Given that the circumference of the right triangle ABC is 4 + 4, the root sign 2, and the length of the center line on the hypotenuse is 2, then the area of the triangle is?

Let the length of one right angle side be x and the other 4 √ 2-x
x^2+(4√2-x)^2=4^2
x^2-4√2x+8=0
(x-2√2)^2-=0
X = 2 √ 2, the other is 4 √ 2-2 √ 2 = 2 √ 2
Area of triangle = 1 / 2 * 2 √ 2 * 2 √ 2 = 4

The angle a = 90 ° BC = 2 in the triangle ABC, the perimeter of the triangle ABC is 2 + root 6, and the area of the triangle ABC is calculated

Solution 1 ∵ BC = 2, the circumference of triangle ABC is 2 + √ 6 ᙽ AB + AC = √ 6, and the square of both sides of the triangle AB ^ 2 + AC ^ 2 + 2 ab.ac=6 According to Pythagorean theorem, AB ^ 2 + AC ^ 2 = BC ^ 2 = 44 + 2 ab.ac=6ab . AC = 1 ν triangle area =（ ab.ac ）/2 = 1 / 2 solution 2 ∵ BC = 2, the circumference of the triangle ABC is 2 + √ 6 ᙽ AB + AC = √ 6, and the angle B is set to be

The area of triangle ABC is (16 root sign 3) / 3 BC = 6 angle a = 60 degree perimeter of triangle ABC

Let the corresponding side of AC be B and the corresponding side of AB be C, then CD = (root 3 / 2) * b ad = B / 2 BD = C-B / 2 in the right triangle CBD [(root 3 / 2) * b] square + [C-B / 2] square = 6 in the right triangle CBD, because the triangle area = (1 / 2) * b * c * sin60 = (16

The area of triangle ABC is 16, 3 / 3, BC = 6, and angle a = 60 degrees. Find the circumference of triangle ABC

First use the area formula
S = 1 / 2B * c * Sina (B, c) is an edge
So we can figure out what B * C is
And then use the cosine
cosA=(b^2+c^2-a^2)÷2bc
We can find out what B ^ 2 + C ^ 2 is
Then (B + C) ^ 2 = B ^ + C ^ 2 + 2BC
You can find the value of B + C by opening the root
Add a = 6 to get the perimeter

The area of △ ABC is known to be 16 Three 3, AC = 6, B = 60 °, then the circumference of △ ABC is______ .

According to the formula of triangle area, 1
2acsin60°=16
Three
3，ac=64
Three
According to the cosine theorem, B2 = A2 + c2-2ac · cos60, i.e. 36 = A2 + C2 AC
∴a2+c2=172
(a + C) 2 = 100,
Then a + C = 10
Therefore, C + 6 = 16
So the answer is: 16

The area of △ ABC is 16 / 3, root sign 3, BC = 6, ∠ a = 60 ° and calculate the circumference of △ ABC

The area of △ ABC is 16 / 3, the root number is 3 = 1 / 2 * AB * ac * Sina ν AB * AC = 64 / 3
BC^2=AB^2+AC^2-2AB*AC*CosA ∴AB^2+AC^2-2AB*AC
∴AB^2+AC^2+2AB*AC=100 ∴AB+AC=10
The circumference of △ ABC = AB + AC + BC = 16

As shown in the figure, in the hexagon ABCDEF, AF ∥ CD, ab ∥ De, and ∠ a = 120 ° and ∠ B = 80 °, find the degrees of ∠ C and ∠ D

Connect AC. ∵ AF ∥ CD,  ACD = 180 ° - ∠ CAF, and  ACB = 180 ° - ∠ B - ∠ BAC, ? BCD = ∠ ACD + ∠ ACB = 180 ° - ∠ CAF + 180 ° - ∠ B - ∠ BAC = 360 ° - 120 ° - 80 ° = 160 °. Connect BD. ? ab ∵ De, ∵ BDE = 180 ° - ∠ abd

As shown in the figure, in the hexagon ABCDEF, AF ∥ CD, ab ∥ De, and ∠ a = 120 ° and ∠ B = 80 °, find the degrees of ∠ C and ∠ D

Given that ad is the median line of the triangle ABC, vector ad = λ AB + μ AC (λ, μ∈ R), find the value of λ + μ If ∠ a = 120 °, vector AB * AC = - 2, find the minimum value of ∣ ad ∣

In the triangle ABC, the vector ad = ab-db and ad = AC-DC are obtained by the triangle rule, and 2ad = AB + AC is obtained by adding the two formulas, so the value of λ + μ is 1

Given that the circumference of the right triangle ABC is 4 + 4, the root sign 2, and the length of the center line on the hypotenuse is 2, then the area of the triangle is?