Question

If 42 cubes of a unit side are stuck together to form a rectangular brick of base perimeter 18 cm, what is the height (in cms) of the brick? (l, b, and h are integers)

Answer 1

Answer:

Step-by-step explanation:

18= 2(width + length)

9= width + length

width and length can be

1+8, 2+7, 3+6 or 4+5

42/ (1x8) = 5.25 which isn't an integer

42/(2x7) = 3 which is an integer

Therefore the height is 3

Answer 2

Given are the volume and base perimeter of a rectangular brick, Find its height.

Explanation:

• The volume of each cube of unit side length is $1\ cm^3$.
• Since, $42$ such cubes are used to make up the brick the volume 'V' of the brick is    $V=42\ cm^3$.
• Now let the length, breadth, and height of the brick be denoted by $l,\ b,\ h$.  
• Hence, from the given base perimeter of the brick we get,
•                       $->P=2(l+b)\\->18=2(l+b)\\->l+b=9$  
• Since the cubes are used to make the brick the dimensions of the brick should be integers.
• Hence, possible values of 'l' and 'b' are, $(l,b)->(1,8),\ (2,7)\ (3,6),\ (4,5)$  
• Now for each pair of lengths and breadths to give the volume of $42\ cm^3$ we get,
•                       $->V=lbh\ \ \ \ \ \ \therefore h=\frac{V}{lb} \\->(l=1,b=8)\ \ \therefore h=\frac{42}{8} = 5.25\ cm \\\\->(l=2,b=7)\ \ \therefore h=\frac{42}{14} = 3\ cm\\\\-> (l=3,b=6)\ \ \therefore h=\frac{42}{18} = 2.33\ cm\\\\->(l=4,b=5)\ \ \therefore h=\frac{42}{20} = 2.1\ cm$  
• Hence the possible value of height of the brick is $3\ cm.$