# Year 7 Pythagoras Theorem Dr J Frost ([email protected])

Year 7 Pythagoras Theorem Dr J Frost ([email protected]) www.drfrostmaths.com Objectives: Apply Pythagoras Theorem to 2D problems. Know common Pythagorean triples. Know what is meant by given an answer in exact form. Find the perpendicular height and area of an equilateral triangle. Last modified: 29th April 2016 STARTER Sketch this right-angled triangle in your book in the centre of a new page. Work out the length of the longest side using a ruler. 3cm 4cm STARTER 5cm 5cm Area = ? Area = ? Now turn each side of the triangle into a square. Can you notice anything about the relationship of the three areas? 3cm 3cm Area = ? 4cm 4cm

32 + 4 2 = 5 2 Pythagoras Theorem ! Write this down Hypotenuse (the longest side) For any right-angled triangle with hypotenuse . Bro Note: notice that its the longest side thats on its own on one side of the equation. The (squared) shorter sides are the ones that are added. Example Step 1: Determine the Reveal > hypotenuse. 2 4 Step 2: Form an equation 2 2 ? 2 2 +4 =

The hypotenuse appears on its own. Step 3: Solve the equation to find the unknown side. or 4.47?to 2dp Further Example 3 Step 1: Determine the Reveal > hypotenuse. Step 2: Form an equation 7 2 2 ? 2 +3 =7 The hypotenuse appears on its own. Step 3: Solve the equation to find the unknown side. or 6.32?to 2dp Surd or decimal? A value written as the

square root of a number is known as a surd. Sometimes its better to leave your answer in surd form (well see why later) rather than as a decimal. to 2dp When we found areas/circumferences of circles, we often left our answer in terms of so that it was exact. Similarly, answers in surd form are exact whereas decimal form answers have to be rounded, and are thus not exact. Test Your Understanding ? Answer: 2 ? Answer: 3 42 1 1 1 5 6

6 4 55 12 8 Answer: ? 4 Answer: 10 Answer: To learn secret way of ninja, find you must. ? ? Pythagoras Mental Arithmetic Weve so far written out the equation , filled in our information, and rearranged to find the missing side. But its helpful to be able to do it in our heads sometimes! If youre looking for the hypotenuse Square root the sum of the squares

If youre looking for another side h Square root the difference of the squares 7 3 5 2 4 2 ? h= 3 +5 = ? 34 2 2 ? 4 = ?33 = 7 Pythagoras Game! Everyone stand up. Each of you will be asked, one at a time, and in your head, to find the missing side of the right-angled triangle. Answer must be in exact form. If you get it wrong, you sit down, and the person who last sat down has the opportunity to steal, where they will be able to stand up again if they correct the answer. Test Run:

5 ? 3 7 ? 2 (Note to teacher: You dont need to specifically click on the green boxes. The next answer will be removed by a mouse/right-arrow press anywhere) Pythagoras Game! ? 3 4 3 4 3 2 1 2 ? 5 ? 10

? 8 ? 7 ? ? 5 8 ? 12 3 2 6 ? 2 4 ? ? 1

5 6 ? 2 3 ? 9 5 2 11 ? ? 4 Pythagoras Game! 5 3 1 2 ? ? 4 4

6 ? ? 5 9 9 7 ? 10 ? 7 ? ? 2 4 1 3 7 2 ? 3 2 ?

10 ? 6 5 ? 8 10 ? 6 ? 3 12 3 ? Exercise 1 Find the side marked with the letter (you do not need to copy the diagrams). 4.5 1 a

7 d b g 23 5.1 6.2 98 this rectangular field to the other. Alice walks round the edge of the field. Bob cuts right across. How much further did Alice walk? 3.6 125 e 19 2 1.8 c 3 Alice and Bob want to get from one corner of

f 9 1.4 7 2.2 80m ? 90 Finish Solutions: (to 3sf) (a) 8.32 ? (b) 3.12 ? (c) 77.6 ? (d) 29.8 ? (e) 5.66 ? (f) 2.61 ? (g) 8.03 ? To rescue a cat I put a ladder of length 10m against a tree, with the foot of the ladder 2.5m away from the tree. How high up the tree is the cat? 9.68m ? 240

Start [Kangaroo Pink 2008 Q3] Four unit squares 4 are placed edge to edge as shown. What is the length of the line ? Solution: ? N [Based on JMO 1996 A6] The length of the shortest diagonal of an octagon is 1. What is the length of the longest diagonal? 11 1 2 Solution: (if this were a proof youd need to justify why its right-angled) ? Starter 29 21 You may have noticed last lesson that sometimes all three sides of the right-angled triangle were integers. These are known as Pythagorean triples. For example: The sides could be 20, 21 and 29, as and thus satisfy Pythagoras Theorem.

How many Pythagorean triples can you find? 20 (3, 4, 5) (20, 21, 29) (11, 60, 61) (13, 84, 85) (5, 12, 13) (12, 35, 37) (16, 63, 65) (36, 77, 85) ? (8, 15, 17) (9, 40, 41) (33, 56, 65) (39, 80, 89) (7, 24, 25) (28, 45, 53) (48, 55, 73) (65, 72, 97) Note that you could also have any multiple of any of these triples as the triangles could be scaled in size. So for example (3, 4, 5) could become (6,8,10) and so on. A final note is that if you changed the powers from 2 to 3, or any higher number, then there would never be any solutions. This is known as Fermats Last Theorem, which was unproven for hundreds of years before being proven in 1995. Harder Questions Theres a variety of ways in which Pythagoras questions could get harder: 9 6

2 2 Area? 2 4 A Multiple triangles chained together. 2 +1 7 12 3 1 3 B Adding lines to form right-angled triangles that werent originally there. C Requiring algebraic manipulation. A :: Multiple Triangles

6 4 What should we do first? Find the central length using the right triangle. ? 3 Then what? Now we can find using the left triangle. ? Notice that . This is why its often important to leave your answers in surd form. Test Your Understanding 4 12 6 2 ?2 = 4 +6 = 52 B :: Adding Lines Sometimes the line(s) you add to form right angled triangle(s) are fairly obvious

7 2 4 2 = 7 +4 = 65 ? 1 And sometimes really not very obvious at all Click to Brosketch > 4 ?3 5? 4? 1 (because the radii of the circles are 4 and 1, and therefore the combined length 5) (because the difference in the distances between the centres of the circle and the bottom is ) By Pythagoras. Therefore . 30-60-90 Triangles 2

30 3 2 60 2 1 We want to find the area of this equilateral triangle. But we first need the perpendicular height. What do we do first? The triangle can be split in half. We get a right-angled triangle with?base 1. So what therefore is the height? Using Pythagoras Theorem, the height is . ? What therefore is the area of the whole triangle? Bro Pro Tip: The height of an equilateral triangle is times half the side length. ? ! The ratio between the lengths of a -- triangle is because it is half an equilateral . e.g. We can halve the hypotenuse to get the shortest length, and multiply the shortest length by to get the other. (This is really important for those who want to do well at the JMO) 2 30 60 1

3 Quickfire Heights! Reminder of Pro Bro Tip: The height of an equilateral triangle is times half the side length. Height Area 2 3 4 2 ? ? Height Area ? ? Height Area ? ? 1 Height Area ? ? Test Your Understanding

4 Medium Difficulty Find the height of this isosceles triangle. Solution: ? 6 Harder Difficulty [JMO 2000 A8] An equilateral triangle is cut out of a square of side 2 cm, as shown. What area of the square remains? Solution: ? 1 3 3 1 Difficult Difficulty [Kangaroo Pink 2012 Q11] Six identical circles fit together tightly in a rectangle of width 6 cm as shown. What is the height, in cm, of the rectangle? Solution: ? C :: Algebraic Triangles 15 3 4

2 2 2 ? (3 ) +(4 ) =15 (You will likely encounter more interesting algebraic Pythagoras problems next year once you cover expanding two brackets) Exercise 2 (exercises on provided sheet) Give answers in exact form unless specified. 2 1 3 12 3 5 2 4 5 ? = 185

? = 13 1 Santa Claus and Rudolph are sitting at the corner of a square swimming pool of 10m by 10m, which has frozen over. The want to get to the other corner of the pool, where Mary Christmas has left some brandy and a carrot. Rudolph runs around the edge of the pool, while Santa, who has recently been on a diet, decides to risk walking diagonally across the ice. Calculate the distance saving, to 2dp. m ? Exercise 2 4 (exercises on provided sheet) Two snowmen are back to back, facing in opposite directions. They each walk 3km forward, turn left and then work a further 4km. How far are the snowmen from each other? 6

2 3 Solution: 10km ? 5 6 6 1 7 6 6 (a) What is the height of this equilateral triangle? Solution: or ? (b) The area? Solution: ? 6 8 25 8 48

25 1 = 8? 3 Find the height of this isosceles triangle. Solution: ? Find the area of this isosceles triangle. ? Solution: 168 Exercise 2 (exercises on provided sheet) 1 9 10 1 1 1 [IMC 2008 Q20] What, in cm2, is the area of this quadrilateral? Solution: 48cm2 ?

1 1 = 6 ? 11 [IMC 2009 Q20] A square, of side two units, is folded in half to form a triangle. A second fold is made, parallel to the first, so that the apex of this triangle folds onto a point on its base, thereby forming an isosceles trapezium. What is the perimeter of this trapezium? A B C D E 5 ?D Solution: Exercise 2 (exercises on provided sheet) 12 27 Determine . 2 ? Exercise 2

(exercises on provided sheet) N1 [JMO 2006 B4] Start with an equilateral triangle of side 2 units, and construct three outward-pointing squares ABPQ, BCTU, CARS and the three sides AB, BC, CA. What is the area of the hexagon PQRSTU? N2 2 3 30 30 120 2 90 90 60 2 2 Diagram ?

2 We can use the angles to show that each of the surrounding triangles can be cut into two and the halves moved to form an equilateral Solutiontriangle ? of side 2. Thus we have 4 equilateral triangles and 3 squares for a total area of 2 [JMO 1999 B4] The regular hexagon has sides of length 2. The point is the midpoint of . is the midpoint of and so on. Find the area of the hexagon . Solution: ? The outer hexagon is composed of 6 equilateral triangles (as indicated). As

the side length is 2, the height is . Thus the inner hexagon consists of 6 equilateral triangles with side length . The height of each is Thus overall area is Exercise 2 N3 [JMO 2007 B5] A window is constructed of six identical panes of glass. Each pane is a pentagon with two adjacent sides of length two units. The other three sides of each pentagon, which are on the perimeter of the window, form half of the boundary of a regular hexagon. Calculate the exact area of the glass in the window. Each pane can be split into 7 equilateral triangles with side length 1. Each has an area of . ? the area is There are of these, thus (exercises on provided sheet) Exercise 2 N4 (exercises on provided sheet) [JMO Mentoring May2012 Q4] A triangle has two angles which measure 30 and 105. The side between these angles has length 2 cm. What is the perimeter of the triangle? (Hint: split the triangle somehow?) Solution: The key here is that can be split into and . Thus we end up with a 30-60-90 triangle (which we know is half an equilateral triangle,

so can work out the sides) and an isosceles right-angled triangle (which we can use ? Pythagoras Theorem on. 2 60 45 1 2 45 1 30 3 Exercise 2 N5 (exercises on provided sheet) N6 9 7 27 3 3 Determine . Solution: We can find the (square of the) central length in two different ways:

? Determine (to 2dp). Solution: Using similar strategy to the previous question: ? One Final Really Cool Thingy h 2 2 + =( h+ ) 2 How far can you see into the horizon when: (a) at sea level, and your height is 1.5m. (b) Sitting in a plane 12km above sea level? It may be helpful to use the radius of the Earth: 6371km.

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