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Question Number 45630 by ajfour last updated on 14/Oct/18

	Answered by tanmay.chaudhury50@gmail.com last updated on 15/Oct/18

	$\frac{1}{2} \times l \times x = \sqrt{s (s - a) (s - b) (s - c)}$ $s = \frac{R + x + x + r + l}{2} = x + \frac{R + r + l}{2}$ $a = R + x b = x + r c = l$ $\frac{1}{2} l x = \sqrt{(x + \frac{R + r + l}{2}) (\frac{r + l + R}{2} - R) (\frac{R + l + r}{2} - r) (x + \frac{R + r + l}{2} - l)}$ $\frac{l^{2} x^{2}}{4} = (x + M) (N) (P) (x + Q) ↑ h e r e i f o r g o t t o p u t x$ $l^{2} x^{2} = 4 (x + M) N P (x + Q) M = \frac{R + r + l}{2} N = (\frac{R + r + l}{2} - R)$ $P = (\frac{R + r + l}{2} - r) Q = \frac{R + r + l}{2} - l$ $\frac{x^{2}}{(x + M) (x + Q)} = \frac{4 N P}{l^{2}}$ $n w p u t M, N, P, Q v a l u e$ $\frac{x^{2}}{(x + M) (x + Q)} = T$ $x^{2} = T (x^{2} + Q x + M x + M Q)$ $x^{2} (T - 1) + x (Q + M) + M Q T = 0$ $x = \frac{- (Q + M) + \sqrt{{(Q + M)}^{2} - 4 (T - 1) (M Q T)}}{2 (T - 1)}$
	Commented by ajfour last updated on 15/Oct/18

	$t h a n k s, T a n m a y S i r .$
	Answered by MrW3 last updated on 15/Oct/18
	$(√((x+r)^2 −x^2 ))+(√((R+x)^2 −x^2 ))=l (√(r(2x+r)))+(√(R(2x+R)))=l r(2x+r)+R(2x+R)+2(√(rR(2x+r)(2x+R)))=l^2 2(√(rR(2x+r)(2x+R)))=(l^2 −R^2 −r^2 )−2(R+r)x let 2e^2 =l^2 −R^2 −r^2 or e=(√((l^2 −R^2 −r^2 )/2)) ⇒(√(rR(2x+r)(2x+R)))=e^2 −(R+r)x rR(2x+r)(2x+R)=e^4 +(R+r)^2 x^2 −2e^2 (R+r)x rR{4x^2 +2(R+r)x+Rr}=e^4 +(R+r)^2 x^2 −2e^2 (R+r)x 4Rrx^2 +2Rr(R+r)x+R^2 r^2 =e^4 +(R+r)^2 x^2 −2e^2 (R+r)x {(R+r)^2 −4Rr}x^2 −2(R+r)(e^2 +Rr)x−(R^2 r^2 −e^4 )=0 (R−r)^2 x^2 −2(R+r)(e^2 +Rr)x−(R^2 r^2 −e^4 )=0 for R=r: ⇒x=((e^4 −R^4 )/(4R(e^2 +R^2 )))=((e^2 −R^2 )/(4R))=((l^2 −4R^2 )/(8R)) for R≠r: ⇒x=((2(R+r)(e^2 +Rr)−2(√((R+r)^2 (e^2 +Rr)^2 +(R−r)^2 (R^2 r^2 −e^4 ))))/((R−r)^2 )) ⇒x=(((R+r)(2e^2 +2Rr)−2(√(Rr(2e^2 +2Rr)(R^2 +r^2 +2e^2 ))))/(2(R−r)^2 )) ⇒x=(((R+r)[l^2 −(R−r)^2 ]−2l(√(Rr[l^2 −(R−r)^2 ])))/(2(R−r)^2 ))$
	$\sqrt{{(x + r)}^{2} - x^{2}} + \sqrt{{(R + x)}^{2} - x^{2}} = l$ $\sqrt{r (2 x + r)} + \sqrt{R (2 x + R)} = l$ $r (2 x + r) + R (2 x + R) + 2 \sqrt{r R (2 x + r) (2 x + R)} = l^{2}$ $2 \sqrt{r R (2 x + r) (2 x + R)} = (l^{2} - R^{2} - r^{2}) - 2 (R + r) x$ $l e t 2 e^{2} = l^{2} - R^{2} - r^{2} o r e = \sqrt{\frac{l^{2} - R^{2} - r^{2}}{2}}$ $\Rightarrow \sqrt{r R (2 x + r) (2 x + R)} = e^{2} - (R + r) x$ $r R (2 x + r) (2 x + R) = e^{4} + {(R + r)}^{2} x^{2} - 2 e^{2} (R + r) x$ $r R {4 x^{2} + 2 (R + r) x + R r} = e^{4} + {(R + r)}^{2} x^{2} - 2 e^{2} (R + r) x$ $4 {R r x}^{2} + 2 R r (R + r) x + R^{2} r^{2} = e^{4} + {(R + r)}^{2} x^{2} - 2 e^{2} (R + r) x$ ${{(R + r)}^{2} - 4 R r} x^{2} - 2 (R + r) (e^{2} + R r) x - (R^{2} r^{2} - e^{4}) = 0$ ${(R - r)}^{2} x^{2} - 2 (R + r) (e^{2} + R r) x - (R^{2} r^{2} - e^{4}) = 0$ $f o r R = r :$ $\Rightarrow x = \frac{e^{4} - R^{4}}{4 R (e^{2} + R^{2})} = \frac{e^{2} - R^{2}}{4 R} = \frac{l^{2} - 4 R^{2}}{8 R}$ $f o r R \neq r :$ $\Rightarrow x = \frac{2 (R + r) (e^{2} + R r) - 2 \sqrt{{(R + r)}^{2} {(e^{2} + R r)}^{2} + {(R - r)}^{2} (R^{2} r^{2} - e^{4})}}{{(R - r)}^{2}}$ $\Rightarrow x = \frac{(R + r) (2 e^{2} + 2 R r) - 2 \sqrt{R r (2 e^{2} + 2 R r) (R^{2} + r^{2} + 2 e^{2})}}{2 {(R - r)}^{2}}$ $\Rightarrow x = \frac{(R + r) [l^{2} - {(R - r)}^{2}] - 2 l \sqrt{R r [l^{2} - {(R - r)}^{2}]}}{2 {(R - r)}^{2}}$
	Commented by ajfour last updated on 15/Oct/18

	$t h a n k y o u S i r!$
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