Question

a teak wood log is first cut in the form of cuboid of length 2.3m ,width 0.7 and of certain thickness its volume is 1.104m how many rectangular planks of size 2.3m*0.75m*0.04m can be cut fro cuboid?

Answer 1

As Volume = l*b*h

⇒1.104 = 2.3*0.7* h

⇒height, h = 1.104/1.61 = 0.6857

Now the cuboid is cut into cubes of size 2.3m*0.75m*0.04m, then Volume(V1) = 0.069m^3

Let the given cuboid is cut into n cuboids then volume of the given cuboid = n*vol of smaller cuboid

   n = V/V1 = 1.104/0.069 = 16

Answer 2

AnsWer:

Let the thickness of the log be x metres. Then,

$\qquad\sf\underline{Volume\:=1.104\:m^3}$

$\sf = 2.3 \times 0.75 \times x = 1.104 \\ \\ = \sf \: x = \frac{1.104}{2.3 \times 0.75} \\ \\ \sf = 0.64 \: m$

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Since the length and breadth of each rectangular plank is the same as that of the cuboid.

$\therefore$$\sf\underline{No.\:of\:rectangular\:park}$

$\sf = \frac{thickness \: of \: cuboid}{thickness \: of \: each \: plank} \\ \\ \sf = \frac{0.64}{0.04} = \frac{64}{4} = 16$

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ALITER :

$\sf\underline{Number\:of\:rectangular\:parks :-}$

$\sf = \frac{volume \: of \: the \: cuboid}{volume \: of \: a \: plank} \\ \\ \sf = \frac{1.104}{0.069} = \frac{1104}{69} = 16$

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