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Coordinate GeometryQuestion and Answers: Page 7

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Find the area of the triangle in the first quadrant between the x and y axis and the tangent to the relationship curve y=(5/x)−(x/5) and x dont equal 0 at (0,5)

$Find the area of the triangle in the$ $first quadrant between the x and y$ $axis and the tangent to the$ $relationship curve y = \frac{5}{x} - \frac{x}{5} and x dont$ $equal 0 at (0, 5)$

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Donnes: AD=1; DB=6; ∡BCD=45° ; ∡BAC=90° Determiner 1) AC ? 2) AE? −−−−−−−−−−−− Solution △BAC ∡BAC=90° BC^2 =AB^2 +AC^2 CD (cote commun aux △BAC et BDC) DB^2 =CD^2 +BC^2 −2BC×CDcos 45° (1) △CAD CD^2 =AD^2 +AC^2 (1) DB^2 =(AD^2 +AC^2 )+(AB^2 +AC^2 )−2(√((AB^2 +AC^2 )(AD^2 +AC^2 ))) cos 45° DB^2 =(AD^2 +AB^2 +2AC^2 −(√(2(AB^2 +AC^2 )(AD^2 +AC^2 ))) 2(AB^2 +AC^2 )(AD^2 +AC^2 )=(AD^2 +AB^2 +2AC^2 −DB^2 )^2 posons: AC=x 2(49+x^2 )(1+x^2 )=(1+49+2x^2 −36)^2 (x^4 +50x^2 +49)=2(x^4 +14x^2 +49) x^4 −22x^2 +49=0 x^2 =11±6(√2) x=(√(11+6(√2) )) AC =4,4142135 2)AB et AC coupent le cercle en (D,B) et( E,C) Nous avons AD×AB=AE×AC AE=((AD×AB)/(AC))=(7/( (√(11+6(√2)))))=((7(√(11+6(√2))))/(11+6(√2))) AE=1,585786

$D o n n e s : AD = 1; DB = 6; ∡ B C D = 45 °; ∡ B A C = 90 °$ $D e t e r m i n e r$ $1) A C ?$ $2) A E ?$ $- - - - - - - - - - - -$ $S o l u t i o n$ $△ B A C ∡ B A C = 90 °$ ${B C}^{2} = {A B}^{2} + {A C}^{2}$ $C D (c o t e c o m m u n a u x △ B A C e t B D C)$ ${D B}^{2} = {C D}^{2} + {B C}^{2} - 2 B C \times C D \cos 45 ° (1)$ $△ C A D {C D}^{2} = {A D}^{2} + {A C}^{2}$ $(1) {D B}^{2} = ({A D}^{2} + {A C}^{2}) + ({A B}^{2} + {A C}^{2}) - 2 \sqrt{({A B}^{2} + {A C}^{2}) ({A D}^{2} + {A C}^{2})} \cos 45 °$ ${D B}^{2} = ({A D}^{2} + {A B}^{2} + 2 {A C}^{2} - \sqrt{2 ({A B}^{2} + {A C}^{2}) ({A D}^{2} + {A C}^{2})}$ $2 ({A B}^{2} + {A C}^{2}) ({A D}^{2} + {A C}^{2}) = {({A D}^{2} + {A B}^{2} + 2 {A C}^{2} - {D B}^{2})}^{2}$ $p o s o n s : A C = x$ $2 (49 + x^{2}) (1 + x^{2}) = {(1 + 49 + 2 x^{2} - 36)}^{2}$ $(x^{4} + 50 x^{2} + 49) = 2 (x^{4} + 14 x^{2} + 49)$ $x^{4} - 22 x^{2} + 49 = 0$ $x^{2} = 11 \pm 6 \sqrt{2} x = \sqrt{11 + 6 \sqrt{2}}$ $A C = 4, 4142135$ $2) A B e t A C c o u p e n t l e c e r c l e e n (D, B) e t (E, C)$ $N o u s a v o n s A D \times A B = A E \times A C$ $A E = \frac{A D \times A B}{A C} = \frac{7}{\sqrt{11 + 6 \sqrt{2}}} = \frac{7 \sqrt{11 + 6 \sqrt{2}}}{11 + 6 \sqrt{2}}$ $A E = 1, 585786$

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The equations of the sides AC, BC and AB of a right−angled triangled with lengths a, b and c are y = −7, x=11 and 4x−3y−5=0 respectively. Find the equation of the inscribed circle of the triangle, if its radius r, is given by r = ((a+b−c)/2).

$The equations of the sides AC, BC$ $and AB of a right - angled triangled$ $with lengths a, b and c are y = - 7,$ $x = 11 and 4 x - 3 y - 5 = 0 respectively .$ $Find the equation of the inscribed$ $circle of the triangle, if its radius r,$ $is given by r = \frac{a + b - c}{2} .$