An equilateral triangle ABC has AB=20√3. P is a point
placed in triangle ABC and D,E and F are the foot of the
perpendiculars from P to AB, BC and AC, respectively. If
PD=9 and PE=10, what is the value of the length of PF? Please work the complete solution.
To find the value of PF, we can use the property of perpendiculars in an equilateral triangle.
First, let's draw the triangle ABC and label the given information:
1. AB = 20√3
2. PD = 9
3. PE = 10
We need to find the length of PF.
Since ABC is an equilateral triangle, all the sides are equal in length. Therefore, AC = BC = AB = 20√3.
Using the property of perpendiculars in an equilateral triangle, we know that the perpendiculars from any point inside the triangle to the sides will divide those sides in the ratio 1:√3:2.
Let's denote the length of PF as x. From the given information, PD = 9 and PE = 10, we can set up the following ratios:
1. PD:PE:PF = 1:√3:2
2. 9:10:x = 1:√3:2
To solve for x, we can set up a proportion:
9/1 = 10/√3 = x/2
First, solve for the value of x/2:
10/√3 = x/2
Cross-multiplying, we get:
√3 * x = 20
Simplifying, we find:
x = 20/√3
To rationalize the denominator, multiply both the numerator and denominator by √3:
x = (20/√3) * (√3/√3)
This gives:
x = (20√3)/3
Therefore, the value of the length of PF is (20√3)/3.