Question 995807
An isosceles triangle has base 7 and height 14. Find the maximum possible area of a rectangle that can be placed inside the triangle with one side on the base of the triangle. (Hint: Similar triangles may be of assistance.) 

Area[max] = 


I had help on solving this but was confused by the answer. I know that an isosceles triangle is one with two equal sides. What I am confused on how this answer was derived. What exactly is happening here? how are "compare ratios" being implemented et cetera

{{{ (x/2)/ (7/2) }}} = {{{ (14-y)/ 14 }}} 
{{{ y = 14-2x }}}
{{{ AREA = xy = 14x-2x^2 }}} 

{{{ For maxm area }}}  {{{ dA/dx=0 }}}

{{{ 14-4x=0 }}} 
{{{ x=3.5 }}}
{{{ y= 7 }}} 

{{{ MaxArea = 7 x 3.5 }}}

{{{ MaxArea = 24.5 square units }}}


Please could someone explain what is happening in each step here. I have no idea how this answer was formed.

Sketch a rectangle INSIDE an isoscles triangle.
Let the rectangle width = x and the rectangle height be y
Sketch the perpendicular hisector of the isosceles
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Notice the upper triangle inside the original triangle.
Notice that the upper triangle has two identical halfs.
Notice that one of those halfs is proportional to half of the isosceles triangle.
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Proportion:
base/base = height/height
(x/2)/(7/2) = (14-y)/y
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Your tutor worked with this proportion to find
and x and y that would give the dimensions of
the rectangle with the maximum area.
Cheers,
Stan H.
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