The question is missing parts. Here is the complete question.
The isosceles triangle below has height AQ of length 3 and base BC of length 2. A point P may be placed anywhere along the line segment AQ.
What is the minimum value of the sum of the lengths of AP, BP and CP?
Answer: The sum is 4.73.
Step-by-step explanation: <u>Height</u> of a triangle is a perpendicualr line linking a vertex and its opposite side.
Because triangle ABC is isosceles, point Q divides the base in 2 equal parts:
BQ = CQ = 1
Suppose QP = x
To calculate minimum value of the sum:
AP = AQ - QP
AP = 3 - x
Since triangles BQP and CQP are congruent and right triangles, use Pythagorean Theorem to figure out the value of BP and CP:
BP = CP =
BP =
Then, sum of AP, BP and CP is
The minimum value is calculated using first derivative:
The value of x is limited: it can assume value of 0, when P=A and x=3, when P=Q. So, interval is [0,3].
x has value:
x = ±
x can't assume negative value because is not in the interval:
x =
To find the minimum value of the sum, substitute x in the function above:
f()=
<u>The </u><u>minimum value</u><u> of the sum of AP, BP and CP is </u><u>4.73.</u>