The right triangle below has legs of length and. The hypotenuse has length.
Four copies of this triangle are arranged as follows.
The hypotenuses form a square of side length and area.
Answer the questions below to find how, and are related.
Part 1: Compute the total combined area of the four triangles:

Part 2: Compute the area of the large (outer) square:
Part 3: Using your answers in Parts 1 and 2, find the area of the small (inner) square.
Part 4: We are given the side lengths and. Compute.
Part 5: Use, or complete the statement below.

Answer + explanation:

Part 1: First, find the area of one triangle using the formula base * height / 2. You have base = 10 and height = 7, so area = 7 * 10 /2 = 70/2 = 35. Since you're finding the combined area of 4, you have to do 35 * 4 = 140.

Part 2: To do this, find the side length of the square. Looking at the picture, you see that all you have to do is 10 + 7 = 17 to find one side. Then, to find the area do 17 * 17 = 289.

Part 3: To find the area of the small square, all you do is area of the large square - combined area of 4 small triangles, so 289 - 140 = 149.

Part 4: a^2 = 7 * 7 = 49. Then, b^2 = 10 * 10 = 100. Adding them together, you get 100 + 49 = 149.

Part 5: Since you already have the a^2 + b^2 part, all you need to do is find what c^2 is. You can see in the picture that c^2 equals the area of the small square, which you found in part 3. c^2 = 149. Then, you have 149 and 149, which are equal so the answer is =.

answer:

it is general formula that is used to write value to its excet position

a11 a12 a13

a21 a22 a23

it is the example of order 2 by 3 matrix in this

a11 means value occur in 1 st row 1st column

a12 means value occurs in 1st row 2nd column

a13 means value occur in 1st row and 3rd column

a21 means value occurs in 2nd row and 1st column

a22 means value occurs in 2nd row and in 2nd column

a23 means value occurs in 2nd row and in 3rd column