each edge of a cube is increased by 10%. find the persantage increase in the surface area of the cube
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Let edge=a
New a =11/10a
TSA=6a^2
New TSA= 6(11/10a)
So
TSA is increased by121%
New a =11/10a
TSA=6a^2
New TSA= 6(11/10a)
So
TSA is increased by121%
Answered by
0
surace area of cube = 6a²
after increasing
a= 11a/10
so new surface area = 6(11a/10)²
6(121a²/100)
726a²/100
7.62a²
percentage= 7.62 a²- 6a²
1.62
162/100
so answer = 162 percent
after increasing
a= 11a/10
so new surface area = 6(11a/10)²
6(121a²/100)
726a²/100
7.62a²
percentage= 7.62 a²- 6a²
1.62
162/100
so answer = 162 percent
aniketkabaddi01:
but .... 6a² let take a=2 so 6*4 =24 now after 10% increase a=2.2 6*(2.2)²= 6*4.84 =29.04
subtracting .. 29.04-24= 5.04
x/100*24=5.04
24x=504
x=504/84
x=21
solved
thanks
this is the right solution actually i thwn conaulted it from my teacher
my pleasure..
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