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?

Yes, It is

Step-by-step explanation:

Demonstrating the Pythagorean Theorem

When you think of each side of a right triangle as also being a side of a square that's attached to the triangle. The area of a square is any given side multiplied by itself. (For example, b x b = b^2).

In order to show that a^2 + b^2 = c^2,

follow these steps:

Get a right triangle on grid paper that you can print. You'll also need scissors, and a ruler.

Cut out the triangle.

Make three squares with sides that are equal to each side of the triangle. Begin with side

a. Measure the length of side a. On the blank piece of grid paper,

- draw a square with sides that are the same length as side a.

- Label this square a2.

- Repeat these steps to create squares for sides b and c. (If you don't have a ruler, just use the triangle as a guide; trace the length of one side, and then draw three more sides of the same length to make a square.)

Cut out the squares. Place each square next to the corresponding sides of the triangle.

Now show that a2 + b2 = c2. Place the squares made from sides a and b on top of square c. You will have to cut one of the squares to get a perfect fit.

Area of Whole Square

The total area of a big square, where each side having a length of a+b, is:

A = (a+b)(a+b)

Area of The Pieces

By adding up the areas of all the smaller pieces:

First, the smaller (tilted) square has an area of:c^2

Each of the four triangles has an area of: ab^2

So all four of them together is: 4ab^2 = 2ab

Adding up the tilted square and the 4 triangles gives:A = c^2 + 2ab

Both Areas Must Be Equal

The area of the large square is equal to the area of the tilted square and the 4 triangles. This can be written as:

(a+b)(a+b) = c^2 + 2ab

By rearrange this, we shall see if we can get the pythagoras theorem:

Start with:(a+b)(a+b) = c^2 + 2ab

Expand (a+b)(a+b):ag2 + 2ab + b^2 = c^2 + 2ab

Subtract "2ab" from both sides:a^2 + b^2 = c^2

Therefore, it's been proven that, a^2 + b^2 = C^2

No, because this proof only works when you can draw grids on the squares. It does not work for all right triangles.

Step-by-step explanation:

the first on is incorect this is prof

9) i can't see the angle so whatever it is just do this: cos(angle) x hypotenuse and then use that answer and they hypotenuse and plug it into this: (answer you just got)^2 + b^2 = (hypotenuse)^2. the first answer will be the answer of the adjacent side and the second answer after this equation will be the opposite side.

10) y = 14 x = √147 or 12.12435565

11) x = 4.8

12) x = 4.2

13) x = 9.3

14) pq: 13.2

qr : 17.6

angle p: 53º

15) angle c: 54.3º

angle a: 35.7º

16) h = 21040

v = 1841

step-by-step explanation:

10) 7/sin(30) = 14 = y

7^2 + b^2 = 14^2

49 + b^2 = 196

b^2 = 196 - 49

√b^2 = √147

x = √147

11) cos(70) x 14 = 4.788282007

x = 4.8

12) tan(25) x 9 = 4.196768923

x = 4.2

13) sin(31) x 18 = 9.270685348

x = 9.3

14) pq: sin(37) x 22 = 13.23993051

pq: 13.2

qr: cos(37) x 22 = 17.56998122

qr : 17.6

angle p: 180 - (37 + 90) = 53º

15) cos-1(7/12) = 54.31466529

angle c: 54.3º

angle a: 180 - (90 + 54.31466529) = 35.68533471

angle a: 35.7º

16) 4 x 5280 = 21120

cos(5) x 21120

h = 21039.63202

h = 21040

sin(5) x 21120

v = 1840.729287

v = 1841