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Emily says she can prove the Pythagorean Theorem using the following diagram. She explains that she can divide the squares on the two shorter sides into grids with equal-sized grid squares. She says she can then rearrange the grid squares to cover the area of the square on the hypotenuse, which proves that the sum of the squares on the two shorter sides equals the square on the hypotenuse.

Is this a valid geometrical proof that a2 + b2 = c2 for all right triangles?

User Ned Howley
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Answer:

The last bit is correct, a2 + b2 = c2 for all right triangles? You do not divide the measurements you can only add up the squares to find the square of the hypotenuse as the sum, and then square root the totaled squares back to a new number. It does not cover the area like the statement said so only the question is true as Pythagoras theorem can be used to find a missing side or to find the missing hypotenuse.

Explanation:

Is this a valid geometrical proof that a2 + b2 = c2 for all right triangles?

Answer yes but only for c2 if we area adding.

This is because if we were looking and missing for a smaller side to the triangle measure and was already given the hypotenuse as one measure, then we can subtract either given alternative side would be; c2 - b2 = a2 to find an alternative side we subtract.

and c2- b2 = a2 works the same as c2 -a2 = b2.

None of the above formula finds the area, for the area we would need to multiply to find the height.

A = 1/2 x a x b

We have a different understanding to area

The area of a triangle is 1/2 given by the equation:

AΔ=(1/2)∗(base)∗(height) which is the smae as A= 1/2 x a x b

Eg)

Given: Measurements of A + B In attachment picture.

A = 3 cm

B = 4 cm

What is the area of the right triangle ABC?

Since the base leg of the given triangle is 4 cm, while the height is 3 cm, this gives:

AΔ=(1/2)∗3∗4=6cm2

So to find area we multiply A by 1/2 and multiply that product by the length of B (base).

Emily says she can prove the Pythagorean Theorem using the following diagram. She-example-1
User MordicusEtCubitus
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