Let a = radical 8-x, B = radical 3x + 4, C = radical x + 2. If a, B, C are the three sides of a right triangle, find the value of X I'm not sure how to determine the maximum side

Let a = radical 8-x, B = radical 3x + 4, C = radical x + 2. If a, B, C are the three sides of a right triangle, find the value of X I'm not sure how to determine the maximum side

It is easy to know that 8-x > 0, and X + 2 > 0, and 3x + 4 > 0. The solution is: - 4 / 3x = - 10. (let) (2) if a + C = b = b] = (8-x) + (8-x + 2) = 3x + 4 (x + 2) = 3x + 4. (3) if B + C = a = a, B + C = a = (3x + 4) + (x + 2) = 8-x. = (x + 2) = 8-x. = > x = 2 / 5. To sum up, x = 2 or x = 2 / 2 / 5, x = 2 or x = 2 / 2 / 5. To sum up, x = 2 or x = 2 / 2 / 5.5

In the right triangle ABC, the angle c is equal to 90 degrees, the opposite sides of angle a, angle B and angle c are a, B, C respectively, and a B = 2, radical 3, C = 3 Find the area of triangle ABC

S=1/2absinC=√3

In the right triangle ABC, the opposite sides of ∠ C = 90 ° and the opposite sides of ∠ a, ∠ B and ∠ C are ABC respectively, and a + B is equal to 2 roots and 3. Calculate the area of ABC

Your topic is not complete. Do you require the extreme value of ABC area
A + B = 2 root sign 3, so (a + b) ^ 2 = 12, that is, a * * 2 + b * * 2 + 2Ab = 12
A * * 2 + b * * 2 + 2Ab > = 4AB, so ab=

The vertex of the parabola y = ax ^ 2 + BX + C is D and intersects with the Y axis at point C. The analytic formula of the straight line CD is y = root 3x + 2 root sign 3C (0, double root sign 3)

Let x = 0,
Substituting the given parabola y = ax 2 + BX + C, the equation is as follows:
y=a×0²+b×0+c
The result is: y = C
That is, the coordinate of point C is (0, c)
By: y = ax 2 + BX + C
It is known that the coordinates of point D are (- B / (2a), (4ac-b 2) / (2a))
By using the two-point formula of the straight line, the analytic formula of the straight line CD is obtained as follows:
(y-c)/(x-0)=[(4ac-b²)/(2a)-c]/[-b/(2a)-0]
The analytical formula of the straight line CD is: y = [(2ac-b 2) / (- b)] x + C
It is known that the analytic formula of the straight line CD is y = (√ 3) x + 2 √ 3
Therefore, there are:
(2ac-b²)/(-b)=√3……………… (1)
c=2√3………………………… (2)
Generation (2) into (1), including:
4a√3=b²-b√3
Two unknowns, one equation, missing conditions,

As shown in the figure, the two different intersections of the parabola y = - 2x ^ 2 + BX and X axis are o and a, and the vertex B is on the line y = root 3x If there is a point P on the parabola, make ∠ OPA = 90 °. If so, ask for the coordinates of point p; if not, please explain the reason

The symmetry axis of the parabola y = - 2x ^ 2 + BX is x = - B / 2A = B / 4,
Substituting the line y = root 3x, then,
P(b/4,√3b/4),
If ∠ OPA = 90 ° satisfies B / 4 = √ 3B / 4,
The solution is b = 0, which is inconsistent, so it does not exist

As shown in the figure, the quadrilateral ABCD is a diamond, the coordinates of point D are (0,2, root 3), and the parabola y = AX2 + BX + C with point C as its vertex just passes through two points a and B on the X axis (1) Find the coordinates of points a, B and C; (2) The analytic formula of parabola passing through a, B and C is obtained; (3) If the parabola is shifted upward along its symmetrical axis and passes through point D, the analytical formula of the parabola after translation is obtained, and how many units are translated? (4) Make am ⊥ CD through two points a and B, respectively. The extension line of BN ⊥ CD crossing CD is at point n. please write down the area enclosed by am, BN and two paraboloids

Because of the diamond ABCD, ad = AB = BC, and because the parabola with point C as the vertex passes through two points a and B on the X axis, AC and BC are symmetrical about CE, AC = AB = BC triangle, ABC is equilateral triangle, angle CBE = 60 degrees, CE = od = radical 3, be = 1, BC = 2, easy to get OA = EB = AE = 1, so ob = 3A (1,0

The vertex of the parabola is C (2, 3) It intersects with X axis at a and B, and their abscissa are two of the equation x2-4x + 3 = 0, then s △ ABC=______ .

∵ from the equation x2-4x + 3 = 0, X1 = 1, X2 = 3,
The coordinates of point a are (1, 0), and those of point B are (3, 0),
∴AB=2，
∴S△ABC=1
2×AB×
3=1
2×2×
3=
3，
That is s △ ABC=
3．
So the answer is:
3．

The parabola y = - 1 / 2x ^ 2 + radical 2 / 2x + 2 and X axis intersect point a, B two points, and Y axis intersect point C. It is proved that △ ABC is a right triangle

x=0,y=2
C（0,2）
Let y = 0
-1/2x²+√2/2x+2=0
x²-√2x-4=0
(x-2√2)(x+√2)=0
X = - √ 2 or 2 √ 2
A（-√2,0）B（2√2,0）
AC slope = (0-2) / (- √ 2-0) = √ 2
BC slope = (0-2) / (2 √ 2-0) = - √ 2 / 2
The product of the two slopes = √ 2 × (- √ 2 / 2) = - 1
So AC vertical BC
ABC is a right triangle

The straight line y = negative third root sign three x plus one and X, Y axes intersect point a and B, take AB as right angle side, make isosceles triangle ABC in the first quadrant, angle ABC = 90 degrees In order to make the area of a triangle equal to the area of a, a, a, a, a, a, a, B, B, a, B, B, B, a, B, a, B, B, a, B, a, B, a, B, a, B, B, a, B, B, B, a, B, B, B, B,

There is a problem with the topic!
If △ BOP is regarded as a triangle with Bo = 1 as the base, then its height is the distance between point P and Y axis. Since the coordinates of point P are (1, a), then the distance from P to y axis is 1. That is to say, no matter how the point P moves along the straight line x = 1, the area of △ BOP can only be 1 × 1 × (1 / 2) = 0.5
Therefore, △ ABC cannot be equal to △ BOP in area

The right angle vertex C of the isosceles triangle ABC is on the Y axis, the hypotenuse AB is on the X axis, the point a is on the left side of the point B, and the right angle side AC = root 2. Try to write out the coordinates of the vertices a, B, C Given the point a (2,0), point B (- half, 0), point C (0,1), draw a parallelogram with three points a, B, C as the vertices, then the fourth vertex cannot be in () A first quadrant B second quadrant C third quadrant D fourth quadrant

1. Let a point coordinate be (- A, 0), (a > 0), B point coordinate as (a, 0), C (0, b) (b > 0),
∵ △ ABC is isosceles RT △,
∴|AC|=|BC|=√2,|AO|=√2/√2=1,
The a coordinate is (- 1,0), B coordinate (1,0), C coordinate (0,1), or C (0, - 1)
2. The fourth vertex cannot be in the third quadrant
Take BC as the diagonal, D in the second quadrant, AC as the diagonal, D in the first quadrant,
Take AB as diagonal, D in the fourth quadrant