Question

if each edge of the cube is decreased by 50 percent , the percentage decreased in its surface area and volume are ​

Answer 1

Answer: A) 125%

Explanation:

Let the edge = a cm

So increase by 50 % = a + a/2 = 3a/2

Total surface Area of original cube = 6a2

TSA of new cube = 6(3a2)2 =6(9a24)= 13.5a2

Increase in area = 13.5a2−6a2 =7.5a2

7.5a2 Increase % =7.5a26a2×100 = 125%

Answer 2

Answer:

The percentage decreased in its surface area and volume are ​$125$ %

Step-by-step explanation:

Given:

Let,

$a \rightarrow a$ Initial edge of the cube

$A \rightarrow$ Initial surface area of the cube

$a^{\prime} \rightarrow$ Increased edge of the cube

$A^{\prime} \rightarrow$ Increased surface area of the cube

Find:

We have to find the percentage increase in the surface area of the cube

Since it's given that

$\begin{aligned}a &=a+a \times \frac{50}{100} \\&=\frac{3}{2} a\end{aligned}$

We have,

$A \prime &=6(a)^{2}$

$&=6\left(\frac{3}{2} a\right)^{2} \quad\left$

$&=\frac{9}{4}\left(6 a^{2}\right)$

$&=\frac{9}{4} A$

Percentage increase in surface area,

$\begin{aligned}&=\frac{A \prime-A}{A} \times 100 \\&=\frac{\frac{9}{4} A-A}{A} \times 100 \\&=\frac{\frac{5}{4} A}{A} \times 100 \\&=125\end{aligned}$

Increase in surface area is $\mathbf{1 2 5 \%} .$