Question 30440
Draw an isosceles triangle and drop a perpendicular from the apex to the mid-point of the base, b. You now have 2 right-angled triangles.


Let the height = h
Let both the angled sides of the isosceles triangle be x, so that the perimeter is x+x+b --> 2x+b=8 which means that x=(8-b)/2.


What we need to do is rewrite the x in terms of b, to satisfy the question.


We have 3 things we can use:
1. Pythagoras' Theorem for one of the right-angled triangles
2. Area of a triangle is (1/2)*base*height
3. the relationship {{{ x=(8-b)/2 }}}
 

1. In either of the 2 rightangled triangles, we have {{{ x^2 = (b/2)^2 + h^2 }}}.


This means that {{{ x^2 = b^2/4 + h^2 }}}
{{{ h^2 = x^2 - b^2/4 }}} -- eqn1


Now, from {{{ x=(8-b)/2 }}}, we have {{{ x = 4 - (b/2) }}}. So {{{ x^2 = (4 - (b/2))^2 }}}. Therefore {{{ x^2 = 16 - 4b + b^2/4 }}}


Right, we now have an equation relating x to b. If we substitute this into eqn1, we get:


{{{ h^2 = x^2 - b^2/4 }}}
{{{ h^2 = (16 - 4b + b^2/4) - b^2/4 }}}
{{{ h^2 = 16 - 4b + b^2/4 - b^2/4 }}}
{{{ h^2 = 16 - 4b }}}
so that {{{ h = sqrt(16 - 4b) }}}


Right, we now have height in terms of just b.


So, from Area of a triangle is (1/2)*base*height, we get


{{{ Area = (1/2)*b*(sqrt(16 - 4b)) }}}
{{{ Area = (b/2)(sqrt(16 - 4b)) }}}
{{{ Area = sqrt((b/2)^2(16 - 4b)) }}} --> this is the useful line of maths to understand. From here, we get:


{{{ Area = sqrt((b^2/4)(16 - 4b)) }}}
{{{ Area = sqrt( ((16b^2)/4) - ((4b^3)/4) ) }}}
{{{ Area = sqrt( 4b^2 - b^3 ) }}}
{{{ Area = sqrt( b^2(4 - b) ) }}}
{{{ Area = b*sqrt(4 - b) }}}


I think this is a correct function for the area of the isosceles triangle when the perimeter is 8cm.


Jon.