Question 71

If three equal cubes are placed adjacently in a row, then the ratio of the total surface area of the new cuboid to that of the sum of the surface areas of the three cubes will be

Solution

Let edge of each cube be $$a=1$$ cm and length, width and height of cuboid be $$l=3$$ cm, $$b=1$$ cm and $$h=1$$ cm

Total surface area of cuboid = $$2(lb+bh+hl)$$

= $$2\times(3+1+3)=14$$ $$cm^2$$

Total surface area of each cube = $$3\times6a^2 = 18\times(1)^2=18$$ $$cm^2$$

$$\therefore$$ Required ratio = $$\frac{14}{18}=7:9$$

=> Ans - (D)

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