My Math Forum Finding Sides and area of trapezoid around two circles.

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 June 25th, 2011, 11:53 PM #11 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 Let $a$ be the length from the vertex of the smaller base to the tangent point of the smaller circle. Draw a line that passes through the centres of two circles. Now shift it $a$ units until the end of the line touches the vertex of smaller base. Use Pythagorean theorem to determine the height $8\sqrt{3}a\text{.}$ Since the line is shifted from where it crosses the two centres, height equals to the sum of two diameters. Hence the height $8\sqrt{3}a\,=\,2(1)\,+\,2(3)\,=\,2(1\,+\,3)\,=\,2\ ,\cdot\,4\,=\,8\text{ i.e. }a\,=\,\frac{\sqrt{3}}{3}\,\text{.}$ Smaller base is $\frac{2\sqrt{3}}{3}\,\text{.}$ Hence to determine the larger base use the fact that larger base:smaller base = 9:1 by using the tangent line of externally tangent circles as mid-base, two trapezoids are similar by 1:3. Smaller base:mid-base = 1:3 and mid-base:larger base = 1:3 = 3:9 therefore smaller base:larger base = 1:9. To determine the larger base, multiply the smaller base by 9 to get $9\,\cdot\,\frac{2\sqrt{3}}{3}\,=\,6\sqrt{3}\text{. }$
 June 26th, 2011, 09:00 AM #12 Newbie   Joined: Jun 2011 Posts: 21 Thanks: 0 Re: Finding Sides and area of trapezoid around two circles. I thought the height was just 8?
 June 26th, 2011, 01:09 PM #13 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 522 Math Focus: Calculus/ODEs Re: Finding Sides and area of trapezoid around two circles. My inclination was to approach this problem with analytic geometry (and like johnny assume an isosceles trapezoid). I oriented the system such that the smallest side of the trapezoid lies on the y-axis and the side opposite lies on the line x = 8. The centers of the circles lie on the x-axis where the smaller circle is then given by (x - 1)² + y² = 1 and the larger circle is given by (x - 5)² + y² = 3². Now, the top side of the trapezoid must be tangent to the two circles. Let $$$x_1,y_1$$$ be the point of tangency with the smaller circle and $$$x_2,y_2$$$ be the point of tangency with the larger circle. We have that the line segments: $(1,0)-$$x_1,y_1$$$ and $(5,0)-$$x_2,y_2$$$ are parallel and both perpendicular to the top side of the trapezoid. Thus, we have: $\frac{x_1-1}{y_1}=\frac{x_2-5}{y_2}$ Square through: $\frac{$$x_1-1$$^2}{y_1^2}=\frac{$$x_2-5$$^2}{y_2^2}$ Recall $$$x_1-1$$^2=1-y_1^2$ and $$$x_2-5$$^2=9-y_2^2$ giving: $\frac{1-y_1^2}{y_1^2}=\frac{9-y_2^2}{y_2^2}$ $\frac{1}{y_1^2}-1=\frac{9}{y_2^2}-1$ Take the positive root: $y_2=3y_1$ Now, substitute for $y_2$: $\frac{x_1-1}{y_1}=\frac{x_2-5}{3y_1}$ $3$$x_1-1$$=x_2-5$ $x_2=3x_1+2$ Thus, the slope of the top side of the trapezoid is: $\frac{y_2-y_1}{x_2-x_1}=\frac{3y_1-y_1}{3x_1+2-x_1}=\frac{2y_1}{2$$x_1+1$$}=\frac{y_1}{x_1+1}$ Recall this slope is perpendicular to the segment $(1,0)-$$x_1,y_1$$$ giving: $\frac{y_1}{1+x_1}=\frac{1-x_1}{y_1}$ $x_1^2+y_1^2=1$ Recall though that: $$$x_1-1$$^2+y_1^2=1$ Equating, we find: $x_1^2+y_1^2=$$x_1-1$$^2+y_1^2$ $x_1^2+y_1^2=x_1^2-2x_1+1+y_1^2$ Combine like terms:  $x_1=\frac{1}{2}$ $y_1=\sqrt{1-x_1^2}=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}$ $x_2=3x_1+2=\frac{7}{2}$ $y_2=3y_1=\frac{3\sqrt{3}}{2}$ Thus the slope m of the top side is: $m=\frac{y_2-y_1}{x_2-x_1}=\frac{\frac{3\sqrt{3}}{2}-\frac{\sqrt{3}}{2}}{\frac{7}{2}-\frac{1}{2}}=\frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}}$ Now, using the point-slope formula, we find the equation of the line representing the top side as: $y-\frac{\sqrt{3}}{2}=\frac{1}{\sqrt{3}}$$x-\frac{1}{2}$$$ Put into slope-intercept form, we have: $y=\frac{1}{\sqrt{3}}x+\frac{\sqrt{3}}{2}-\frac{1}{2\sqrt{3}}=\frac{1}{\sqrt{3}}x+\frac{1}{\ sqrt{3}}$ Thus: The smaller base b is $b=2\cdot y(0)=\frac{2}{\sqrt{3}}$ The larger base B is $B=2\cdot y(=\frac{18}{\sqrt{3}}=6\sqrt{3}" /> The two sides s are  The area A is $A=\frac{1}{2}(B+b)h=\frac{1}{2}$$\frac{20}{\sqrt{3 }}$$8=\frac{80}{\sqrt{3}}$ Note that these results agree completely with the results given by johnny, although I did not bother to rationalize the denominators.
June 26th, 2011, 01:18 PM   #14
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Re: Finding Sides and area of trapezoid around two circles.

Quote:
 Originally Posted by Spaghett I thought the height was just 8?
There are at least two orientations. The surrounding trapezoid might be an isosceles trapezoid with height = 8 or it might be a trapezoid having two consecutive angles that are right angles (as I drew earlier). In this case I think the height is $4+2\sqrt{3} \approx 7.464$. I think these are the two extreme cases and that as you roll the small circle around the large one you can different trapezoids with heights ranging from $4+2\sqrt{3}$ up to 8.

 June 29th, 2011, 11:52 PM #15 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 522 Math Focus: Calculus/ODEs Re: Finding Sides and area of trapezoid around two circles. I thought it might be fun (at least I thought so at first until I spent a great deal of time chasing down silly mistakes! ) to generalize a bit, and let the smaller circle have radius r and the larger circle have radius kr where k ? 1. As before, we will assume an isosceles trapezoid and orient the system such that the smallest side of the trapezoid lies on the y-axis and the side opposite lies on the line x = 2r(k + 1). The centers of the circles lie on the x-axis where the smaller circle is then given by (x - r)² + y² = r² and the larger circle is given by (x - r(2 + k))² + y² = (kr)². Now, the top side of the trapezoid must be tangent to the two circles. Let $$$x_1,y_1$$$ be the point of tangency with the smaller circle and $$$x_2,y_2$$$ be the point of tangency with the larger circle. We have that the line segments: $(r,0)-$$x_1,y_1$$$ and $(r(2+k),0)-$$x_2,y_2$$$ are parallel and both perpendicular to the top side of the trapezoid. Thus, we have: $\frac{x_1-r}{y_1}=\frac{x_2-r(2+k)}{y_2}$ Square through: $\frac{$$x_1-r$$^2}{y_1^2}=\frac{$$x_2-r(2+k)$$^2}{y_2^2}$ Recall $$$x_1-r$$^2=r^2-y_1^2$ and $$$x_2-r(2+k)$$^2=(kr)^2-y_2^2$ giving: $\frac{r^2-y_1^2}{y_1^2}=\frac{(kr)^2-y_2^2}{y_2^2}$ $\frac{r^2}{y_1^2}-1=\frac{(kr)^2}{y_2^2}-1$ Take the positive root: $y_2=ky_1$ Now, substitute for $y_2$: $\frac{x_1-r}{y_1}=\frac{x_2-r(2+k)}{ky_1}$ $k$$x_1-r$$=x_2-r(2+k)$ $x_2=k$$x_1-r$$+r(2+k)=kx_1+2r$ Thus, the slope of the top side of the trapezoid is: $\frac{y_2-y_1}{x_2-x_1}=\frac{ky_1-y_1}{kx_1+2r-x_1}=\frac{y_1(k-1)}{x_1(k-1)+2r}$ Recall this slope is perpendicular to the segment $(r,0)-$$x_1,y_1$$$ giving: $\frac{y_1(k-1)}{x_1(k-1)+2r}=\frac{r-x_1}{y_1}$ Cross-multiplication yields: $y_1^2(k-1)=$$r-x_1$$$$x_1(k-1)+2r$$$ Recall that $y_1^2=r^2-$$x_1-r$$^2=2rx_1-x_1^2$ so we have: $$$2rx_1-x_1^2$$(k-1)=$$r-x_1$$$$x_1(k-1)+2r$$$ Expansion, simplification, and solving for $x_1$ yields: $x_1=\frac{2r}{k+1}$ $y_1=\sqrt{2rx_1-x_1^2}=\frac{2r\sqrt{k}}{k+1}$ $x_2=kx_1+2r=\frac{2r(2k+1)}{k+1}$ $y_2=ky_1=\frac{2k^{\frac{3}{2}}r}{k+1}$ Thus, the slope m of the top side is: $m=\frac{r-x_1}{y_1}=\frac{k-1}{2\sqrt{k}}$ Now, using the point-slope formula, we find the equation of the line representing the top side as: $y-\frac{2r\sqrt{k}}{k+1}=\frac{k-1}{2\sqrt{k}}$$x-\frac{2r}{k+1}$$$ Put into slope-intercept form, we have: $y=\frac{k-1}{2\sqrt{k}}x+\frac{r}{\sqrt{k}}$ Thus: The smaller base b is $b=2\cdot y(0)=\frac{2r}{\sqrt{k}}$ The larger base B is $B=2\cdot y$$2r(k+1)$$=2k^{\frac{3}{2}}r$ The two sides s are  The area A is $A=\frac{1}{2}(B+b)h=\frac{1}{2}$$2rk^{\frac{3}{2}} +\frac{2r}{\sqrt{k}}$$$$2r(k+1)$$=\frac{2r^2$$k^2+ 1$$(k+1)}{\sqrt{k}}$
 June 30th, 2013, 11:09 AM #16 Newbie   Joined: Jun 2013 Posts: 2 Thanks: 0 Re: Finding Sides and area of trapezoid around two circles. Ive seen people say the smaller base is sqrt(3) / 3 by using the pythagorean theorem. Can someone go into a little more depth and explain how to use the pythagorean theorem on this problem when you only know the height, and you are assuming its an isoceles trapezoid? You only have the one side of a right triangle, thats the height of 8, so i'm highly confused how people leap to the square root of three over three. Thanks,
June 30th, 2013, 11:13 AM   #17
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Quote:
 Originally Posted by johnny Let $a$ be the length from the vertex of the smaller base to the tangent point of the smaller circle. Draw a line that passes through the centres of two circles. Now shift it $a$ units until the end of the line touches the vertex of smaller base. Use Pythagorean theorem to determine the height $8\sqrt{3}a\text{.}$ Since the line is shifted from where it crosses the two centres, height equals to the sum of two diameters. Hence the height $8\sqrt{3}a\,=\,2(1)\,+\,2(3)\,=\,2(1\,+\,3)\,=\,2\ ,\cdot\,4\,=\,8\text{ i.e. }a\,=\,\frac{\sqrt{3}}{3}\,\text{.}$ Smaller base is $\frac{2\sqrt{3}}{3}\,\text{.}$ Hence to determine the larger base use the fact that larger base:smaller base = 9:1 by using the tangent line of externally tangent circles as mid-base, two trapezoids are similar by 1:3. Smaller base:mid-base = 1:3 and mid-base:larger base = 1:3 = 3:9 therefore smaller base:larger base = 1:9. To determine the larger base, multiply the smaller base by 9 to get $9\,\cdot\,\frac{2\sqrt{3}}{3}\,=\,6\sqrt{3}\text{. }$
I dont understand your reasoning on how the base equals sqrt{3} / 3. You said "using pythagorean theorem" But to this point you only have one side to work with. where does the squart root of 3 come from?

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