Question

A Volume of cuboidal box is 250cm^2. If its length is 20cm and height is 5cm, find its breadth​

Answer 1

Given -

• Volume of cuboidal box = 250cm²

• Length of box = 20cm

• Height of box = 5cm

To find -

• Breadth of the Cuboidal box.

Formula used -

• Volume of cuboid = L × B × H

Solution -

Here, we are given with the volume, height and length of cuboidal box and we need to find the height, for that we will use the formula of volume of cuboid.

Let the breadth of box be termed as B

Now,

Volume of cuboid = L × B × H

where,

L = length

B = Breadth

H = Height

On substituting the values -

Volume = L × B × H

250cm² = 20cm × B × 5cm

250cm² = 100cm × B

B = $\sf\cancel\dfrac{250cm}{100cm}$

B = 2.5cm

Verification -

Volume = L × B × H

250cm² = 20cm × 2.5cm × 5cm

250cm² = 250cm²

$\therefore$ Breadth of the Cuboidal box is 2.5cm

_____________________________________________

Answer 2

${\bold{\sf{\underline{Understanding \: the \: concept}}}}$

❉ This question says that there is a cuboidal shape box given its volume is 250 cm² And now this question says that it's length is 20 cm and it's height is 5 cm We have to find it's breadth.

$\setlength{\unitlength}{0.74 cm}\begin{picture}\thicklines\put(5.6,5.4){\bf A}\put(11.1,5.4){\bf B}\put(11.2,9){\bf C}\put(5.3,8.6){\bf D}\put(3.3,10.2){\bf E}\put(3.3,7){\bf F}\put(9.25,10.35){\bf H}\put(9.35,7.35){\bf G}\put(3.5,6.1){\sf 2.5\:cm}\put(7.7,6.3){\sf 20\:cm}\put(11.3,7.45){\sf 5\:cm}\put(6,6){\line(1,0){5}}\put(6,9){\line(1,0){5}}\put(11,9){\line(0,-1){3}}\put(6,6){\line(0,1){3}}\put(4,7.3){\line(1,0){5}}\put(4,10.3){\line(1,0){5}}\put(9,10.3){\line(0,-1){3}}\put(4,7.3){\line(0,1){3}}\put(6,6){\line(-3,2){2}}\put(6,9){\line(-3,2){2}}\put(11,9){\line(-3,2){2}}\put(11,6){\line(-3,2){2}}\end{picture}$

${\bold{\sf{\underline{Given \: that}}}}$

❇ Volume of cuboidal box = 250 cm²

❇ Length of cuboidal box = 20 cm.

❇ Height of cuboidal box = 5 cm.

${\bold{\sf{\underline{To \: find}}}}$

❇ Breadth of cuboidal box.

${\bold{\sf{\underline{Solution}}}}$

❇ Breadth of cuboidal box = 2.5 cm

${\bold{\sf{\underline{Using \: concept}}}}$

❇ Volume of cuboid formula

${\bold{\sf{\underline{Using \: formula}}}}$

❇ V = L × H × B

${\bold{\sf{\underline{These \: also \: means}}}}$

❉ Breadth as B

❉ Height as H

❉ Length as L

❉ Volume as V

${\bold{\sf{\underline{Full \: solution}}}}$

❇ ${\bold{\sf{V}}}$ = ${\bold{\sf{L \times B \times H}}}$

❇ ${\bold{\sf{250}}}$ = ${\bold{\sf{20 \times B \times 5}}}$

❇ ${\bold{\sf{250}}}$ = ${\bold{\sf{100 \times B}}}$

❇ ${\bold{\sf{\dfrac{250}{100}}}}$ = ${\bold{\sf{B}}}$

❇ ${\bold{\sf{\dfrac{25}{10}}}}$ = ${\bold{\sf{B}}}$

❇ ${\bold{\sf{\dfrac{5}{2}}}}$ = ${\bold{\sf{B}}}$

❇ ${\bold{\sf{2.5}}}$ = ${\bold{\sf{B}}}$

❇ ${\bold{\sf{B}}}$ = ${\bold{\sf{2.5 \: cm}}}$

${\bold{\sf{\underline{Knowledge \: booster}}}}$

Diagram of a cuboid :

$\setlength{\unitlength}{0.74 cm}\begin{picture}\thicklines\put(5.6,5.4){\bf A}\put(11.1,5.4){\bf B}\put(11.2,9){\bf C}\put(5.3,8.6){\bf D}\put(3.3,10.2){\bf E}\put(3.3,7){\bf F}\put(9.25,10.35){\bf H}\put(9.35,7.35){\bf G}\put(3.5,6.1){\sf x\:cm}\put(7.7,6.3){\sf y\:cm}\put(11.3,7.45){\sf z\:cm}\put(6,6){\line(1,0){5}}\put(6,9){\line(1,0){5}}\put(11,9){\line(0,-1){3}}\put(6,6){\line(0,1){3}}\put(4,7.3){\line(1,0){5}}\put(4,10.3){\line(1,0){5}}\put(9,10.3){\line(0,-1){3}}\put(4,7.3){\line(0,1){3}}\put(6,6){\line(-3,2){2}}\put(6,9){\line(-3,2){2}}\put(11,9){\line(-3,2){2}}\put(11,6){\line(-3,2){2}}\end{picture}$

Request :

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