Question

The width of a box is two-third of its length and height is one-third of its length. If the volume of the box is 48cm^3, find the area and base of the box.

Answer 1

FINAL ANSWER:-

area of box = $88cm^{2}$

base of box $=l*b$ ⇒ $24cm^{2}$

$l=6cm\\b=4cm\\h=2cm$

GIVEN:-

width of a box is two-third of its length=$\frac{2}{3} l$

height is one-third of its length=$\frac{l}{3}$

TO FIND:-

• area

• base

FORMULAS USED:-

• $volume \:of\:cuboid=l*b*h$

• $TSA\:of\:cuboid=2(lb+lh+bh)$

SOLUTION:-

width of box = $\frac{2l}{3}$

height of box =$\frac{l}{3}$

length of box = $l$

$volume \:of\:box=l*b*h$

$48= l*\frac{2}{3} l*\frac{1}{3} l\\48=\frac{l*2l*l}{9} \\48=\frac{2l^{3}}{9} \\2l^{3}=48*9\\2l^{3}=432\\l^{3}=\frac{432}{2} \\l^{3}=216\\l=\sqrt[3]{216} \\l=6\:cm$

hence as we have find out length so let's find width and height

width of a box is two-third of its length=$\frac{2}{3} l$ ⇒ $\frac{2*6}{3}$ ⇒ $4 \:cm$

height is one-third of its length=$\frac{l}{3}$ ⇒ $\frac{6}{3}$ ⇒ $2\:cm$

here,

l=6cm

b=4cm

h=2cm

area of box = $TSA\:of\:cuboid=2(lb+lh+bh)$

$=2( 6 * 4 +6* 2+ 4 * 2 )\\= 2 ( 24 +12 + 8 )\\= 2(44) \\=88cm^{2}$

base of the box = $l*b$

$=6*4\\=24cm^{2}$

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some important formulas are:-

$TSA\:of\:cylinder=2\pi rh +2\pi r^{2}\\TSA\:of\:hemisphere=3\pi r^{2}\\TSA\:of\:sphere=4\pi r^{2}\\TSA\:of\:cone=\pi r(r+l)\:\:or \:\:\pi rl + \pi r^{2}\\TSA\:of\:cube=6s^{2}\\TSA\:of\:cuboid=2(lb+lh+bh)\\\\CSA\:of\:cylinder=2\pi rh\\CSA\:of\:hemisphere=2\pi r^{2}\\CSA\:of\:sphere=4\pi r^{2} (same\:as\:TSA\:of\:sphere)\\CSA\:of\:cone=\pi rl\\CSA\:of\:cube=4s^{2}\\CSA\:of\:cuboid=2(lb+bh)$

Answer 2

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