Question 1207840
The vertices of triangle ABC are A(1, 1), B(9, 3), and C(3, 5).
(a) Find the perimeter of triangle ABC.

l(AB)+l(BC)+l(AC) = perimeter
Use distance formula 
{{{l(AB) = sqrt((1-9)^2+(1-3)^2)}}}
l(AB) = 2sqrt(17)

{{{l(BC) =sqrt((9-3)^2+(3-5)^2)}}}
l(AB) = 2sqrt(10)

{{{l(C,A) =sqrt((3-1)^2+(5-1)^2)}}}
l(C,A) =2sqrt(5)

Perimeter of   ABC: = {{{2sqrt(17) + 2sqrt(10) + 2sqrt(5)}}}

(b) Find the perimeter of the triangle that is formed by joining the midpoints of the three sides of triangle ABC.

Use mid point formula let M(AB),M(BC),M(AC) be the three points


mid point (AB)= (1+9)/2 = 5  , (3+1)/2 = 2  => (5,2)

mid point (BC) = (9+3)/2)= 6 and (3+5)/ 2=4  => (6,4)

mid point (AC)= (1+3)/2 =2 and (1+5)/2 = 3 => (2,3)

We know the co ordinates of mid points of the triangle which forms a triangle when joined.

use the same formula as above to find the the three lengths joining the mid points

I used calculator to find the three lengths  (M stands for mid point.)

M(AB) +M(BC)+M(AC) = {{{sqrt(5)+sqrt(17)+sqrt(10)}}}

(c) Compute the ratio of the perimeter in part (a) to the perimeter in part (b)./

 {{{2sqrt(17) + 2sqrt(10) + 2sqrt(5)}}}/ {{{sqrt(5)+sqrt(17)+sqrt(10)}}} =  2

(d) What theorem from geometry provides the answer for part (c) without using the results in (a) and (b)?

The midpoint theorem states that if a segment is formed by connecting the midpoints of two of the sides of a triangle, then that segment must be parallel to the third side. the segment is half the length of the third side.