Team Geography Page 1

15

Analytical Geometry

Page 21. Example 1)
horizontal line Right triangle abc with line from point b to the hypotenuse
Let b = The base of the triangle
Let d = the difference
Let x = the hypotinuse
Let x-d = the perpendicular

We know b = [square root]x²-(x-d)²; or b[/square root] = [square root]x²-(x²-2dx+d²)[/square root]; or [squaring]
both numbers b²=x²-x²+2dx-d²; or 2dx=b²+d²; or x = (b²+d²)/(2d)
horizontal line

(Exm. 2)
Rectangle defh with area x is inscribed in triangle abc, which has a height of h.

Let ab' = Area of the triangle Rectangle
Let x = Less side of the rectangle
[therefore] ab = Greater side of the rectangle
b = The base of the triangle
h = altitude

Then we have b:h::(ab')/x:h-x; hence bh-bx = (ahb')/x; or
bhs-bx² = ahb; or bx²-bhx = -ahb'. Solving we have
bx²-bhx+((^bh)/2)²; or x²-hx+(h/2)² = -ahb'/b + (h/2)²
x = h/2 ± [square root](h/2)² - ab'h/b[/square root]