It is known that B and C on both sides of △ ABC are the two roots of the equation x2 KX + 40 = 0, the area of which is 3cm2 and the perimeter is 20cm

b. C is the two roots of the equation x & sup2; - KX + 40 = 0,
According to Weida's theorem: B + C = k, BC = 40, then B & sup2; + C & sup2; = (B + C) & sup2; - 2BC = K & sup2; - 80
The circumference is 20cm, the third side a = 20 - (B + C) = 20-k, a & sup2; = (20-k) & sup2;
S△ABC=(1/2)bcsinA=10√3,
So (1 / 2) * 40 * Sina = 10 √ 3, Sina = √ 3 / 2, a = 60 ° or a = 120 °
When a = 60 °, according to the cosine theorem (20-k) & sup2; = K & sup2; - 80-2 * 40 * cos60 °, k = 13
When a = 120 °, according to the cosine theorem (20-k) & sup2; = K & sup2; - 80-2 * 40 * cos120 °, k = 11,
But when k = 11, the discriminant (- K) & sup2; - 4 * 1 * 40 = - 39 < 0 of the equation x & sup2; - KX + 40 = 0 has no solution
So a = 60 ° and K = 13

If the length of two diagonals of a diamond is 10 cm and 8 cm respectively, what is its area?

According to the diamond area formula: S = (E * f) / 2 (E and F in the formula represent the length of two diagonal lines respectively)
Can be obtained: its area = (root 10 * root 8) / 2
=(4 radical 5) / 2
=2 radical 5 (CM) ^ 2

It is known that the area of the rectangle is 6 root sign 3, the acute angle formed by the intersection of two diagonals is 60 ° and the length of the shorter side of the rectangle is calculated

Let the shorter side length be X
Because the acute angle formed by the intersection of two diagonals is 60 degrees
So the diagonal is 2x long,
The long side is: √ (4x & # 178; - X & # 178;) &# 178; = √ 3x
So area s = √ 3x & # 178; = 6 √ 3
x =√6
A: the shorter side length is √ 6
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If there is a rectangular piece of paper with root number 3 7 times the length and root number 12 times the width, now cut out a square with width as the side length. Please calculate it
Find the perimeter of the remaining rectangle

√12=2√3
After cutting the square, the length of the remaining rectangle is: 7 √ 3-2 √ 3 = 5 √ 3
The width is 2 √ 3
Therefore, the perimeter is: 2 (5 √ 3 + 2 √ 3) = 2 × 7 √ 3 = 14 √ 3
In fact, the length of the remaining rectangle is the length width of the original rectangle, and the width of the remaining rectangle remains unchanged
So the length + width of the remaining rectangle = the length of the original rectangle
So we can use 7 √ 3 × 2
As long as the cut is a square, no matter how long the side of the square is, the perimeter of the remaining rectangle is 2 times of 7 √ 3

It is known that B and C on both sides of △ ABC are the two roots of the equation x2 KX + 40 = 0, the area of which is 3cm2 and the perimeter is 20cm