To find the distances OA, OB, and AB, we use the distance formula:
1. Distance OA:
OA = √((3 - 0)^2 + (3√3 - 0)^2) = √(9 + 27) = √36 = 6
2. Distance OB:
OB = √((2√3 - 0)^2 + (0 - 6)^2) = √(12 + 36) = √48 = 4√3
3. Distance AB:
AB = √((3 - 2√3)^2 + (3√3 - 0)^2) = √(9 - 12 + 12 + 27) = √36 = 6
To find the area S of triangle ABO, we can use the formula for the area of a triangle given the lengths of the sides:
S = √(s(s-AB)(s-OA)(s-OB))
where s is the semi-perimeter of the triangle:
s = (OA + OB + AB)/2 = (6 + 4√3 + 6)/2 = 8 + 2√3
Now plug in the values:
S = √((8 + 2√3)(2 + 2√3)(2 - 2√3)(8 - 2√3))
= √((64 - 12)(4 - 12)(16 - 12)(64 - 12))
= √(52 * (-8) * 4 * 52)
= √(108 * 208)
= √(22464)
= 6√394
Therefore, the area of triangle ABO is 6√394.
When lines y=√3x and y= negative√3x +6 intersect at point A, and line y= negative√3x +6 and the x axis intersect at point B, solve the following problems.
Find the distances OA,OB and AB using point A (3, 3√3) and point B (2√3,0)
Find the area S of triangle ABO
1 answer