If the length of a diagonal line of diamond ABCD is 6 sides and the length of AB is a root of the square of equation x - 7x + 12 = 0, then the perimeter of diamond ABCD is In the two known equations x ^ 2 + ax + (A-1) = 0, the absolute value of negative root is greater than that of positive root, and the value range of a? It is known that the equation x ^ 2-mx + M-1 = 0 has two real roots x1, X2 and 1 / X1 + 1 / x2 = m, then M =? One diagonal of the diamond ABCD is 3

If the length of a diagonal line of diamond ABCD is 6 sides and the length of AB is a root of the square of equation x - 7x + 12 = 0, then the perimeter of diamond ABCD is In the two known equations x ^ 2 + ax + (A-1) = 0, the absolute value of negative root is greater than that of positive root, and the value range of a? It is known that the equation x ^ 2-mx + M-1 = 0 has two real roots x1, X2 and 1 / X1 + 1 / x2 = m, then M =? One diagonal of the diamond ABCD is 3

1、x1+x2

If the length of a diagonal of the diamond is 6 and the length of side AB is a root of the square of equation x - 7x + 12 = 0, what is the area of the diamond ABCD?

If you use your mobile phone, I can't draw a picture for you. You can understand it yourself! X-7x + 12 = (x-3) (x-4) = 0, so AB = 3 or AB = 4 because the diagonal is 6, so half of the diagonal is 3, so AB = 3 doesn't hold. So AB = 4 according to Pythagorean theorem: the other half of the diagonal is the root 7, then the face of the triangle with ab as the hypotenuse

If the length of a diagonal of diamond ABCD is 6 and the length of side AB is a root of the equation x-7x + 12 = 0, then the area of diamond ABCD is 0

The equation is (x-3) (x-4) = 0
We get x = 3, or 4
Because there is a diagonal line of 6, half of it is 3, the length of side and the half diagonal line form a right triangle, and the length of side is hypotenuse, so it needs to be greater than 3
So AB = 4, the other diagonal = 2 √ (4 ^ 2-3 ^ 2) = 2 √ 7
ABCD area = 1 / 2 * 6 * 2 √ 7 = 6 √ 7

The diagonal length of the diamond ABCD is 6, AB is the solution of the equation A & # 178; - 7a + 12 to find the perimeter of the diamond
QAQ

x²－7x＋12＝0
(x-3)(x-4)=0
X = 3 or x = 4
The side length is 3 or 4, and because a diagonal line is 6, the diagonal line and both sides can form a triangle, so 3 is omitted
The side length is 4
Perimeter = 4x4 = 16

Given the rectangle ABCD, ab = 4, BC = 3, then a and B are the focus, and the eccentricity of the ellipse passing through C and D is 0___ ．

∵ AB = 4, BC = 3, a and B are the focus, ∵ C = 2, B2A = 3, ∵ B2 = 3A, ∵ A2-4 = 3A ∵ a = 4, ∵ e = CA = 24 = 12

Given the rectangle ABCD, ab = 4, BC = 3, then a and B are the focus, and the eccentricity of the ellipse passing through C and D is 0___ ．

∵ AB = 4, BC = 3, a and B are the focus, ∵ C = 2, B2A = 3, ∵ B2 = 3A, ∵ A2-4 = 3A ∵ a = 4, ∵ e = CA = 24 = 12

As shown in the figure, in rectangle ABCD, ab = 3, BC = 5. Through the diagonal intersection o, make OE ⊥ AC and intersect ad with E, then the length of AE is ()
A. 1.6B. 2.5C. 3D. 3.4

Let AE = x, then ed = ad-ae = 5-x. in RT △ EDC, according to Pythagorean theorem, we can get EC2 = de2 + DC2, that is, X2 = (5-x) 2 + 32, and the solution is x = 3.4

As shown in the figure, ▱ ABCD, ab = m, the parabola y = AX2 + BX + C with point C as the vertex passes through points a and B on the x-axis, then the coordinates of point B are___ ．

∵ in the parallelogram ABCD, CD ‖ AB and CD = AB = m, the abscissa of point C is m, let the symmetry axis of parabola intersect with X axis at point h, then ah = BH = m2, and the coordinates of point B are B (3M2, 0). So the answer is: (3M2, 0)

In the vertex of square ABCD, point a and B are on the parabola y = x ^ 2, point C and D are on the straight line y = x-4, find the perimeter of the square
In the vertex of square ABCD, point a and B are on the parabola y = x ^ 2, and point C and D are on the straight line y = x-4
Let me understand,

Let AB equation be y = x + B and y = x & sup2
x1=1/2+1/2√1+4b;y1=1/2+1/2√1+4b +b
x1=1/2-1/2√1+4b;y1=1/2-1/2√1+4b +b
So AB = √ 2 + 8b = BC = 1 / √ 2 · (B + 4) gives B1 = 2, B2 = 6
Ab1 = 3 √ 2, perimeter 12 √ 2; AB2 = 5 √ 2; perimeter 20 √ 2

It is known that the four vertices of the quadrilateral ABCD are a (2,3), B (1, - 1), C (- 1, - 2), D (- 2,2). The slope of the straight line of the quadrilateral ABCD is calculated

AB slope is (- 1-3) / (1-2) = 4 BC slope is (- 2 + 1) / (- 1-1) = 1 / 2 CD slope equals AB slope Da slope equals BC slope