A geometry problem by A Former Brilliant Member

Geometry Level 1

The edge of a cube is increased by 40%, by how much percentage will its surface area increased?

Try some of my problems: Math Problems - Set 1 .
40% 96% 140% 69%

A Former Brilliant Member by A Former Brilliant Member

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2 solutions

S 1 S 2 = \frac{S_1}{S_2}= a 2 ( 1.4 a ) 2 = \frac{a^2}{(1.4a)^2}= 1 1.96 \frac{1}{1.96}

1.96 S 1 = S 2 1.96S_1=S_2

% i n c r e a s e d = increased= ( 1.96 − 1 ) ( 100 ) = 96 (1.96-1)(100)=96 %

A l t e r n a t e S o l u t i o n : Alternate Solution:

s e t set a = 1 a=1

S 1 = 6 ( 1 ) 2 = 6 S_1=6(1)^2=6

S 2 = 6 ( 1.4 ) 2 = 11.76 S_2=6(1.4)^2=11.76

% i n c r e a s e d = ( 11.76 − 6 6 ) ( 100 ) = 96 increased=(\frac{11.76-6}{6})(100)=96 %

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This question is confusing. I only increased the size of one edge, making the larger shape a cuboid. This gave me 40%

Justin Arun - 3 years ago

Let s s be the side length of the original cube, then the side length of the larger cube is 1.4 s 1.4s .

The surface area of the original cube is 6 s 2 6s^2 .

The surface area of the larger cube is 6 ( 1.4 s ) 2 = 11.76 s 2 6(1.4s)^2=11.76s^2 .

% i n c r e a s e d i n s u r f a c e a r e a = ( 11.76 s 2 − 6 s 2 6 s 2 ) ( 100 % ) = 96 % \%~increased~in~surface~area=\left(\dfrac{11.76s^2-6s^2}{6s^2}\right)(100\%)=\color{#69047E}\boxed{96\%}

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