Question

5. Two vertices of a triangle lie at (4, 0) and (8, 0). The perimeter of the triangle is 12 units. What are all the possible locations of the third vertex? How do you know you have found them all? Can you determine which of these vertices will produce a triangle with the largest area?

Answer 1

Answer: x=8 and y=±3

Step-by-step explanation:

Perimeter= Sum of length of sides

Let A= (4,0)

B= (8,0)

C= (x,y)

So

AB+BC+CA = 12

$\sqrt{(8-4)^{2} -(0-0)^{2} } + \sqrt{(x-8)^{2}+(y-0)^{2} } + \sqrt{(4-x)^2+(0-y)^{2} }$= 12

Solving it, the possible outcomes comes out to be

x= 4, y=±3

x=8, y=±3

x=2, y=0

x=10, y=0

For the largest area, we would use

x=8 and y=±3