The areas of the three surfaces meeting at one vertex of a cuboid are $700 \, cm^2$,$300 \, cm^2$,and $525 \, cm^2$ respectively. Find the dimensions and volume of the cuboid.

(N/A) Let the length,breadth,and height of the cuboid be $l \, cm$,$b \, cm$,and $h \, cm$ respectively.
The areas of the three adjacent faces meeting at a vertex are given by $lb$,$bh$,and $hl$.
Given:
$lb = 700 \, cm^2$ --- $(i)$
$bh = 300 \, cm^2$ --- $(ii)$
$hl = 525 \, cm^2$ --- $(iii)$
Multiplying $(i)$,$(ii)$,and $(iii)$:
$(lb) \times (bh) \times (hl) = 700 \times 300 \times 525$
$l^2 b^2 h^2 = 110,250,000$
$(lbh)^2 = (10,500)^2$
$lbh = 10,500 \, cm^3$ (This is the volume of the cuboid).
To find the dimensions:
From $(i)$ and $(iii)$,$lb = 700$ and $hl = 525$. Dividing $(i)$ by $(iii)$ gives $b/h = 700/525 = 4/3$,so $b = 4h/3$.
Substitute $b$ into $(ii)$:
$(4h/3) \times h = 300$
$h^2 = 300 \times 3 / 4 = 225$
$h = 15 \, cm$.
Now,$b = 4(15)/3 = 20 \, cm$.
And $l = 700/20 = 35 \, cm$.
Thus,the dimensions are $35 \, cm$,$20 \, cm$,and $15 \, cm$,and the volume is $10,500 \, cm^3$.