A box is in the form of a cube. Its edge is 5 m long. Find
(a) the total length of the edges
(b) the cost of painting the outside of the box, on all the surfaces, at the rate of Rs. 5 per m sq.
(c) the volume of liquid which the box can hold.

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Solution

(a) Length of edges = Number of edeges of base $\times$ 3 $\times$ Length of each edge
$= 4 \times 3 \times 5 = 60$ $m$

(b) TSA = $6 a^{2} = 6 \times 25 = 150$ $m^{2}$
Therefore, cost of painting = $150 \times 5 = R s . 750$

(c) Volume = $a^{3} = 5^{3} = 125$ $m^{3}$

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A box is in the form of a cube. Its edge is 5 m long. Find(a) the total length of the edges(b) the cost of painting the outside of the box, on all the surfaces, at the rate of Rs. 5 per m sq.(c) the volume of liquid which the box can hold.

(a) Length of edges = Number of edeges of base × 3 × Length of each edge=4×3×5=60 m(b) TSA = 6a2=6×25=150 m2Therefore, cost of painting = 150×5=Rs.750(c) Volume = a3=53=125 m3

A box is in the form of a cube. Its edge is 5 m long. Find
(a) the total length of the edges
(b) the cost of painting the outside of the box, on all the surfaces, at the rate of Rs. 5 per m sq.
(c) the volume of liquid which the box can hold.

(a) Length of edges = Number of edeges of base $\times$ 3 $\times$ Length of each edge
$= 4 \times 3 \times 5 = 60$ $m$

(b) TSA = $6 a^{2} = 6 \times 25 = 150$ $m^{2}$
Therefore, cost of painting = $150 \times 5 = R s . 750$

(c) Volume = $a^{3} = 5^{3} = 125$ $m^{3}$