The express of two cubes are 6 cm and 8 cm. There is a third cube whose total surface area equals the sum of the total surface area of first two cubes. Then the edge of the third cube is_____ cm.
[tex] \sf{}Surface \: area \: of \: first \: cube \: = 6a² \\ \\ \sf{}Surface \: area \: of \: first \: cube \: = 6 ( 6 )² \\ \\ \sf{}Surface \: area \: of \: first \: cube = 6 (36) \\ \\ \sf{}Surface \: area \: of \: first \: cube = 216 cm²[/tex]
[tex]\\[/tex]
Now surface area of second cube,
[tex] \sf{}Surface \: area \: of \: second\: cube \: = 6a² \\ \\ \sf{}Surface \: area \: of \: second \: cube \: = 6 ( 8 )² \\ \\ \sf{}Surface \: area \: of \: second \: cube = 6 (81) \\ \\ \sf{}Surface \: area \: of \: second \: cube = 426 cm²[/tex]
[tex]\\[/tex]
As it us given that sum of the total surface arwa of first and secind cube us equals to the surface area of third cube therefore ,
[tex] \sf{}Surface \: area \: of \: third \: cube \: = ( 216 + 476 ) cm²\\\\ \sf{} Surface \: area \: of \: third \: cube \: =702 {cm}^{2} [/tex]
Answers & Comments
Answer:
14cm
Step-by-step explanation:
6+8-=14 so its 14cm OK bro
Verified answer
Solution :
[tex] \sf{}Surface \: area \: of \: first \: cube \: = 6a² \\ \\ \sf{}Surface \: area \: of \: first \: cube \: = 6 ( 6 )² \\ \\ \sf{}Surface \: area \: of \: first \: cube = 6 (36) \\ \\ \sf{}Surface \: area \: of \: first \: cube = 216 cm²[/tex]
[tex]\\[/tex]
Now surface area of second cube,
[tex] \sf{}Surface \: area \: of \: second\: cube \: = 6a² \\ \\ \sf{}Surface \: area \: of \: second \: cube \: = 6 ( 8 )² \\ \\ \sf{}Surface \: area \: of \: second \: cube = 6 (81) \\ \\ \sf{}Surface \: area \: of \: second \: cube = 426 cm²[/tex]
[tex]\\[/tex]
As it us given that sum of the total surface arwa of first and secind cube us equals to the surface area of third cube therefore ,
[tex] \sf{}Surface \: area \: of \: third \: cube \: = ( 216 + 476 ) cm²\\\\ \sf{} Surface \: area \: of \: third \: cube \: =702 {cm}^{2} [/tex]
[tex]\\[/tex]
And we have to find its edge therefore ,
[tex] \sf{}surface \: area \: = 6 {(a)}^{2} \\ \\ \sf{}702 {cm}^{2} = 6 {(a)}^{2} \\ \\ \sf{}( \frac{702}{6} ) {cm}^{2} = {a}^{2} \\ \\ \sf{} 117 {cm}^{2} = {a}^{2} \\ \\ \sf{}a = \sqrt{117} \\ \\ \boxed{\sf{}a = 10.8 \: cm}[/tex]
Therefore ,edge of the third cube whose surface arwa is eqyals to sum of first two cubes us equals to 10.8 cm.
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[tex]\underline{ \underline{ \sf{ \red{ \bold{ more \: Formulas}}}}} \\ \\ \: \sf{}volume \: of \: cube \: = {a}^{3} \\ \\ \sf{}volume \: of \: cylinder = \pi {r}^{2} h \\ \\ \sf{}volume \: of \: cone = \frac{1}{3} \pi \: {r}^{2} h \\ \\ \sf{}volume \: of \: sphere \: = \frac{4}{3} \pi \: {r}^{3} \\ \\ \sf{} volume \: of \: hemisphere = \frac{2}{3} \pi {r}^{3}[/tex]