the following measurements are known in the given sketch
- AD = 3x-1
- BD = 7x-6
- AE = 3
- CE = x+2
1. Determine the integral values value(s) of x for which DE||BC ?
To determine the integral values of x for which DE is parallel to BC, we need to use the property that if two lines are parallel, then their corresponding angles are equal.
From the given sketch, we can see that angle DAE is corresponding to angle CEB.
Therefore, we have:
angle DAE = angle CEB
We know that DE is parallel to BC if and only if angle DAE = angle CEB.
Now, we can find the value of angle DAE and angle CEB:
angle DAE = tan^(-1)(AE/AD)
angle CEB = tan^(-1)(CE/BD)
Plugging in the given measurements, we get:
angle DAE = tan^(-1)(3/(3x-1))
angle CEB = tan^(-1)((x+2)/(7x-6))
For DE to be parallel to BC, angle DAE must be equal to angle CEB. Therefore, we have:
tan^(-1)(3/(3x-1)) = tan^(-1)((x+2)/(7x-6))
Taking tan on both sides, we get:
3/(3x-1) = (x+2)/(7x-6)
Cross multiplying, we get:
3(7x-6) = (x+2)(3x-1)
21x - 18 = 3x^2 + 6x - 3
Rearranging and simplifying the equation, we get:
3x^2 + 15x - 15 = 0
Dividing by 3, we get:
x^2 + 5x - 5 = 0
This is a quadratic equation that can be solved using the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / 2a
where a = 1, b = 5, and c = -5
Plugging in the values, we get:
x = (-5 ± sqrt(5^2 - 4*1*(-5))) / 2*1
x = (-5 ± sqrt(25 + 20)) / 2
x = (-5 ± sqrt(45)) / 2
x = (-5 ± 3√5) / 2
Therefore, the integral values of x for which DE is parallel to BC are:
x = (-5 + 3√5) / 2
x = (-5 - 3√5) / 2