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15

Analytical Geometry

Page 21. Example 1)

We know b = [square root]x^{2}-(x-d)^{2}; or b[/square root] = [square root]x^{2}-(x^{2}-2dx+d^{2})[/square root]; or [squaring] both numbers b^{2}=x^{2}-x^{2}+2dx-d^{2}; or 2dx=b^{2}+d^{2}; or x = (b^{2}+d^{2})/(2d)

(Exm. 2)

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-bx^{2} = ahb; or bx^{2}-bhx = -ahb'. Solving we have ~~b~~x^{2}-~~b~~hx+((^{b}h)/2)^{2}; or x^{2}-hx+(h/2)^{2} = -ahb'/b + (h/2)^{2} x = h/2 ± [square root](h/2)^{2} - ab'h/b[/square root]

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16 Analytical Geometry Page 31

(Exm 3)

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

But P = x + y +

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Analytical Geometry {(Examh} 4)(Zage 31) {17} [Ger] a+b+c = Perimeter of the Triangle {x}= The radius of the [???] circle {Then (} by Case III [merseieation](zage 276) {we hav}e the [x]y(a+b+c) y/2 = 1

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