Can You Pivot 2?

Geometry Level 4

A trapezoid 10' long with parallel sides of 12" and 1" is drawn on a coordinate plane with one corner on the origin and 2 sides placed on the x x and y y axis.

Find the centroid of the trapezoid and state the x x value of the centroid to the nearest .001 inch.


The answer is 43.077.

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2 solutions

Guiseppi Butel
Dec 13, 2015

Soln Soln

Trapezoid ABCDE is made up of a triangle ABC and a rectangle BCDE.
Area of ABC= 1 2 11 120 = 660 , t h a t o f B C D E = 1 120 = 120 \color{#D61F06}{\frac 1 2 *11*120=660,~~ that~ of ~BCDE= 1*120=120} .
G t G_t of triangle is one third from base.
So its horizontal distance from vertical side is 1 3 120 = 40 , a n d f r o m H o r i z o n t a l s i d e 11 3 \frac 1 3 *120=40,~ and~ from ~Horizontal ~side~\dfrac {11} 3
But the horizontal side is y=1, G t ( 40 , 11 3 + 1 ) G t ( 40 , 14 3 ) . \therefore ~ G_t (40,\dfrac {11} 3 + 1) ~\implies ~\color{#D61F06}{G_t (40,\dfrac {14} 3)}.
Area of BCDE has its C.G. at its center, So G r ( 60 , 1 2 ) . \color{#D61F06}{G_r(60,\dfrac 1 2 ).}
Let G be the C.G. of the trapezoid ABCDE. Considering horizontal components,
X G r X G t = 60 40 = 20. L e t X G X G t = t a n d X G X G r = r . T a k i n g m o m e n t s a b o u t X G , 660 t = 120 r r t = 11 2 , b u t t + r = 20. t = 20 2 11 + 2 . S o X G t = 40 + 40 13 = 43.077 We can solve for the Y-component in the same way. X_{G_r}\! -\! X_{G_t}=60-40=20.~ ~ Let ~X_GX_{G_t}=t~ and~ X_GX_{G_r}=r.~ ~ Taking~ moments ~ about~ X_G, \\ 660*t=120*r~ ~ \implies ~\dfrac r t=\dfrac {11} 2,~ but~ t+r=20. \therefore~ t=20*\dfrac2 {11+2}. \\So~ X_{G_t}=40+\dfrac{40}{13}= \Large \color{#3D99F6}{43.077} \\ \text{We can solve for the Y-component in the same way.}



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