Question

8. Three cubes having edges 18 cm, 24 cm and 30 cm respectively are melted and made into a new cube. Find the edge of the new cube so formed. ​

Answer 1

Answer

Volume of cube of length 18 cm = 18 × 18 × 18 = 5832

Volume of cube of length 24 cm = 24 × 24 × 24 = 13824

Volume of cube of length 30 cm = 30 × 30 × 30 = 27000

Therefore, total Volume of three cubes = 5832 + 13824 + 27000 = 46656

Now , cube root of 46656 will give us the edge of new cube. cube root of 46656 = 36 cm

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

Three cubes having edges 18 cm, 24 cm and 30 cm respectively are melted and made into a new cube.

Let assume that,

• Edge of first cube, x = 18 cm

• Edge of second cube, y = 24 cm

• Edge of third cube, z = 30 cm

Now, further assume that

• The edge of new cube is a cm

Now, we know that, when any object is melted and recast into a another object, then volume of first object is equals to volume of other object.

So, using this

$\rm \: Volume_{(Cube \: 1)} + Volume_{(New\:Cube \: 2)} + Volume_{(New\:Cube \: = )} = Volume_{(New\:Cube)}$

$\rm \: {18}^{3} + {24}^{3} + {30}^{3} = {a}^{3}$

$\rm \: {(6 \times 3)}^{3} + {(6 \times 4)}^{3} + {(6 \times 5)}^{3} = {a}^{3}$

$\rm \: {6}^{3} \: [ {3}^{3} + {4}^{3} + {5}^{3}] = {a}^{3}$

$\rm \: {6}^{3} \: [ 27 + 64 + 125] = {a}^{3}$

$\rm \: {6}^{3} \times [ 216] = {a}^{3}$

$\rm \: {6}^{3} \times {6}^{3} = {a}^{3}$

$\rm \: {(6 \times 6)}^{3} = {a}^{3}$

$\rm \: {36}^{3} = {a}^{3}$

$\rm\implies \:a = 36 \: cm$

Hence, the edge of the new cube thus formed = 36 cm

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{CSA_{(cylinder)} = 2\pi \: rh}\\ \\ \bigstar \: \bf{Volume_{(cylinder)} = \pi {r}^{2} h}\\ \\ \bigstar \: \bf{TSA_{(cylinder)} = 2\pi \: r(r + h)}\\ \\ \bigstar \: \bf{CSA_{(cone)} = \pi \: r \: l}\\ \\ \bigstar \: \bf{TSA_{(cone)} = \pi \: r \: (l + r)}\\ \\ \bigstar \: \bf{Volume_{(sphere)} = \dfrac{4}{3}\pi {r}^{3} }\\ \\ \bigstar \: \bf{Volume_{(cube)} = {(side)}^{3} }\\ \\ \bigstar \: \bf{CSA_{(cube)} = 4 {(side)}^{2} }\\ \\ \bigstar \: \bf{TSA_{(cube)} = 6 {(side)}^{2} }\\ \\ \bigstar \: \bf{Volume_{(cuboid)} = lbh}\\ \\ \bigstar \: \bf{CSA_{(cuboid)} = 2(l + b)h}\\ \\ \bigstar \: \bf{TSA_{(cuboid)} = 2(lb +bh+hl )}\\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$