When lines y=square root by 3x and y=-square root by 3x +6 intersect at point A, and line y=- square root by 3x +6 and the x axis intersect at point B.

Question

Find the area S of triangle ABO

Answer 1

First, we need to find the coordinates of points A and B.

To find point A, we need to solve for the intersection of y = sqrt(3x) and y = -sqrt(3x) + 6.

sqrt(3x) = -sqrt(3x) + 6

2sqrt(3x) = 6

sqrt(3x) = 3

3x = 9

x = 3

Plugging x = 3 back into y = sqrt(3x):

y = sqrt(9) = 3

Therefore, A = (3, 3).

To find point B, we solve for the intersection of y = -sqrt(3x) + 6 and the x-axis (y = 0).

0 = -sqrt(3x) + 6

sqrt(3x) = 6

3x = 36

x = 12

Therefore, B = (12, 0).

Now, we can calculate the length of sides AB, AO, and BO:

AB = sqrt((12-3)^2 + (0-3)^2) = sqrt(81 + 9) = sqrt(90) = 3sqrt(10)

AO = sqrt(3^2 + 3^2) = 3sqrt(2)

BO = 12

Now we can use Heron's formula to find the area of triangle ABO:

S = sqrt(s(s-AB)(s-AO)(s-BO))

where s = (AB + AO + BO)/2

Plugging in the values:

s = (3sqrt(10) + 3sqrt(2) + 12)/2 = (3sqrt(10) + 3sqrt(2) + 24)/2 = 3sqrt(10) + 3sqrt(2) + 12

S = sqrt((3sqrt(10) + 3sqrt(2) + 12)(3sqrt(10) + 3sqrt(2) + 12 - 3sqrt(10))(3sqrt(10) + 3sqrt(2) + 12 - 3sqrt(2))(3sqrt(10) + 3sqrt(2) + 12 - 12))

S = sqrt((3sqrt(10) + 3sqrt(2) + 12)(3sqrt(10))(3sqrt(2))(12))

S = sqrt((108 + 36sqrt(20) + 36sqrt(6) + 36)(72sqrt(3)))

S = sqrt(4500sqrt(3))

So, the area of triangle ABO is 30sqrt(3) square units.