Question

The volume of a cuboid is 2520 cm3

. Find its height if its length is 15 cm and breadth is 21 cm​

Answer 1

Answer :-

• The height of the cuboid is 8cm.

Step-by-step explanation:

To Find :

• The height of a cuboid

Solution:

Given that,

• The volume of a cuboid = 2520cm³
• The length of cuboid = 15cm
• The breadth of cuboid = 21cm

$\large\underline{\frak{As \: we \: know \: that,}}$

$\dag \: \: \underline{\boxed{\bf{\pink{Volume \: of \: cuboid = lbh}}}}$

Where,

• l = Length
• b = Breadth
• h = Height

$\longmapsto \: \bf{lbh = Volume} \\ \\ \\ \\ \longmapsto \: \bf{l \times b \times h = 2520} \\ \\ \\ \\ \longmapsto \: \bf{15 \times b \times h = 2520} \\ \\ \\ \\ \longmapsto \: \bf{15 \times 21 \times h = 2520} \\ \\ \\ \\ \longmapsto \: \bf{315 \times h = 2520} \\ \\ \\ \\ \longmapsto \: \bf{h = \dfrac{2520}{315}} \\ \\ \\ \\ \longmapsto \: \bf{h = \cancel{ \dfrac{2520}{315}}} \\ \\ \\ \\ \longmapsto \: \boxed{\bf{ \red{h = 8}}}$

Hence, The height of the cuboid is 8cm.

Answer 2

$\frak{ \bigstar \: Diagram}$

$\setlength{\unitlength}{1cm}\begin{picture}\thicklines\multiput(0,0)(2.6,0){2}{\line(0,1){5}}\multiput(0,0)(0,5){2}{\line(1,0){2.6}}\multiput(1.4,1.4)(0,5){2}{\line(1,0){2.6}}\multiput(1.4,1.4)(2.6,0){2}{\line(0,1){5}}\multiput(0,0)(0,5){2}{\line(1,1){1.4}}\multiput(2.6,0)(0,5){2}{\line(1,1){1.4}}\put(-0.5,-0.4){\sf A}\put(1.4,0.85){\sf D}\put(2.8,-0.4){\sf B}\put(4.1,1){\sf C}\put(1.4,6.55){\sf H}\put(4.1,6.5){\sf G}\put(-0.4,5.1){\sf E}\put(2.85,4.7){\sf F}\put(1,-0.5){\sf 15 \: cm}\put(3.5,0.35){\sf 21 \: cm}\put(4.3,3.5){\sf  {?} \: cm }\end{picture}$

$\frak{ \bigstar \: Required \: Solution}$

Given:

• Volume of cuboid = 2520 cm³

• Length = 15 cm

• Breadth = 21 cm

To find:

• height of cuboid

We know:

$\bigstar \boxed{ \rm{}volume \: of \: cuboid = length \times width \times height}$

By using this formula we can find value of volume of cuboid

$\dashrightarrow \sf{}volume \: of \: cuboid = length \times width \times height$

$\dashrightarrow \sf{}2520= 15 \times 21 \times height$

$\dashrightarrow \sf{} 15 \times 21 \times height = 2520$

$\dashrightarrow \sf{} height =\frac{2520}{15 \times 21}$

$\dashrightarrow \sf{} height =\dfrac{2 \times 2 \times 3 \times 3 \times 7 \times 5 \times 2}{5 \times 3 \times 7 \times 3}$

$\dashrightarrow \sf{} height =\dfrac{2 \times 2 \times\cancel 3 \times \cancel 3 \times \cancel 7 \times \cancel5 \times 2}{ \cancel5 \times \cancel 3 \times \cancel 7 \times \cancel 3}$

$\dashrightarrow \sf{} height =2 \times 2 \times 2$

$\dashrightarrow \underline{ \boxed{ \sf{} height =8 \: cm}}$

$\therefore \underline{\sf{} \gray{ height \: of \: cuboid =}8 \: cm}$

$\textbf{Formulas related to SA and Volume :}$

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