This solution includes to the problem of Pythagoras Perimeters. For the first part of the solution: A right-angles triangle has a perimeter of 12 and an area of 36-6c, to fix by putting the steps into correct orders. For the second one, by adapting the method of the first solution to this for the new perimeter of 30 and an area of 225-15c. For the extension part, doing the same method with the second part and by replacing the perimeter into p and the hypotenuse into c.
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First part of solution
perimeter of 12 and an area of 36-6c
Putting the steps into order - g,c,a,f,d,e,h,b
1. (g) >> a+b+c=12
2. (c) >> a+b=12−c
3. (a) >> Squaring both sides: a²+2ab+b²=144−24c+c²
4. (f) >> By Pythagoras' Theorem, a²+b²=c²
5. (d) >> So 2ab=144−24c
6. (e) >> Area of the triangle =ab/2
7. (h) >> Dividing by 2: ab=72−12c
8. (b) >> So Area of the triangle =36−6c
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First question
Explanation of each steps
1. (g) >> a+b+c=12
Firstly, we can know that the perimeter is 12 so it means that the three sides of the triangle add together equals 12. The three sides of the triangle are hypotenuse (which equals c), opposite (which equals b) and adjacent (which equals c).
2. (c) >> a+b=12−c
Secondly, move the letter c to the other side, in order to do it, move the c by subtracting it to the other side so it forms a+b=12-c
3. (a) >> Squaring both sides: a²+2ab+b²=144−24c+c²
Thirdly, square both sides for the equation by using expansion.
>> (a+b)² = (12-c)²
>> a²+2ab+b²=144−24c+c²
4. (f) >> By Pythagoras' Theorem, a²+b²=c²
Fourthly, by using the Pythagoras' Theorem to solve the equation. In this theorem, a²+b²=c², which means that square up the opposite and the adjacent side and then these two sides square the square of the hypotenuse side.
5. (d) >> So 2ab=144−24c
Fifthly, by applying the theorem into the equation of a²+2ab+b²=144−24c+c². We know that a²+b²=c², which means the both sides is equal so we can remove a²+b²=c² off. Then, the equation remains into 2ab=144−24c.
6. (e) >> Area of the triangle =ab/2
Sixthly, we know that the area of the triangle equals ab/2 which means ab equals the area of a rectangle/square which has the similar sides of the triangle so that, we need to divide the equation by 2 so we can get the area of the triangle.
>> 2ab=144−24c
>> ab=72-12c
7. (h) >> Dividing by 2: ab=72−12c
Seventhly, by applying the division the equation (2ab=144−24c) by 2 and get the result.
>> ab=72-12c
>> ab/2=36-6c
8. (b) >> So Area of the triangle =36−6c
Lastly, we finally get the result!
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Second question
perimeter of 30 and an area of 225-15c
Putting the steps into order - g,c,a,f,d,e,h,b
1. (g) >> a+b+c=30
2. (c) >> a+b=30−c
3. (a) >> Squaring both sides: a²+2ab+b²=900−60c+c²
4. (f) >> By Pythagoras' Theorem, a²+b²=c²
5. (d) >> So 2ab=900−60c
6. (e) >> Area of the triangle =ab/2
7. (h) >> Dividing by 2: ab=450−30c
8. (b) >> So Area of the triangle =225-15c
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Third question
by replacing the perimeter into p and the hypotenuse into c
1. (g) >> a+b+c=p
2. (c) >> a+b=p−c
3. (a) >> Squaring both sides: a²+2ab+b²=p²−2pc+c²
4. (f) >> By Pythagoras' Theorem, a²+b²=c²
5. (d) >> So 2ab=p²−2pc
6. (e) >> Area of the triangle =ab/2
7. (h) >> Dividing by 2: ab=p²/2−pc
8. (b) >> So Area of the triangle =p²/4−pc/2
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And here is my solution of the problem - Pythagoras Perimeters