In the RT triangle ABC, the right angle side AB, the oblique side C, and a, B, C form the arithmetic sequence, area = 12, find the perimeter?

In the RT triangle ABC, the right angle side AB, the oblique side C, and a, B, C form the arithmetic sequence, area = 12, find the perimeter?

Let a = B-D, C = B + D
1 / 2Ab = 12, ab = 24, then: B (B-D) = 24,
From a ^ 2 + B ^ 2 = C ^ 2, we get: (B-D) ^ 2 + B ^ 2 = (B + D) ^ 2, B = 4D
The root number is 2
Perimeter = a + B + C = B-D + B + B + D = 3B = 12 root sign 2

Given that the three sides of the right triangle ABC are isometric series, and the area of the triangle is 6, calculate the circumference of the triangle

Since you mentioned the arithmetic sequence, I think you should belong to a senior high school student. I will answer this question directly. If you can't, continue to ask. The three sides of the triangle are 3, 4, 5, and the circumference is 12

If C is the three sides of the right triangle ABC, a, B, C form an arithmetic sequence, then a: B: C = several,

3:4:5
Right triangle by Pythagorean Theorem A ^ 2 + B ^ 2 = C ^ 2
So C - B = B - A - > b = (c + a) / 2 generation is obtained as the first formula
a :c = 3 :5
Let a = 3x, C = 5x, and then let Pythagorean theorem B = 4x
So a: B: C = 3:4:5

Given that the circumference of the right triangle ABC is 4 + 4, the root sign 3, and the median line of the hypotenuse is 2, what is the area of the triangle ABC? Why can't the equation be solved by setting a right angle side as X and another right angle side as 4 root sign 3-x?

Bevel = 4
The sum of two right angles is 4 √ 3
If two right angles are a and B respectively, then:
a²+b²=16
a+b=4√3
a²+b²+2ab=48
16+2ab=48
2ab=32
ab=16
The area of triangle ABC is: AB / 2 = 8

Right triangle ABC, the midpoint to the hypotenuse is equal to 2, the perimeter is equal to 4 + 4, and the root sign 3 is used to calculate the area of the triangle

Δ ABC area = 8
Let two right angles be x, y
From the midpoint to the hypotenuse equals 2, we know that the hypotenuse = 4
Then, from the circumference, we can get:
x+y+4=4+4√3…… (1)
And x 2 + y 2 = 16 II.
Simultaneous equations
According to ①: (x + y) 2 = 16 * 3
The formula of expansion minus ② is: 2XY = 16 * 2
The area of the triangle is equal to 1 / 2 (x * y) = 8
Note: √ is root sign

Given that the circumference of a right triangle is a fixed value L, find its maximum area

Let the side lengths of the triangle be a, B and C respectively, where C is the hypotenuse
Known: a + B + C = L
Pythagorean Theorem A ^ 2 + B ^ 2 = C ^ 2
The area s = 1 / 2 * a * B. in order to maximize s, a × B must be maximized
Because a ^ 2 + B ^ 2 > = 2Ab, when a = B (indicating isosceles right triangle), AB takes the maximum value, that is, ab = (a ^ 2 + B ^ 2) / 2 = C ^ 2 / 2
So Max s = C ^ 2 / 4
Because a + B + C = L, a = B, C = root 2 × a, C = 2 ^ 0.5 * L / (2 + 2 ^ 0.5)
Then the maximum value of S is L ^ 2 / (12 + 8 * 2 ^ 0.5)
Note: C ^ 2 is the second power of C, and 2 ^ 0.5 is the 0.5 power of 2, which means the root sign 2

Given that the circumference of a right triangle is 4, find the maximum area of the right triangle and the length of each side

Let the two right sides of a right triangle be a and B, and the hypotenuse is C. using the Pythagorean theorem, we can get that: a 2 + B 2 = C 2 A + B + C = 4. From a + B + C = 4, 4-C = a + B, and both sides are squared at the same time, the result is: 16-8c + C? = a? + B? + 2Ab 16-8c + C? = C? + 2Ab 16-8c = 2Ab ≤ A & S

The maximum area of a right triangle with a perimeter L (constant) is (using inequality,

If the length of three sides is a, B, C (bevel), then a ^ 2 + B ^ 2 = C ^ 2
Because a + B + C = L = a + B + (a ^ 2 + B ^ 2) ^ (1 / 2) ≥ 2 (AB) ^ (1 / 2) + (2Ab) ^ (1 / 2) = (2 + √ 2) (AB) ^ (1 / 2)
So ab ≤ [(3-2 √ 2) / 2] * L ^ 2
The maximum value of S = AB / 2 is [(3-2 √ 2) / 4] * L ^ 2

The triangle ABC is a right triangle, ABC = 90, ab = 10 cm, AC = 8 cm, BC = 6 cm. Take AC, BC, AB as the diameter of semicircle, calculate the area of shadow part The blank part is a semicircle with a base of 10 cm

By condition:
Shadow area = small semicircle area + middle semicircle area + triangle area - large semicircle area
=1/2×（6/2）²π+1/2×（8/2）²π+1/2×6×8-1/2×（10/2）²π
=9π/2+4π+24-25π/2
=24.

As shown in the figure, in the triangle ABC, ad is perpendicular to BC, CE is perpendicular to AB, ad = 8 cm, CE = 7 cm, AB + BC = 21 cm. What is the area of triangle ABC in square centimeter?

Let AB length be x cm. 7x × 12 = (21-x) × 8 × 12 x = 11.2