Question

please help me to solve both the questions






please

Answer 1

this pic is of sol2

sol 3.

Assume the initial value of edge was xx
So, Initial Surface Area, A1=6x2A1=6x2
The edge increased by 50%, so, the final edge length is, 3x/23x/2
So, final Surface Area, A2=6(3x/2)2=6(9x2/4)=6x2(9/4)=9/4(A1)A2=6(3x/2)2=6(9x2/4)=6x2(9/4)=9/4(A1)
So, Increase in Surface Area, dA=A2−A1=9/4(A1)−A1=A1(9/4−1)=A1(5/4)dA=A2−A1=9/4(A1)−A1=A1(9/4−1)=A1(5/4)
So, the area increased by 5/45/4
Now, percentage increase will be ((5/4)∗100=1255/4)∗100=125%
So, the percentage increase in the Surface Area of the cube will be 125%.
I hope my answer was helpful.