Question

find lateral surface area and total surface area and volume of cube whose side are 10 cm.

Answer properly guys ​

Answer 1

Answer:

lsa =4a²

4×10 ×10 = 400 cm²

tsa = 6a²

6×10×10 =600cm²

volume =a³

10×10×10 =1000cm³

pls mark me brainliest

Answer 2

Answer:

Step-by-step explanation:

Question :-

• Find lateral surface area, total surface area, and volume of cube whose side is 10 cm.

To Find :-

• LSA of cube.
• TSA of cube.
• Volume of cube.

Solution :-

Given :

• Side = 10 cm

According to the question,

By using the formula,

${\sf{\longrightarrow LSA\ of\ cube\ =\ 4a^2}}$

Where,

• a = length of the side

Finding LSA of cube :

${\sf{\longrightarrow LSA\ of\ cube\ =\ 4a^2}}$

${\sf{\longrightarrow 4(10)^2}}$

${\sf{\longrightarrow 4(100)}}$

${\sf{\longrightarrow 4 \times 100}}$

${\sf{\longrightarrow 400\ cm^2}}$

$\therefore$ Hence, LSA of cube is 400 cm².

Now, let's find the TSA of cube.

$\rule{300}{2}$

By using the formula,

${\sf{\longrightarrow TSA\ of\ cube\ =\ 6a^2}}$

Where,

• a = length of the side

Finding TSA of cube :

${\sf{\longrightarrow TSA\ of\ cube\ =\ 6a^2}}$

${\sf{\longrightarrow 6(10)^2}}$

${\sf{\longrightarrow 6(100)}}$

${\sf{\longrightarrow 6 \times 100}}$

${\sf{\longrightarrow 600\ cm^2}}$

$\therefore$ Hence, TSA of cube is 600 cm².

Finally, let's find the volume of cube.

$\rule{300}{2}$

By using the formula,

${\sf{\longrightarrow Volume\ of\ cube\ =\ a^3}}$

Where,

• a = length of the side

Finding volume of cube :

${\sf{\longrightarrow Volume\ of\ cube\ =\ a^3}}$

${\sf{\longrightarrow (10)^3}}$

${\sf{\longrightarrow 10 \times 10 \times 10}}$

${\sf{\longrightarrow 1,000\ cm^3}}$

$\therefore$ Hence, volume of cube is 1,000 cm³.

More To Know :-

$\begin{array}{|c|c|c|}\cline{1-3}\bf Shape&\bf Volume\ formula&\bf Surface\ area formula\\\cline{1-3}\sf Cube&\tt l^3}&\tt 6l^2\\\cline{1-3}\sf Cuboid&\tt lbh&\tt 2(lb+bh+lh)\\\cline{1-3}\sf Cylinder&\tt {\pi}r^2h&\tt 2\pi{r}(r+h)\\\cline{1-3}\sf Hollow\ cylinder&\tt \pi{h}(R^2-r^2)&\tt 2\pi{rh}+2\pi{Rh}+2\pi(R^2-r^2)\\\cline{1-3}\sf Cone&\tt 1/3\ \pi{r^2}h&\tt \pi{r}(r+s)\\\cline{1-3}\sf Sphere&\tt 4/3\ \pi{r}^3&\tt 4\pi{r}^2\\\cline{1-3}\sf Hemisphere&\tt 2/3\ \pi{r^3}&\tt 3\pi{r}^2\\\cline{1-3}\end{array}$