If the sum of the two right sides of a right triangle is 6, the minimum area of a square with the hypotenuse of the right triangle as the side length is 0______ ．

If the sum of the two right sides of a right triangle is 6, the minimum area of a square with the hypotenuse of the right triangle as the side length is 0______ ．

∵ the sum of the two right sides of a right triangle is 6, ∵ if one right side of a right triangle is x and the length of the hypotenuse is y, then the length of the other right side is 6-x, ∵ s = x (6-x), that is, s = - x2 + 6x, ∵ smin = − 624 × (− 1) = − 36 − 4 = 9

The area of an isosceles right triangle is 8 square centimeters. What is the area of a square with its hypotenuse as its side?

It's very simple
Let the right side of an equal right triangle be a, then according to the Pythagorean theorem, the length of its oblique side is (√ 2) a
If the area of a right triangle is 8 square centimeters, then a & sup2 / 2 = 8
So a = 4
So the area of the square with the hypotenuse as the side is [(√ 2) a] & sup2; = 2A & sup2; = 32

In a right triangle, the area of the two squares with the right angle side as the side length is 30 and 20 respectively, and the area of the square with the oblique side as the side length is ()
A. 25B. 50C. 100D. 60

Let two right angle sides be a and B respectively, and the hypotenuse be c. then we know from the meaning of the title: A2 = 30, B2 = 20. According to the Pythagorean theorem: C2 = A2 + B2 = 30 + 20 = 50, so we choose B

The perimeter of a right triangle is 18 cm and the hypotenuse is 8 cm. Find its area
Why haven't you received it yet?

Let the other two sides be a and B
Then a + B = 10
a2+b2=64
(a+b)^2=100=a2+b2+2ab=64+2ab
ab=18
Area = 1 / 2Ab = 9

A right triangle has a right side with a length of 16, and the length of the other two sides is also a positive integer. What is its perimeter?

Let the other two sides ask a and C respectively
Then we get a * a + 16 = C * C, which is rewritten as (c + a) (C-A) = 16;
Since the other two sides are positive integers, then c + A and C-A are both positive integers;
So there are two sets of results
C + a = 16
c-a=1
And C + a = 8
c-a=2
In fact, only the second group is correct, so a = 5, C = 3;
The perimeter is 3 + 4 + 5 = 12;

1. A right triangle has a right side with a length of 11, and the length of the other two sides is also a positive integer. What is the perimeter of the triangle?
One more question,
It is known that the height of the hypotenuse ab of the right triangle ABC is 2.4cm, and the right sides AC = 4cm, BC = 3cm?
(if this problem is OK, it can be explained with the attached figure.)

According to the theorem, the sum of the squares of the two right sides of a right triangle = the square of the third side, that is, a ^ 2 + 11 ^ 2 = C ^ 2
Solving the equation C ^ 2-A ^ 2 = 121
We get (c + a) x (C-A) = 121
Because both a and C are positive integers, if we decompose the common divisor by 121, we can get 1,11121. Because a is not equal to C, and a, C are not 0, so C-A = 1, C + a = 121
Perimeter = a + B + C = 121 + 11 = 132
The geometry is two cones when the line AB is rotated one circle
N is the degree of the center angle of the circle, R is the generatrix, and R is the radius of the bottom circle
Side area = n π R & # 178 / 360
AB=5,CD=12/5,
The first side area is 2 π * 12 / 5 * 3 / 2 = 36 π / 5,
Similarly, the second side area is 2 π * 12 / 5 * 4 / 2 = 48 π / 5,
Total lateral area = 36 π / 5 + 48 π / 5 = 84 π / 5

The length of one right side of a right triangle is 11, and the length of the other two sides is also a positive integer. What is its perimeter?
↖ (^ω^) ↗

Let the hypotenuse be x and the other right angle be y
According to Pythagorean theorem: x ^ 2 = y ^ 2 + 121
x^2-y^2=121
(x+y)(x-y)=121
Because 121 is prime, only 121 * 1 = 121
So x + y = 121
x-y=1
Solving the equations, x = 61, y = 60
The perimeter of the triangle is: 11 + 60 + 61 = 132

If a right triangle has a right side with a length of 11 and the other two sides are also positive integers, the perimeter of the triangle is ()
A. 120B. 121C. 132D. 123

Let another right edge be x and the hypotenuse be y. according to the Pythagorean theorem, it is obtained that y2-x2 = 112, ∵ x (y + x) (Y-X) = 121 = 11 × 11 = 121 × 1, ∵ X and y are integers, y ∵ x, ∵ x + y ∵ Y-X, that is, only x + y = 121, Y-X = 1. The solution is: x = 60, y = 61, ∵ the perimeter of triangle is 11 + 60 + 61 = 132, so C

12. If the length of one right side of a right triangle is 11 and the length of the other two sides are also positive integers, the perimeter of the right triangle is 11--

The length of three sides is 61,60,11
The following attached VB code, right, the second right angle side to 10000
Private Sub Command1_ Click()
For a = 1 To 10000
c = Sqr(121 + a ^ 2)
If c = Int(c) Then
Print a
End If
Next a

As shown in the figure, there is a right triangle paper with two right angles AC = 6cm, BC = 8cm and ab = 10cm. Now fold the edge AC along the straight line ad so that it falls on the edge AB and coincides with Ae. What is the length of CD?

According to the title, if ad is the angle cab bisector, then the triangle ACB is equal to the triangle AED
If CD is x, then De is X
Meanwhile, the triangle bed is similar to the triangle BCA
Then BD: Ba = de: CA
That is (8-x): 10 = x: 6
X=3