Solve each of the following problems fully. Some may require several sheets depe

Question

Solve each of the following problems fully. Some may require several sheets depending on the number of methods you demonstrate. Make sure you have looked over the grading rubric for homework so that you understand I am looking for more than answers to the questions I ask. You should always be using various strategies and explaining them. Detailed work should accompany each problem 1. Two sides of a right triangle have lengths 360 lengths for the third side. and 480 Find the possible 2. The two part diagram below, which shows two different dissections of the same square, was designed to help prove the Pythagorean Theorem. Provide the missing details.

Explanation / Answer

1. Two side of a triangle are given:

360 and 480.

Here there are 2 possibilities:

i) both are just sides and not Hypotenuse

ii)480 is the Hypotenuse.

Length of third side:(Using Pythagoras Theorem)

i) length=(4802+3602)1/2=600units

ii) length=(4802-3602)1/2=317.49units

2.Here you will notice that the first square of side (a+b) units is made of a square of side 'a' , a square of side 'b' , and four right angled triangles of sides which are not the hypotenuse a,b.

therefor equating are we get

(a+b)2=a2(ie area of square of side 'a')+b2(ie area of square of side 'b')+4(area of the triangle)

but we know,

(a+b)=a2+b2+2ab

=> are of a right angled triangle with side (not hypotenuse) a and b is=0.5ab.

Moving on to the second square.

This is made of a square of side=hypotenuse of the triangle of side a and b, and 4 right angled triangle with sides(not hypotenuse) a,b.

Let the side the hypotenuse of the square be 's',

therefor we have:(equating areas)

(a+b)2=4(0.5ab)+s2(ie are of the square in the middle)

=>a2+b2+2ab=2ab+s2

=> a2+b2=s2

where a and b are the sides of a right angled triangle and s their hypotenuse.

Therefor Pythagoras theorem has been proved.

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