Question

find the unknown measurements of these cuboids l=10cm,h=8cm,volume=440 cm3,find b.​

Answer 1

Step-by-step explanation:

Volume of cuboid= l*b*h

440= 10*8*b

5.5= b

Hence, breadth= 5.5cm

Answer 2

Answer:

Diagram :

$\setlength{\unitlength}{0.74 cm}\begin{picture}\thicklines\put(5.6,5.4){\bf}\put(11.1,5.4){\bf}\put(11.2,9){\bf}\put(5.3,8.6){\bf}\put(3.3,10.2){\bf}\put(3.3,7){\bf}\put(9.25,10.35){\bf}\put(9.35,7.35){\bf}\put(3.5,6.1){\sf{5.5\:cm}}\put(7.7,6.3){\sf{10\:cm}}\put(11.3,7.45){\sf{8\:cm}}\put(6,6){\line(1,0){5}}\put(6,9){\line(1,0){5}}\put(11,9){\line(0,-1){3}}\put(6,6){\line(0,1){3}}\put(4,7.3){\line(1,0){5}}\put(4,10.3){\line(1,0){5}}\put(9,10.3){\line(0,-1){3}}\put(4,7.3){\line(0,1){3}}\put(6,6){\line(-3,2){2}}\put(6,9){\line(-3,2){2}}\put(11,9){\line(-3,2){2}}\put(11,6){\line(-3,2){2}}\end{picture}$

• ⇢ See the Diagram from website Brainly.in.

$\begin{gathered}\end{gathered}$

Given :

• ⇢ Lenght of cuboid = 10 cm
• ⇢ Height of cuboid = 8 cm
• ⇢ Volume of cuboid = 440 cm³.

$\begin{gathered}\end{gathered}$

To Find :

• ⇢ Breadth of cuboid

$\begin{gathered}\end{gathered}$

Using Formula :

${\longrightarrow{ \underline{\boxed{\sf{Volume \: of \: cuboid = lbh}}}}}$

• ⇢ l = length
• ⇢ b = breadth
• ⇢ h = height

$\begin{gathered}\end{gathered}$

Solution :

Finding the breadth of cuboid by substituting the values in the formula :

$\begin{gathered}\quad{\dashrightarrow{\sf{Volume \: of \: cuboid = lbh}}}\\\\ \quad{\dashrightarrow{\sf{Volume \: of \: cuboid = l \times b \times h}}} \\ \\ {\dashrightarrow{\sf{440 = 10 \times b \times 8}}} \\ \\ {\dashrightarrow{\sf{440 = 80\times b}}} \\ \\ {\dashrightarrow{\sf{440 = 80b}}} \\ \\ {\dashrightarrow{\sf{b = \dfrac{440}{80} }}} \\ \\ {\dashrightarrow{\sf{b = \cancel{\dfrac{440}{80}}}}} \\ \\ {\dashrightarrow{\sf{b = 5.5 \: cm}}} \\ \\ {\bigstar{\underline{\boxed{\sf{\purple{Breadth = 5.5 \:cm}}}}}}\end{gathered}$

Hence, the breadth of cuboid is 5.5 cm.

$\begin{gathered}\end{gathered}$

Vefication :

$\begin{gathered}\quad{\dashrightarrow{\sf{Volume \: of \: cuboid = lbh}}}\\\\ \quad{\dashrightarrow{\sf{Volume \: of \: cuboid = l \times b \times h}}} \\ \\ {\dashrightarrow{\sf{440 = 10 \times 5.5 \times 8}}} \\ \\{\dashrightarrow{\sf{440 = 10 \times \dfrac{55}{10} \times 8}}} \\ \\ {\dashrightarrow{\sf{440 = \cancel{10} \times \dfrac{55}{\cancel{10}} \times 8}}} \\ \\ {\dashrightarrow{\sf{440 = 1 \times 55 \times 8}}} \\ \\ {\dashrightarrow{\sf{440 = 55 \times 8}}} \\ \\{\dashrightarrow{\sf{440 \: {cm}^{2} = 440 \: {cm}^{2} }}} \\ \\ {\bigstar{\underline{\boxed{\textsf{\purple{LHS = RHS}}}}}}\end{gathered}$

Hence Verified!

$\begin{gathered}\end{gathered}$

Learn More :

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

$\rule{220pt}{3.5pt}$