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 October 20th, 2011, 07:48 PM #1 Senior Member   Joined: Sep 2011 From: New York, NY Posts: 333 Thanks: 0 Quiz problem that I got killed on Today I had a quiz with the following problem: Start with a rectangle with sides $L_1$ and $L_2$, draw a line of height $h$ straight up from one vertex of the rectangle. Then, from the top of that line draw a line down to the other three vertices of the rectangle. Find the volume of this irregular triangle. Any ideas? I'm trying to make the figure to post.
 October 20th, 2011, 09:30 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,208 Thanks: 516 Math Focus: Calculus/ODEs Re: Quiz problem that I got killed on Since it is mentioned to draw lines "down" to the remaining three vertices of the rectangle ($L_1$ vertical, $L_2$ horizontal), I used the upper left vertex to draw h from. Based on the diagram I sketched, I find the following areas to be equivalent: $L_1L_2+\frac{h}{2}L_2=\frac{L_2}{2}$$2L_1+h$$$ The term $\frac{h}{2}L_2$ represents the triangle that immediately stands out. Simplifying, we find this is an identity. I want to comment that the triangle I am using is a right triangle, as the resulting polygon is a quadrilateral. If there is another triangle you have in mind, I sure your forthcoming image will clear this up. I see the line from the upper vertex to the opposing lower vertex divides this right triangle into a smaller right triangle and a scalene triangle. Perhaps it is the area of this scalene triangle you want to find? If this is the case, then let x and y be the two bases, such that: $x+y=L_2$ By similarity, we have: $\frac{h}{x}=\frac{L_1+h}{L_2}$ From this, we conclude: $y=\frac{L_1L_2}{L_1+h}$ Thus, the area of the scalene triangle is: $\frac{hL_1L_2}{2$$L_1+h$$}$ And for good measure, the area of the smaller right triangle: $\frac{h^2L_2}{2$$L_1+h$$}$
October 21st, 2011, 07:11 AM   #3
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Re: Quiz problem that I got killed on

Quote:
 Originally Posted by aaron-math Today I had a quiz with the following problem: Start with a rectangle with sides $L_1$ and $L_2$, draw a line of height $h$ straight up from one vertex of the rectangle. Then, from the top of that line draw a line down to the other three vertices of the rectangle. Find the volume of this irregular triangle. Any ideas? I'm trying to make the figure to post.
May I ask who gave you this quiz?

 October 22nd, 2011, 02:46 AM #4 Global Moderator   Joined: Dec 2006 Posts: 19,994 Thanks: 1855 I suspect that "tetrahedron" was intended, not "triangle", the line of length h being drawn perpendicular to the rectangle.
 October 25th, 2011, 09:24 PM #5 Senior Member   Joined: Sep 2011 From: New York, NY Posts: 333 Thanks: 0 Re: Quiz problem that I got killed on Sorry, I shouldn't said triangle. This is what the figure would look like: Side view $\hspace{600}$ Front view This is what a friend came up with.
 October 29th, 2011, 07:17 AM #6 Member   Joined: Oct 2011 Posts: 45 Thanks: 0 Re: Quiz problem that I got killed on It looks like the triangles are not the same dimensions this would make the quiz rather difficult yes no maybe?
 October 29th, 2011, 08:07 AM #7 Global Moderator   Joined: Dec 2006 Posts: 19,994 Thanks: 1855 The above proof is correct, but it's well-known anyway that the volume of any pyramid or cone is a third of the product of its height and the area of its base.
 October 29th, 2011, 08:36 AM #8 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Quiz problem that I got killed on Because of the straight lines, the rectangle forming a horizontal cross section will always have area proportional to the distance from the vertex. Since that height is "h", the rectangle at the base has area $L_1L_2$ and at the vertex the area is 0, the area of a rectangle at height y will be $L_1L_2(h- y)/h$. Taking each cross section as a three dimensional rectangle of height "dy", each will have volume $\frac{L_1L_2(h- y)}{h}dy$. The total volume will be $\frac{L_1L_2}{h}\int_{y=0}^h (h- y)dy$
October 29th, 2011, 11:34 AM   #9
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Quote:
 Originally Posted by HallsofIvy . . . will always have area proportional to the distance from the vertex.
Incorrect - should be square of the distance.

 October 30th, 2011, 03:26 PM #10 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Quiz problem that I got killed on Thanks, that is exactly right.

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