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Question-20149




Question Number 20149 by ajfour last updated on 22/Aug/17
Commented by ajfour last updated on 22/Aug/17
The roof is even a flat plane.Lengths  a, b, c, y are perpendicular to the  bottom rectangular face with sides  h and l . Find height y and the  volume  enclosed by the surfaces.
$${The}\:{roof}\:{is}\:{even}\:{a}\:{flat}\:{plane}.{Lengths} \\ $$$$\boldsymbol{{a}},\:\boldsymbol{{b}},\:\boldsymbol{{c}},\:\boldsymbol{{y}}\:{are}\:{perpendicular}\:{to}\:{the} \\ $$$${bottom}\:{rectangular}\:{face}\:{with}\:{sides} \\ $$$$\boldsymbol{{h}}\:{and}\:\boldsymbol{{l}}\:.\:{Find}\:{height}\:\boldsymbol{{y}}\:{and}\:{the}\:\:{volume} \\ $$$${enclosed}\:{by}\:{the}\:{surfaces}. \\ $$
Answered by mrW1 last updated on 22/Aug/17
((a+y)/2)=((b+c)/2)  ⇒y=b+c−a    V=l×h×((a+b+c+y)/4)=l×h×((a+b+c+b+c−a)/4)  =((lh(b+c))/2)
$$\frac{\mathrm{a}+\mathrm{y}}{\mathrm{2}}=\frac{\mathrm{b}+\mathrm{c}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{y}=\mathrm{b}+\mathrm{c}−\mathrm{a} \\ $$$$ \\ $$$$\mathrm{V}=\mathrm{l}×\mathrm{h}×\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{y}}{\mathrm{4}}=\mathrm{l}×\mathrm{h}×\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{b}+\mathrm{c}−\mathrm{a}}{\mathrm{4}} \\ $$$$=\frac{\mathrm{lh}\left(\mathrm{b}+\mathrm{c}\right)}{\mathrm{2}} \\ $$
Commented by ajfour last updated on 22/Aug/17
thank you Sir, you make it so simple!
$${thank}\:{you}\:{Sir},\:{you}\:{make}\:{it}\:{so}\:{simple}! \\ $$
Answered by ajfour last updated on 23/Aug/17
let eqn. of plane be     (r^� −a^� ).n^� =0  let origin be the bottom left corner.  n^� =[li^� +(b−a)k^�^�  ]×[hj^� +(c−a)k^� ]     =−h(b−a)i^� −l(c−a)j^� +lhk^� ]  ⇒  −h(b−a)x−l(c−a)y+lhz=alh  z=a+(((b−a)x)/l)+(((c−a)y)/h)       at x=l, y=h , z=y (asked)  so   y=a+b−a+c−a          y+a=b+c    ⇒   y=b+c−a  V=∫_0 ^(  l) (∫_0 ^(  h) zdy)dx     =∫_0 ^(  l) [ah+(((b−a)hx)/l)+(((c−a)h)/2)]dx    =ahl+(((b−a)lh)/2)+(((c−a)lh)/2)    =((lh)/2)(2a+b−a+c−a)    Volume V=((lh(b+c))/2) .
$${let}\:{eqn}.\:{of}\:{plane}\:{be} \\ $$$$\:\:\:\left(\bar {{r}}−\bar {{a}}\right).\hat {{n}}=\mathrm{0} \\ $$$${let}\:{origin}\:{be}\:{the}\:{bottom}\:{left}\:{corner}. \\ $$$$\bar {{n}}=\left[{l}\hat {{i}}+\left({b}−{a}\right)\hat {{k}}\right]×\left[{h}\hat {{j}}+\left({c}−{a}\right)\hat {{k}}\right] \\ $$$$\left.\:\:\:=−{h}\left({b}−{a}\right)\hat {{i}}−{l}\left({c}−{a}\right)\hat {{j}}+{lh}\hat {{k}}\right] \\ $$$$\Rightarrow\:\:−{h}\left({b}−{a}\right){x}−{l}\left({c}−{a}\right){y}+{lhz}={alh} \\ $$$${z}={a}+\frac{\left({b}−{a}\right){x}}{{l}}+\frac{\left({c}−{a}\right){y}}{{h}} \\ $$$$\:\:\:\:\:{at}\:{x}={l},\:{y}={h}\:,\:{z}=\boldsymbol{{y}}\:\left({asked}\right) \\ $$$${so}\:\:\:\boldsymbol{{y}}={a}+{b}−{a}+{c}−{a} \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{{y}}+{a}={b}+{c}\:\:\:\:\Rightarrow\:\:\:\boldsymbol{{y}}=\boldsymbol{{b}}+\boldsymbol{{c}}−\boldsymbol{{a}} \\ $$$${V}=\int_{\mathrm{0}} ^{\:\:{l}} \left(\int_{\mathrm{0}} ^{\:\:{h}} {zdy}\right){dx} \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{\:\:{l}} \left[{ah}+\frac{\left({b}−{a}\right){hx}}{{l}}+\frac{\left({c}−{a}\right){h}}{\mathrm{2}}\right]{dx} \\ $$$$\:\:={ahl}+\frac{\left({b}−{a}\right){lh}}{\mathrm{2}}+\frac{\left({c}−{a}\right){lh}}{\mathrm{2}} \\ $$$$\:\:=\frac{{lh}}{\mathrm{2}}\left(\mathrm{2}{a}+{b}−{a}+{c}−{a}\right) \\ $$$$\:\:{Volume}\:\boldsymbol{{V}}=\frac{\boldsymbol{{lh}}\left(\boldsymbol{{b}}+\boldsymbol{{c}}\right)}{\mathrm{2}}\:. \\ $$

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