each edge of cube is increased by 50%.finds the percentage increase in surface areas.
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Let the original edge of the cube = a
now the edge is increased by 50 %
so 50% of a = 50/100 ×a= a/2
now the new edge = a/2 + a = a/2+ 2a/2
=3a/2
now the surface area of the original cube = 6a²
and the surface area of the new cube = 6a² = 6 × (3a/2)²
= 27a²/2
now the increase in surface area = 27a²/2 - 6a²
= 27a²/2 - 12a²/2
= 15a²/2
now the percentage increase in surface area = 15a²/2/6a² × 100
= 15a² / 2× 6a² ×100
= 125 %
I hope I am correct !!
now the edge is increased by 50 %
so 50% of a = 50/100 ×a= a/2
now the new edge = a/2 + a = a/2+ 2a/2
=3a/2
now the surface area of the original cube = 6a²
and the surface area of the new cube = 6a² = 6 × (3a/2)²
= 27a²/2
now the increase in surface area = 27a²/2 - 6a²
= 27a²/2 - 12a²/2
= 15a²/2
now the percentage increase in surface area = 15a²/2/6a² × 100
= 15a² / 2× 6a² ×100
= 125 %
I hope I am correct !!
VaibhavBobby:
it is completely wrong
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0
Let the original edge be 'a'
T.S.A of cube = 6a(a)
new edge = 3a/2
T.S.A of the original cube = 6(3a/2)(3a/2)
increase in T.S.A = 21(a)(a)
Increase % = 21(100)/6
= 350%
T.S.A of cube = 6a(a)
new edge = 3a/2
T.S.A of the original cube = 6(3a/2)(3a/2)
increase in T.S.A = 21(a)(a)
Increase % = 21(100)/6
= 350%
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