Question #509cd

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Question #509cd

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Answer 1 · 2016-01-26T18:54:01

$196 f t .^{2}$ $196 ft.^2$

Explanation:

Refer to figures below

I created this figure using MS Excel

From the given data we have
$D_{1} = 6 f t .$ $D_1=6 ft.$ => $r_{1} = 3 f t .$ $r_1=3ft.$
$D_{2} = 4 f l .$ $D_2=4 fl.$ => $r_{2} = 2 f t .$ $r_2=2ft.$
$h = 12 f t .$ $h=12ft.$

As we can see in Fig. (a)
$\frac{r_{2}}{H - h} = \frac{r_{1}}{H}$ $r_2/(H-h)=r_1/H$
$\frac{2}{H - 12} = \frac{3}{H}$ $2/(H-12)=3/H$ => $2 H = 3 H - 36$ $2H=3H-36$ => $H = 36$ $H=36$

Since $s = \sqrt{2} \cdot r$ $s=sqrt(2)*r$ (see Fig.(b))
$s_{1} = 3 \sqrt{2}$ $s_1=3sqrt(2)$
$s_{2} = 2 \sqrt{2}$ $s_2=2sqrt(2$

Area of bases
$S_{b 1} = s_{1}^{2} = {(3 \sqrt{2})}^{2} = 18$ $S_(b1)=s_1^2=(3sqrt(2))^2=18$
$S_{b 2} = s_{2}^{2} = {(2 \sqrt{2})}^{2} = 8$ $S_(b2)=s_2^2=(2sqrt(2))^2=8$

Side of the slanted trapezoid (see Fig.(c))
$x^{2} = h^{2} + {(r_{1} - r_{2})}_{1}^{2}$ $x^2=h^2+(r_1-r_2)^2$
$x = \sqrt{12^{2} + 1^{2}} = \sqrt{145}$ $x=sqrt(12^2+1^2)=sqrt(145)$

Height of the slanted trapezoid (see Fig.(d))
$y^{2} = x^{2} - {(\frac{s_{1} - s_{2}}{2})}^{2}$ $y^2=x^2-((s_1-s_2)/2)^2$
$y^{2} = {(\sqrt{145})}^{2} - {(\frac{3 \sqrt{2} - 2 \sqrt{2}}{2})}^{2}$ $y^2=(sqrt (145))^2-((3sqrt(2)-2sqrt(2))/2)^2$
$y^{2} = 145 - \frac{1}{2} = \frac{289}{2}$ $y^2=145-1/2=289/2$ => $y = \frac{17 \sqrt{2}}{2}$ $y=(17sqrt(2))/2$

Area of each slanted trapezoid
$S_{t r a p e z o i d} = \frac{s_{1} + s_{2}}{2} \cdot y = \frac{3 \sqrt{2} + 2 \sqrt{2}}{2} \cdot \frac{17 \sqrt{2}}{2}$ $S_(trapezoid)=(s_1+s_2)/2*y=(3sqrt(2)+2sqrt(2))/2*(17sqrt(2))/2$ => $S_{t r a p e z o i d} = \frac{85}{2}$ $S_(trapezoid)=85/2$

Total Area
$S_{T} = S_{b 1} + S_{b_{2}} + 4 \cdot S_{t r a p e z o i d} = 18 + 8 + 4 \cdot \frac{85}{2} = 196 f t .^{2}$ $S_T=S_(b1)+S_(b_2)+4*S_(trapezoid)=18+8+4*85/2=196 ft.^2$

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Question #509cd

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