How to find the length of the hypotenuse when we know the length of the two right sides of a right triangle Don't use Pythagorean theorem and function method, stand on the position of fifth grade students, use their method

How to find the length of the hypotenuse when we know the length of the two right sides of a right triangle Don't use Pythagorean theorem and function method, stand on the position of fifth grade students, use their method

The composition method was used
Take four identical right triangles and make a big square with a small square in the middle,
If the two right sides of a right triangle are known, the area of a large square and the area of four triangles can be calculated directly,
So we can calculate the area of a small square,
Then calculate the side length of the small square, which is the oblique side length of the right triangle

Given that the length of two right angle sides of a right triangle is 160m, 108m, find the length of the other hypotenuse Tell me what you know

Let the hypotenuse of the RT triangle be X
Then x ^ 2 = 160 ^ 2 + 108 ^ 2
The length of the third side is x under the root sign

The length of a right angle side of a right triangle is fixed to 8, and the length of the hypotenuse is not more than 100 Want VB language Hurry!

Just use a loop: dim I%, J%, l š for I = 1 to int (SQR (100 ^ 2 - 8 ^ 2)) l = SQR (I ^ 2 + 8 ^ 2) if l = int (L) thenprint "the three sides of a right triangle are: 8," & I &, "& lend ifnext iPrint" respectively

Given that the two right sides of a right triangle are 100 and 70 respectively, find the length of the hypotenuse

Square of beveled edge = 100? + 70? = 14900
Hypotenuse = 10 times root number 149 = 122.066

The length of a right angle is 1.7, and the length of one side of a right angle is one Determinant

If a right angle side length is a (a > 0), then the other right angle side length is a + 1, a ^ 2 + (a + 1) ^ 2 = 7 ^ 2, the solution is a = (- 1 + √ 97) / 2 or a = (- 1 - √ 97) / 2 (less than 0, round off), so the two right angle side lengths are (- 1 + √ 97) / 2 and (1 + √ 97) / 2

The difference between the two right angles of a triangle is 5cm and the area is 7cm 2

Let the shorter right angle side be xcm, and the longer one is (x + 5) cm,
One
2x•(x+5)=7,
The result is: x2 + 5x-14 = 0,
∴(x+7)(x-2)=0,
Ψ x = 2 or x = - 7 (omitted)
5+2=7cm,
22+72=
53cm.
The length of the hypotenuse is
53cm.

The hypotenuse length of a right triangle is 8 cm. The difference between the two sides is 2 cm

15 square centimeter, a = x B = X-2
You're wrong upstairs. Check it out

The length of the hypotenuse of a right triangle is 8 cm. The difference between the lengths of the two right sides is 2 cm. Find the area of the triangle Is it possible to solve it arithmetically

Yes, the sum of the squares of the two right sides is equal to the square of the hypotenuse. If the short right angle side is x, then the other side is x + 2. Therefore, the equation x + (x + 2) = 8 = 64 is 2x + 4x-60 = 0. The solution of the equation x = - 12 or x = 5. Because x is a positive integer, x = 5, so the area is 5 × 7 △ 2 = 17.5

The length of the hypotenuse of a right triangle is 25 cm. The difference between the two right sides is 17

Solving equations
If the short right angle side is x, then the long right angle side is x + 17
According to the Pythagorean theorem, x ^ 2 + (x + 17) ^ 2 = 25 ^ 2 = 625
X = 7
So the two right angles are 7, 24

How to bend a right triangle with 24 cm long wire? How to bend one right angle side 2 cm longer than the other? It seems that you can't solve the equation of degree one! It seems that there is no solution to the bivariate linear equation?

If the shorter right angle side length is a and the longer right angle side is (a + 2), then the oblique side length is (22-2a). According to (22-2a) * (22-2a) = a * a + (a + 2) * (a + 2), a * a-46a + 240 = 0 is simplified, and a = 6 or 40 (omitted). Therefore, the three sides of the right triangle are 6, 8 and 10
I will not write the unit. In fact, these three data are a typical set of right triangle data, just like 3, 4, 5, the bivariate linear equation seems to have no solution? You should be wrong