My Math Forum Finding altitude of a part of trapezoid with given area...

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April 30th, 2012, 03:48 AM   #1
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Finding altitude of a part of trapezoid with given area...

Is there a formula for this one?

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 April 30th, 2012, 03:58 AM #2 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 Re: Finding altitude of a part of trapezoid with given area. Extend to right to form right triangle. Then work with similar right triangles.
 April 30th, 2012, 01:01 PM #3 Global Moderator   Joined: May 2007 Posts: 6,805 Thanks: 716 Re: Finding altitude of a part of trapezoid with given area. y = ax + b. When x = 0, y = 10, when x = 10, y = 5. Plug into equation and solve for a and b.
 April 30th, 2012, 07:13 PM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond Re: Finding altitude of a part of trapezoid with given area. Area of a trapezoid $A\,=\,h\,\cdot\,\frac{a\,+\,b}{2}$, where $h$ is height and $a$ and $b$ are the parallel sides. So we've got $23\,=\,x\,\cdot\,\frac{y\,+\,10}{2}$ as the area of the blue part and $52\,=\,(10\,-\,x)\,\cdot\,\frac{y\,+\,5}{2}$ as area of the red part, where $x$ is the width of the blue area and $y$ is the length of the segment between the blue and red areas. From the first equation, $x\,=\,\frac{46}{y\,+\,10}$ Substitute that into the second equation, $52\,=\,$$10\,-\,\frac{46}{y\,+\,10}$$\,\cdot\,\frac{y\,+\,5}{2} \\ \frac{(5y\,+\,27)(y\,+\,5)}{y\,+\,10}\,-\,52\,=\,0 \\ (5y\,+\,27)(y\,+\,5)\,-\,52(y\,+\,10)\,=\,0 \\ 5y^2\,+\,25y\,+\,27y\,+\,135\,-\,52y\,-\,520\,=\,0 \\ 5y^2\,=\,385 \\ y\,=\,\sqrt{77}$
 May 1st, 2012, 03:57 AM #5 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 Re: Finding altitude of a part of trapezoid with given area. Extending to form right triangle: triangle has legs 10 and 20; total area 100; new red area = 100-23 = 77: so y(20 - x) = 154 [1] Similar right triangles means y / (20 - x) = 10/20: so x = 20 - 2y [2] [2] in [1]: y[20 - (20 - 2y)] = 154; solve: y = SQRT(77)

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