When each edge of a cube is increased by 50% by what percent is the surface area of the cube increased?

If each edge of a cube is increased by 50%, the percentage increase in the surface area is

 125%Let the original edge of the cube be a units.

Then, the original surface area of the cube = 6a2 units

New edge of the cube = 150% of a 

`= (150"a")/100`

`= (3"a")/2`

Hence, new surface area `= 6  × ((3"a")/2)^2`

`=(27a^2)/2`

Increase in area` =((27a^2)/2 - 6"a"^2) `

                         `=(15"a"^2)/2`

% increasein surface area `= ("15a"^2/2xx1/(6"a"^2xx100))%`

=125 %

Concept: Concept of Surface Area, Volume, and Capacity

  Is there an error in this question or solution?

Text Solution

Solution : Let the edge of cube be `2` units.<br> Increased side = `3` units<br> Here, side of a cube is increased by `50%`=`1/2`<br> So, original S.A = `6xx2xx2`=`24` sq. units<br> Increased surface area=`6xx3xx3` = `54` sq. units<br> % increase in area `(54-24)/24xx100` = `125%`<br> Hence, option(d) is correct.

Each edge of a cube is increased by 27/2x2 6x20ex0ex.Find the percentage increase in the surface area of the cube.

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