It is known that the sum of the lengths of two sides of a triangle is 10, if the cosine of the angle between the two sides is exactly a root of the equation 2x & # 178; - 3x-2 = 0 Find the minimum perimeter of the triangle

It is known that the sum of the lengths of two sides of a triangle is 10, if the cosine of the angle between the two sides is exactly a root of the equation 2x & # 178; - 3x-2 = 0 Find the minimum perimeter of the triangle


From the equation, cos θ = - 1 / 2 or 2 (rounding) ∧ C ^ 2 = a ^ 2 + B ^ 2-2abcos θ = (a + b) ^ 2-ab.c ≥ 5 √ 5. A + B + C ≥ 10 + 5 √ 5



Given that the lengths of the two sides of a triangle are 4 and 5 respectively, and the cosine of their angle is the root of the equation 2x2 + 3x-2 = 0, then the length of the third side is ()
A. 20B. 21C. 22D. 61


Solve the equation 2x2 + 3x-2 = 0 to get x = - 2, or x = 12 ∵ the cosine of the angle θ between the two sides of the triangle is the root of the equation 2x2 + 3x-2 = 0, so cos θ = 12, then the length of the third side is 42 + 52 − 2.4.5.12 = 21, so choose B



If we know that the lengths of the two sides of a triangle are 4 and 5 respectively, and the cosine value of their angle is the root of the equation 2x square + 3x-2 = 0, then the length of the third side is


Just a moment



If the perimeter of the triangle is even, and the two sides are 2 and 7 respectively, the length of the third side should be______ .


According to the trilateral relation of a triangle, it is obtained that the third side should be greater than 7-2 = 5 and less than 7 + 2 = 9, and the perimeter is even. If the known sum of two sides is 9, then the third side should be odd, so the third side should be equal to 7



It is known that the two sides of a triangle are 1 and 2 respectively, and the value of the third side is the root of the square of the equation 2x - 5 + 3 = 0
I'm in a hurry


The perimeter is 4.5
2x²-5+3=0
(x-1)(2x-3)=0
The solution is as follows
X = 1 or x = 3 / 2
Because the value of X is the third side of the triangle,
According to the sum of the two sides of the triangle is greater than the third side,
The difference between the two sides is less than that of the third side
We can see that x = 1 does not satisfy the condition,
We should give up
So x = 3 / 2
So the perimeter of the triangle is 1 + 2 + 3 / 2 = 4.5



Given that a.b.c. is the trilateral length of △ ABC, B.C satisfies (b-2) &# 178; + ic-3i = 0, and a is the solution of the equation ia-4i = 2, find the perimeter of △ ABC
Judging the shape of △ ABC


b=2
c=3
A = 2 or 6
When a = 2
2 + 2 is greater than 3
The triangle ABC is an isosceles triangle with a circumference of 7
When a = 6
If 2 + 3 is less than 6, it does not form a triangle



Let a, B, C be the trilateral length of △ ABC, B, C satisfy the square of | B-2 | + | C-3 | = 0, and a is the solution of the equation | x-4 | = 2, and find the perimeter of △ ABC


|X-4 | = 2X-4 = positive and negative, 2x = positive and negative 2 + 4x1 = 6, X2 = 2 | B-2 | + | C-3 | = 0 (b-2) square + | C-3 | = 0 (b-2) square ≥ 0, | C-3 | ≥ 0, so (b-2) square = 0, | C-3 | = 0, so B-2 = 0, C-3 = 0b = 2, C = 3, and because the sum of two sides is greater than the third side, if a = 6, then B + C = 2 + 3 = 5, a ≥ B + C



It is known that a, B and C are the side lengths of △ ABC, B and C satisfy (b-2) 2 + C − 3 = 0, and a is the solution of the equation | A-4 | = 2. Find the perimeter of △ ABC and judge the shape of △ ABC


∵ (b-2) 2 + | C-3 | = 0, ∵ B-2 = 0, C-3 = 0, ∵ B = 2, C = 3, ∵ A-4 | = 2, ∵ a = 6 or 2. When a = 6, B = 2, C = 3, it can not form a triangle. When a = 2, B = 2, C = 3, it is an isosceles triangle with perimeter of 7



Given that ABC is the side length of triangle ABC, BC satisfies (b-2) ^ 2 + / C-3 / = 0, and a is the solution of equation / x-4 / = 2, find the perimeter of triangle ABC
And judge the shape of the triangle


(b-2)^2+/c-3/=0
b=2
c=3
/x-4/=2
X = 6 (not rounding off)
x=2
Perimeter: 2 + 2 + 3 = 7
The shape of a triangle is an isosceles triangle



The perimeter of an isosceles triangle is 16cm. The middle line of a waist divides the triangle into two triangles with a perimeter difference of 2cm, and calculates the sides of the isosceles triangle


The vertex of the triangle is a, and the two waists are AB and AC respectively, so the bottom edge is BC, and the high Cd on the edge of ab. I won't draw the picture. I'll draw it myself,
Let AB = X,
So AB = AC = 2ad = 2dB
(1) If triangle ADC > triangle BDC, the following results can be obtained:
AD+AC+DC-(DB+BC+DC)=2
That is AD + AC + dc-ad-bc-dc = 2
So ac-bc = 2
That is BC = X-2
So the perimeter of triangle ABC = AB + AC + BC + 16
That is, x + X + X-2 = 16
X=6
The waist length is 6
(2) If triangle BDC > triangle ADC, we can get
DB+BC+DC-(AD+AC+DC)=2
That is DB + BC + dc-ad-ac-dc = 2
That is BC-AC = 2
That is BC = 2 + X
So the perimeter of triangle ABC = AB + AC + BC + 16
That is, x + X + X + 2 = 16
X=14/3