The perimeter of a triangle is 60 cm. it's height is 17.3. what is its area?

1 Answer
Jun 27, 2016

Answer:

#0.0173205#[#"m"^2#]

Explanation:

Adopting side #a# as the triangle base, the upper vertice describes the ellipse

#(x/r_x)^2+(y/r_y)^2=1#

where

#r_x = (a+b+c)/2# and #r_y = sqrt(((b+c)/2)^2-(a/2)^2)#

when #y_v = h_0# then #x_v = (sqrt[a^2 - (b + c)^2 + 4 h_0^2] p_0)/(2 sqrt[a^2 - (b + c)^2])#. Here #p_v={x_v,y_v}# are the upper vertice coordinates #p_0=a+b+c# and #p=p_0/2#.

The ellipse focuses location are:

#f_1 = {-a/2,0}# and #f_2 = {a/2,0}#

Now we have the relationships:

1) #p (p-a) (p-b) (p-c) = (a^2 h_0^2)/4# Henon´s formula

2) From #a + norm(p_v-f_1)+norm(p_v-f_2) = p_0# we have

#a + sqrt[h_0^2 + 1/4 (a - (sqrt[a^2 - (b + c)^2 + 4 h_0^2] p_0)/sqrt[ a^2 - (b + c)^2])^2] + sqrt[ h_0^2 + 1/4 (a + (sqrt[a^2 - (b + c)^2 + 4 h_0^2] p_0)/sqrt[ a^2 - (b + c)^2])^2] = p_0#

3) #a+b+c=p_0#

Solving 1,2,3 for #a,b,c# gives

#( a = ( p_0^2-4 h_0^2)/(2 p_0), b= (4 h_0^2 + p_0^2)/(4 p_0), c= (4 h_0^2 + p_0^2)/(4 p_0) )#

and substituting #h_0=0.173, p_0=0.60#

#{a = 0.200237, b = 0.199882, c = 0.199882}#

with an area of #0.0173205#