Team Geometry

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]

16 Analytical Geometry Page 31

(Exm 3) Circle with radius R is inscribed in triangle abc

Let P = The perimeter of the triangle [Let] R = [The] radius of the inscribed circle [Let] x = [The] altitude of triangle [Let] y = [The] base [of triangle]

Then, since the area of a triangle is equal to its perimeter into half the radius of the inscribed circle, (Book B. Prop IX) we have this equation , or xy = Pa, [never] y = Pr/x

But P = x + y + ; or ; and squaring x² + y² = P² - 2Px + 2xy - 2Py + x² + y²; or 2 Px - 2xy = P² - 2Py; or (2p - 2y)x = P² - 2Py: hence x = . Now substituting Pr/x for y; and we have . [Or] we have 2Px² - 2Prx = P²x - 2P²r or P(2x² - 2rx - Px) = -2Pr) or or or or

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