveal solutions… …and click again to reveal Qs 7-10…