Answer:
a cubical box has lateral the greater lateral surface area and cuboical box has the smaller total surface area
Step-by-step explanation:
cuz......
10×12.5=125
and....
10×8=80
so therefore,the cubical box has 125 CM and the cuboidal box has 80 cm
[tex]\large\underline{\sf{Solution-}}[/tex]
Dimensions of cubical box:
Edge of cubical box, a = 10 cm
So,
[tex]\sf\: Lateral\:surface\:area_{(Cube)} = {4a}^{2} = {4(10)}^{2} = 400 \: {cm}^{2} \\ [/tex]
[tex]\sf\: Total\:surface\:area_{(Cube)} = {6a}^{2} = {6(10)}^{2} = 600 \: {cm}^{2} \\ [/tex]
Dimensions of cuboidal box:
Length of cuboidal box, l = 12.5 cm
Breadth of cuboidal box, b = 10 cm
Height of cuboidal box, h = 8 cm
[tex]\sf\: Lateral\:surface\:area_{(Cuboid)} \\ [/tex]
[tex]\sf\: = \: 2(l + b)h \\ [/tex]
[tex]\sf\: = \: 2(12.5 + 10) \times 8 \\ [/tex]
[tex]\sf\: = \: 16 \times 22.5 \\ [/tex]
[tex]\sf\: = \: 360 \: {cm}^{2} \\ [/tex]
Now,
[tex]\sf\: Total\:surface\:area_{(Cube)} \\ [/tex]
[tex]\sf\: = \: 2(lb + bh + hl) \\ [/tex]
[tex]\sf\: = \: 2(12.5 \times 10 + 10 \times 8 + 8 \times 12.5) \\ [/tex]
[tex]\sf\: = \: 2(125 + 80 + 100) \\ [/tex]
[tex]\sf\: = \: 2 \times 305 \\ [/tex]
[tex]\sf\: = \: 610 \: {cm}^{2} \\ [/tex]
Thus, From above calculations, we have
[tex]\sf\: Lateral\:surface\:area_{(Cube)} = 400 \: {cm}^{2} \\ [/tex]
[tex]\sf\: Total\:surface\:area_{(Cube)} = 600 \: {cm}^{2} \\ [/tex]
[tex]\sf\: Lateral\:surface\:area_{(Cuboid)} = 360 \: {cm}^{2} \\ [/tex]
[tex]\sf\: Total\:surface\:area_{(Cuboid)} = 610 \: {cm}^{2} \\ [/tex]
So, from above calculations, we concluded that
[tex]\sf\: Lateral\:surface\:area_{(Cube)} > Lateral\:surface\:area_{(Cuboid)} \\ [/tex]
[tex]\sf\: Total\:surface\:area_{(Cube)} < Total\:surface\:area_{(Cuboid)} \\ [/tex]
(i) Cubical box has the greater lateral surface area and by 40 cm².
(ii) Cubical box has the smaller total surface area and by 10 cm²
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Answers & Comments
Answer:
a cubical box has lateral the greater lateral surface area and cuboical box has the smaller total surface area
Step-by-step explanation:
cuz......
10×12.5=125
and....
10×8=80
so therefore,the cubical box has 125 CM and the cuboidal box has 80 cm
[tex]\large\underline{\sf{Solution-}}[/tex]
Dimensions of cubical box:
Edge of cubical box, a = 10 cm
So,
[tex]\sf\: Lateral\:surface\:area_{(Cube)} = {4a}^{2} = {4(10)}^{2} = 400 \: {cm}^{2} \\ [/tex]
[tex]\sf\: Total\:surface\:area_{(Cube)} = {6a}^{2} = {6(10)}^{2} = 600 \: {cm}^{2} \\ [/tex]
Dimensions of cuboidal box:
Length of cuboidal box, l = 12.5 cm
Breadth of cuboidal box, b = 10 cm
Height of cuboidal box, h = 8 cm
So,
[tex]\sf\: Lateral\:surface\:area_{(Cuboid)} \\ [/tex]
[tex]\sf\: = \: 2(l + b)h \\ [/tex]
[tex]\sf\: = \: 2(12.5 + 10) \times 8 \\ [/tex]
[tex]\sf\: = \: 16 \times 22.5 \\ [/tex]
[tex]\sf\: = \: 360 \: {cm}^{2} \\ [/tex]
Now,
[tex]\sf\: Total\:surface\:area_{(Cube)} \\ [/tex]
[tex]\sf\: = \: 2(lb + bh + hl) \\ [/tex]
[tex]\sf\: = \: 2(12.5 \times 10 + 10 \times 8 + 8 \times 12.5) \\ [/tex]
[tex]\sf\: = \: 2(125 + 80 + 100) \\ [/tex]
[tex]\sf\: = \: 2 \times 305 \\ [/tex]
[tex]\sf\: = \: 610 \: {cm}^{2} \\ [/tex]
Thus, From above calculations, we have
[tex]\sf\: Lateral\:surface\:area_{(Cube)} = 400 \: {cm}^{2} \\ [/tex]
[tex]\sf\: Total\:surface\:area_{(Cube)} = 600 \: {cm}^{2} \\ [/tex]
[tex]\sf\: Lateral\:surface\:area_{(Cuboid)} = 360 \: {cm}^{2} \\ [/tex]
[tex]\sf\: Total\:surface\:area_{(Cuboid)} = 610 \: {cm}^{2} \\ [/tex]
So, from above calculations, we concluded that
[tex]\sf\: Lateral\:surface\:area_{(Cube)} > Lateral\:surface\:area_{(Cuboid)} \\ [/tex]
[tex]\sf\: Total\:surface\:area_{(Cube)} < Total\:surface\:area_{(Cuboid)} \\ [/tex]
Now,
(i) Cubical box has the greater lateral surface area and by 40 cm².
(ii) Cubical box has the smaller total surface area and by 10 cm²