Pythagoras' Theorem ExerciseA self marking exercise on the application of Pythagoras' Theorem. 
Here are some questions which can be answered using Pythagoras' Theorem. You can earn a trophy if you get at least 14 questions correct. Each time you finish a question click the 'Check' button lower down the page to see if you got it right!
[Don't forget to include the units in your answers after question one]
InstructionsTry your best to answer the questions above. Type your answers into the boxes provided leaving no spaces. As you work through the exercise regularly click the "check" button. If you have any wrong answers, do your best to do corrections but if there is anything you don't understand, please ask your teacher for help. When you have got all of the questions correct you may want to print out this page and paste it into your exercise book. If you keep your work in an ePortfolio you could take a screen shot of your answers and paste that into your Maths file. 



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Mathematicians are not the people who find Maths easy; they are the people who enjoy how mystifying, puzzling and hard it is. Are you a mathematician? Comment recorded on the 9 April 'Starter of the Day' page by Jan, South Canterbury: "Thank you for sharing such a great resource. I was about to try and get together a bank of starters but time is always required elsewhere, so thank you." Comment recorded on the 3 October 'Starter of the Day' page by Fiona Bray, Cams Hill School: "This is an excellent website. We all often use the starters as the pupils come in the door and get settled as we take the register." 
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AnswersThere are answers to this exercise but they are available in this space to teachers, tutors and parents who have logged in to their Transum subscription on this computer. A Transum subscription unlocks the answers to the online exercises, quizzes and puzzles. It also provides the teacher with access to quality external links on each of the Transum Topic pages and the facility to add to the collection themselves. Subscribers can manage class lists, lesson plans and assessment data in the Class Admin application and have access to reports of the Transum Trophies earned by class members. If you would like to enjoy adfree access to the thousands of Transum resources, receive our monthly newsletter, unlock the printable worksheets and see our Maths Lesson Finishers then sign up for a subscription now: Subscribe 

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Level 1  Finding the hypotenuse
Level 2  Finding a shorter side
Level 3  Mixed questions
Level 4  Pythagoras coordinates
Level 5  Pythagoras' Theorem exercise
Exam Style questions requiring an application of Pythagoras' Theorem and trigonometric ratios to find angles and lengths in rightangled triangles.
Three Dimensions  Three dimensional Pythagoras and trigonometry questions
More on this topic including lesson Starters, visual aids, investigations and selfmarking exercises.
Answers to this exercise are available lower down this page when you are logged in to your Transum account. If you don’t yet have a Transum subscription one can be very quickly set up if you are a teacher, tutor or parent.
See the National Curriculum page for links to related online activities and resources.
The area of the square on the hypotenuse of a right angled triangle is equal to the sum of the areas of the squares on the two shorter sides.
You may have learned the theorem using letters to stand for the lengths of the sides. The corners (vertices) of the rightangled triangle is labelled with capital (upper case) letters. The lengths of the sides opposite them are labelled with the corresponding small (lower case) letters.
Alternatively the sides of the rightangled triangle may me named using the capital letters of the two points they span.
As triangle can be labelled in many different ways it is probably best to remember the theorem by momorising the first diagram above.
To find the longest side (hypotenuse) of a rightangled triangle you square the two shorter sides, add together the results and then find the square root of this total.
To find a shorter side of a rightangled triangle you subtract the square of the other shorter side from the square of the hypotenuse and then find the square root of the answer.
AB^{2} = AC^{2}  BC^{2}
AB^{2} = 4.7^{2}  4.1^{2}
AB^{2} = 22.09  16.81
AB^{2} = 5.28
AB = √5.28
AB = 2.3m (to one decimal place)
The diagrams aren't always the same way round. They could be rotated by any angle.
The rightangled triangles could be long and thin or short and not so thin.
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