My Math Forum Volume of a Water Trough (Isosceles Trapezoid)

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 October 20th, 2011, 09:45 PM #1 Newbie   Joined: Oct 2011 Posts: 1 Thanks: 0 Volume of a Water Trough (Isosceles Trapezoid) A water trough is 6 m long and its cross-section is an isosceles trapezoid which is 100 cm wide at the bottom and 200 cm wide at the top, and the height is 50 cm. The trough is not full. Give an expression for V, the volume of water in the trough in cm3, when the depth of the water is d cm.
 October 21st, 2011, 04:54 AM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,879 Thanks: 1087 Math Focus: Elementary mathematics and beyond Re: Volume of a Water Trough (Isosceles Trapezoid) Making a diagram will help. 50/50 = 1, so there is a 1:1 ratio between the depth of water in the tank and the 'overhang' (where the waterline extends past the bottom of the tank). Two of these 'overhangs' + 100 equals the length of the waterline and 100 is the base. The formula for the area of a trapezoid is height * (top + bottom)/2, so volume is V = Length * depth * (waterline + base)/2, so we have 600d((2d + 100) + 100)/2 = 600d(d + 100), where d is the depth of water in the trough, in centimeters.
 October 21st, 2011, 03:55 PM #3 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,204 Thanks: 511 Math Focus: Calculus/ODEs Re: Volume of a Water Trough (Isosceles Trapezoid) We could generalize greg1313's method (you still want to make a/another diagram...) a bit to first set: $W_1$ = width at top of trough $W_2$ = width at base of trough $W_2 $L$ = length of trough $0 $h$ = height of trough $0 $d$ = depth of water $0\le d\le h$ $W_d$ = width of water at surface $W_2\le W_d\le W_1$ $W_x$ = horizontal overhang of water at the surface. We see then: $W_d=W_2+2W_x$ A cross-section of the volume along the length has area A ( and then using the standard formula for the area of a trapezoid) of: $A=\frac{d}{2}$$W_d+W_2$$$ From geometric similarity using the triangular overhangs: $\frac{d}{W_x}=\frac{h}{\frac{W_1-W_2}{2}}$ $W_x=\frac{d$$W_1-W_2$$}{2h}$ $A=\frac{d}{2}$$W_2+\frac{d\(W_1-W_2$$}{h}+W_2\)$ $A=\frac{d}{2h}$$2W_2h+d\(W_1-W_2$$\)$ $A=\frac{d$$\(2h-d$$W_2+dW_1\)}{2h}$ Then: $V=AL=\frac{d\cdot L}{2h}$$\(2h-d$$W_2+dW_1\)$ Let's factor on d on the right, the unknown: $V=\frac{d\cdot L}{2h}$$\(W_1-W_2$$d+2hW_2\)$ Using the data given for the problem: L = 600 cm h = 50 cm W_1 = 200 cm W_2 = 100 cm we may now state: $V=\frac{600d}{100}$$\(200-100$$d+100\cdot100\)=600d$$d+100$$\text{ cm^3}$

 Tags isosceles, trapezoid, trough, volume, water

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