Question

If each of the dimensions of a cubiod is increased by 50%, then find increase in the volume of a cubiod. Also find the ratio of volumes before and after the dimensions are increased​

Answer 1

Answer:

option is B)

Old Surface Area =2(lb+bh+lh)

New dimensions =(l+50%l)×(b+50%b)×(h+50%h)

=

2

3

l×

2

3

b×

2

3

h

New Surface Area =2(

2

3

l×

2

3

b+

2

3

h×

2

3

b+

2

3

l×

2

3

h)

=2×

4

9

(lb+bh+lh)

Change in surface area = New S.A.− Old S.A.

=2×

4

5

(lb+bh+lh)

Percentage change in surface area =

2×(lb+bh+lh)

2×

4

5

(lb+bh+lh)

×100 %

=125 %