Pythagoras' theorem?

Question

I know that the pythagoras' theorem is A squared plus B squared equals C squared. Can anyone tell me why this is true please put diagrams and also explanation. must be in Year 8 Maths.

Answer 1

Sorry to interrupt you guys.

But, i have a different approach.

I will prove it using trigonometry.

From the figure shown in the link,

hypotenuse = c

Let the angle between 'b' and 'c' be θ.

so, in terms of angle θ,

perpendicular = a
base = b

We know,
Sin θ = perpendicular / hypotenuse
Cos θ = base / hypotenuse

And,

(Sin θ)^2 + (Cos θ)^2 = 1
or, (perpendicular / hypotenuse)^2 + (base / hypotenuse)^2 = 1
or, (a/c)^2 + (b/c)^2 = 1
or, a^2/c^2 + b^2/c^2 = 1
or, (a^2 + b^2)/c^2 = 1
or, a^2 + b^2 = c^2

(base)^2 + (perpendicular)^2 = (hypotenuse)^2

Thats the pythagorean theorem.

[I forgot the name of the side that is perpendicular to another side. So, i called it perpendicular. Sorry about that.]

Answer 2

Draw a square. Divide each side into one long segment (call it "b") and one short segment (call it "a"), so that b and a segments alternate as you go around the square. Add four lines to create four right triangles around the square, each one having one leg of length a and one leg of length b; call the hypotenuse "c". Shade the triangles. You now have an unshaded square in the middle, and it has side length c and thus area c^2. The large, original square has area equal to (a + b)^2 = a^2 + 2ab + b^2. Each shaded triangle has area equal to (1/2)ab, so all four together have area equal to 2ab. But the shaded triangles and the unshaded square must together have the same area as the large square. Thus, a^2 + 2ab + b^2 = c^2 + 2ab. Subtracting 2ab from both sides, a^2 + b^2 = c^2.

Answer 3

Nobody can tell you why this is true, although we can draw pictures and prove that it's true. Something amazing about Math is that it's inherent in our world - we didn't invent it! The relationship between the sides of a triangle has always been the same, we just didn't know how to describe or calculate it until Euclid and Pythagoras discovered it.

Answer 4

Draw a right angled triangle with right angle at C.
CA is horizontal of length 4 cm
CB is vertical of length 3 cm
Join A to B
AB is hypotenuse and will be of length 5 cm

Construct a square on CA ( area = 16 cm ² )
Construct a square on CB ( area = 9 cm ² )
Construct a square on AB ( area = 25 cm ² )
25 = 16 + 9
ie in a right angled traingle, the square on the hypotenuse = sum of squares on other two sides

Answer 5

All these answers are good. I'm not going to steel their thunder, but you might be interested to know that even Euclid found it difficult to get a solid proof of the Pythagorean theorem. Euclid had a good proof in the first book of the Elements, his textbook on plainer geometry.

this is a list of 74 different proofs.

http://www.cut-the-knot.org/pythagoras/index.shtml

Answer 6

draw and cut a square any size and label it A
draw and cut a square any size (Different than A) and call it B
draw a little bigger square larger than A and B and call it C


lay them down on the table put them together so they form a triangle in the middle

a squared + b squared = c squared

therefore :
c squared - a squared = b squared

do it on math paper with the squares on it
count the squares on each square A , B, C, and confirm.

Answer 7

look at the diagram below on the link. This should help you, it shows that the squares of each of the sides A and B are equal to the squared of side C.

Hope this helps!

Answer 8

Dear , acc. to pythagorus thm.
In aright angled triangle , the sq. of perpendicular side plus the sq. of its base is equal to the sq. of its hypotinuse side .
A !\
! \
! \ i.e = (AB)^2 + (BC)^2 = (AC)^2
! \
B ! __ \ C

The diagram might not be clear in it but i hope what i wanted to say