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505-555 (Easy)|   Geometry|      
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Bunuel
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In the cube above, the length of line segment AB is 8. The surface are 05 Dec 2018, 05:53
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In the cube above, the length of line segment AB is 8. The surface area of the cube equals what?


A. 64

B. \(64\sqrt{2}\)

C. 128

D. 192

E. \(192\sqrt{2}\)


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In the cube above, the length of line segment AB is 8. The surface are 05 Dec 2018, 07:33
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Area of a square = Diagonal^2/2
So, 64/2 = 32

Surface area = 32 x 6 = 192

D
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ArjunJag1328
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In the cube above, the length of line segment AB is 8. The surface are 05 Dec 2018, 08:05
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Let me admit that I am quiet weak at Geometry. So feel free to correct me if my method is incorrect.

Surface area of a cube = 6\(a^2\)
Diagonal of a cube = \(\sqrt{3}\)a

d = \(\sqrt{3}\)a

\(\sqrt{3}\)a =8

a = \(\frac{8}{\sqrt{3}a}\)

Surface area of a cube = 6\(a^2\)

=> 6 \(*\) \((8/\sqrt{3})^2\)
=>6 \(*\) \(\frac{64}{3}\)
=> 128

IMO C
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In the cube above, the length of line segment AB is 8. The surface are 05 Dec 2018, 09:15
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Bunuel


In the cube above, the length of line segment AB is 8. The surface area of the cube equals what?


A. 64

B. \(64\sqrt{2}\)

C. 128

D. 192

E. \(192\sqrt{2}\)


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side of cube would be 2a2= 64
a= 4 \sqrt{2}

TSA of cube 6s^2
6*16*2= 192 IMO D
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In the cube above, the length of line segment AB is 8. The surface are 05 Dec 2018, 09:18
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ArjunJag1328
Let me admit that I am quiet weak at Geometry. So feel free to correct me if my method is incorrect.

Surface area of a cube = 6\(a^2\)
Diagonal of a cube = \(\sqrt{3}\)a

d = \(\sqrt{3}\)a

\(\sqrt{3}\)a =8

a = \(\frac{8}{\sqrt{3}a}\)

Surface area of a cube = 6\(a^2\)

=> 6 \(*\) \((8/\sqrt{3})^2\)
=>6 \(*\) \(\frac{64}{3}\)
=> 128

IMO C
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ArjunJag1328 how did you calculate diagonal of cube as \sqrt{3}a ?
it should be \sqrt{2}a in fact as ; it would be 45:45:90 triangle .. x:x:x\sqrt{2} ..
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In the cube above, the length of line segment AB is 8. The surface are 05 Dec 2018, 09:24
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Archit3110
ArjunJag1328
Let me admit that I am quiet weak at Geometry. So feel free to correct me if my method is incorrect.

Surface area of a cube = 6\(a^2\)
Diagonal of a cube = \(\sqrt{3}\)a

d = \(\sqrt{3}\)a

\(\sqrt{3}\)a =8

a = \(\frac{8}{\sqrt{3}a}\)

Surface area of a cube = 6\(a^2\)

=> 6 \(*\) \((8/\sqrt{3})^2\)
=>6 \(*\) \(\frac{64}{3}\)
=> 128

IMO C

ArjunJag1328 how did you calculate diagonal of cube as \sqrt{3}a ?
it should be \sqrt{2}a in fact as ; it would be 45:45:90 triangle .. x:x:x\sqrt{2} ..
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I was confused whether to take diagonal of a square or cube but I chose to work it out with the diagonal of a cube because the question was to find the surface area of a cube.
My bad!!! So we are supposed to take the diagonal of a square for these types of questions. right?

Thanks indeed for taking your time to help me out.

Regards
Arjun
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In the cube above, the length of line segment AB is 8. The surface are 05 Dec 2018, 09:35
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ArjunJag1328
Archit3110
ArjunJag1328
Let me admit that I am quiet weak at Geometry. So feel free to correct me if my method is incorrect.

Surface area of a cube = 6\(a^2\)
Diagonal of a cube = \(\sqrt{3}\)a

d = \(\sqrt{3}\)a

\(\sqrt{3}\)a =8

a = \(\frac{8}{\sqrt{3}a}\)

Surface area of a cube = 6\(a^2\)

=> 6 \(*\) \((8/\sqrt{3})^2\)
=>6 \(*\) \(\frac{64}{3}\)
=> 128

IMO C

ArjunJag1328 how did you calculate diagonal of cube as \sqrt{3}a ?
it should be \sqrt{2}a in fact as ; it would be 45:45:90 triangle .. x:x:x\sqrt{2} ..

I was confused whether to take diagonal of a square or cube but I chose to work it out with the diagonal of a cube because the question was to find the surface area of a cube.
My bad!!! So we are supposed to take the diagonal of a square for these types of questions. right?

Thanks indeed for taking your time to help me out.

Regards
Arjun
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ArjunJag1328 : i did not understand your line "So we are supposed to take the diagonal of a square for these types of questions. "

If you see the picture , you would see that its a triangle 45:45:90 , you can then find sides accordingly or else use relation a^2=b^2 +c^2 , since b=c so a^2=2b^2 or else side a= b \sqrt{2}
given diagnoal is 8 say 'a'
so 8= b \sqrt{2}

so each side of cube would be , b = 8/sqrt2 or 4 \sqrt{2}..

hope this helps...
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