AB and CD bisect each other at E. CE = x^2 - 3, AE = 6x - 10 and CD = 66. Find AB
CE = 33 = x^2 - 3
x^2 = 36
x = 6
...
well, 33 + 3 used to be 69 for unusually high values of 3.
(I was wondering about the irrational value of x that I ended up with)
Go with Steve's solution,
To find the length of AB, we can use the fact that AB and CD bisect each other at point E. This means that CE = DE and AE = BE.
Given that CE = x^2 - 3 and CD = 66, we can set up the following equation since CE = DE:
CE + DE = CD
Substituting the values, we get:
(x^2 - 3) + (x^2 - 3) = 66
Simplifying the equation, we have:
2x^2 - 6 = 66
Adding 6 to both sides of the equation:
2x^2 = 72
Dividing both sides by 2:
x^2 = 36
Taking the square root of both sides:
x = ±6
Now that we have the value of x, we can find the lengths of CE and AE:
CE = x^2 - 3
CE = (6)^2 - 3 = 33
AE = 6x - 10
AE = 6(6) - 10 = 26
Since AE = BE, we have:
BE = 26
And since CE = DE, we have:
DE = 33
To find the length of AB, we can use the Pythagorean theorem, which states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Applying this to triangle ABE, we get:
AB^2 = AE^2 + BE^2
Substituting the values, we have:
AB^2 = 26^2 + 33^2
Calculating the squares:
AB^2 = 676 + 1089
Adding the numbers:
AB^2 = 1765
Taking the square root of both sides:
AB = √1765
So, the length of AB is √1765.
Since E is the midpoint, CE = 33 = x^2 - 3
x^2 = 69
x = √69
so AB = 2AE = 2(6x - 10)
= 12x - 20
= 12√69 - 20