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 C

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?

The sum is 4.73.

Step-by-step explanation: Height 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()=

The minimum value of the sum of AP, BP and CP is 4.73.

13

step-by-step explanation:

169-1=168

168-3=165

165-5=160

160-7=153

153-9=144

144-11=133

133-13=120

120-15=105

105-17=88

88-19=69

69-21=48

48-23=25

25-25=0

since it took 13times to completely subtract it hence its square root is 13