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

Jun 27, 2016

$0.0173205$[${\text{m}}^{2}$]

#### Explanation:

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

${\left(\frac{x}{r} _ x\right)}^{2} + {\left(\frac{y}{r} _ y\right)}^{2} = 1$

where

${r}_{x} = \frac{a + b + c}{2}$ and ${r}_{y} = \sqrt{{\left(\frac{b + c}{2}\right)}^{2} - {\left(\frac{a}{2}\right)}^{2}}$

when ${y}_{v} = {h}_{0}$ then ${x}_{v} = \frac{\sqrt{{a}^{2} - {\left(b + c\right)}^{2} + 4 {h}_{0}^{2}} {p}_{0}}{2 \sqrt{{a}^{2} - {\left(b + c\right)}^{2}}}$. Here ${p}_{v} = \left\{{x}_{v} , {y}_{v}\right\}$ are the upper vertice coordinates ${p}_{0} = a + b + c$ and $p = {p}_{0} / 2$.

The ellipse focuses location are:

${f}_{1} = \left\{- \frac{a}{2} , 0\right\}$ and ${f}_{2} = \left\{\frac{a}{2} , 0\right\}$

Now we have the relationships:

1) $p \left(p - a\right) \left(p - b\right) \left(p - c\right) = \frac{{a}^{2} {h}_{0}^{2}}{4}$ Henon´s formula

2) From $a + \left\lVert {p}_{v} - {f}_{1} \right\rVert + \left\lVert {p}_{v} - {f}_{2} \right\rVert = {p}_{0}$ 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 ${h}_{0} = 0.173 , {p}_{0} = 0.60$

$\left\{a = 0.200237 , b = 0.199882 , c = 0.199882\right\}$

with an area of $0.0173205$