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Question-45930




Question Number 45930 by Tawa1 last updated on 18/Oct/18
Answered by MJS last updated on 19/Oct/18
for any shape  length (or radius, height, width...) ×k  ⇒ area×k^2   ⇒ volume ×k^3   l×k ⇒ a×k^2 , v×k^3   a×k ⇒ l×(√k), v×(√k^3 )  v×k ⇒ l×(k)^(1/3) , a×(k^2 )^(1/3)
$$\mathrm{for}\:\mathrm{any}\:\mathrm{shape} \\ $$$$\mathrm{length}\:\left(\mathrm{or}\:\mathrm{radius},\:\mathrm{height},\:\mathrm{width}…\right)\:×{k} \\ $$$$\Rightarrow\:\mathrm{area}×{k}^{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{volume}\:×{k}^{\mathrm{3}} \\ $$$${l}×{k}\:\Rightarrow\:{a}×{k}^{\mathrm{2}} ,\:{v}×{k}^{\mathrm{3}} \\ $$$${a}×{k}\:\Rightarrow\:{l}×\sqrt{{k}},\:{v}×\sqrt{{k}^{\mathrm{3}} } \\ $$$${v}×{k}\:\Rightarrow\:{l}×\sqrt[{\mathrm{3}}]{{k}},\:{a}×\sqrt[{\mathrm{3}}]{{k}^{\mathrm{2}} } \\ $$
Commented by Tawa1 last updated on 19/Oct/18
Am trying to study it
$$\mathrm{Am}\:\mathrm{trying}\:\mathrm{to}\:\mathrm{study}\:\mathrm{it} \\ $$
Commented by Tawa1 last updated on 19/Oct/18
God bless you sir but i don′t get it.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}\:\mathrm{but}\:\mathrm{i}\:\mathrm{don}'\mathrm{t}\:\mathrm{get}\:\mathrm{it}. \\ $$
Commented by Tawa1 last updated on 19/Oct/18
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by MJS last updated on 19/Oct/18
square  side=a ⇒ area=a^2   side=k×a ⇒ area=(k×a)^2 =k^2 ×a^2   same for triangle  side=a; height=h_a  ⇒  area=((a×h_a )/2)  side=k×a; height=k×h_a  ⇒ area=(((k×a)×(k×h_a ))/2)=k^2 ×((a×h_a )/2)  same for circle  radius=r ⇒ area=πr^2   radius=k×r ⇒ area=π(k×r)^2 =k^2 πr^2     similar for volumes
$$\mathrm{square} \\ $$$$\mathrm{side}={a}\:\Rightarrow\:\mathrm{area}={a}^{\mathrm{2}} \\ $$$$\mathrm{side}={k}×{a}\:\Rightarrow\:\mathrm{area}=\left({k}×{a}\right)^{\mathrm{2}} ={k}^{\mathrm{2}} ×{a}^{\mathrm{2}} \\ $$$$\mathrm{same}\:\mathrm{for}\:\mathrm{triangle} \\ $$$$\mathrm{side}={a};\:\mathrm{height}={h}_{{a}} \:\Rightarrow\:\:\mathrm{area}=\frac{{a}×{h}_{{a}} }{\mathrm{2}} \\ $$$$\mathrm{side}={k}×{a};\:\mathrm{height}={k}×{h}_{{a}} \:\Rightarrow\:\mathrm{area}=\frac{\left({k}×{a}\right)×\left({k}×{h}_{{a}} \right)}{\mathrm{2}}={k}^{\mathrm{2}} ×\frac{{a}×{h}_{{a}} }{\mathrm{2}} \\ $$$$\mathrm{same}\:\mathrm{for}\:\mathrm{circle} \\ $$$$\mathrm{radius}={r}\:\Rightarrow\:\mathrm{area}=\pi{r}^{\mathrm{2}} \\ $$$$\mathrm{radius}={k}×{r}\:\Rightarrow\:\mathrm{area}=\pi\left({k}×{r}\right)^{\mathrm{2}} ={k}^{\mathrm{2}} \pi{r}^{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{similar}\:\mathrm{for}\:\mathrm{volumes} \\ $$

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