Question

Find the areas of the rectangles with the following pairs of monomials as their length and breadth respectively.
$
  \left( i \right)\left( {p,q} \right) \\
  \left( {ii} \right)10m,5n \\
  \left( {iii} \right)20{x^2},5{y^2} \\
  \left( {iv} \right)\left( {4x,3{x^2}} \right) \\
  \left( v \right)3mn,4np \\
$

Answer 1

Hint: Use the area of the rectangle formula when the length and breadth of the rectangle is given.
The area of the rectangle is always in square units.
Formula used: Area of a rectangle= $l \times b$ square units, where ‘l’ is the length and ‘b’ is the breadth of the rectangle.

Complete step-by-step solution:
$\left( i \right)\left( {p,q} \right)$
Here, given the length and the breadth of the rectangle is ‘p’ and ‘q’ respectively.
Area of the rectangle= $l \times b$ , where $l = p,b = q$
Area Of the rectangle= $p \times q$
$ = pq$ square units
\[\left( {ii} \right)10m,5n\]
Here, given the length and the breadth of the rectangle is ‘10m’ and ‘5n’ respectively.
Area of the rectangle= $l \times b$ , where $l = 10m,b = 5n$
Area= $10m \times 5n$
‘10m’ can be written as $10 \times {m^1} \times {n^0}$
‘5n’ can be written as $5 \times {m^0} \times {n^1}$
$
  10m \times 5n = 10 \times {m^1} \times {n^0} \times 5 \times {m^0} \times {n^1} \\
   = 10 \times 5 \times {m^1} \times {m^0} \times {n^0} \times {n^1} \\
   = 50 \times {m^{1 + 0}} \times {n^{0 + 1}} \\
  \left( {\because {a^m} \times {a^n} = {a^{m + n}}} \right) \\
   = 50 \times {m^1} \times {n^1} \\
   = 50mn \\
$
Therefore, the area of the rectangle with sides \[10m,{\text{ }}5n\] is $50mn$ square units.
$\left( {iii} \right)20{x^2},5{y^2}$
Here, given the length and the breadth of the rectangle is $20{x^2},5{y^2}$ respectively.
Area of the rectangle= $l \times b$ , where $l = 20{x^2},b = 5{y^2}$
    $
   = 20{x^2} \times 5{y^2} \\
   = 20 \times {x^2} \times 5 \times {y^2} \\
   = 20 \times 5 \times {x^2} \times {y^2} \\
   = 100{x^2}{y^2} \\
$
Therefore, the area of the rectangle with sides $20{x^2},5{y^2}$ is \[100{x^2}{y^2}\] square units.
$\left( {iv} \right)4x,3{x^2}$
Here, we are given with the length and breadth of the rectangle as $4x,3{x^2}$ respectively.
Area of the rectangle= $l \times b$ , where $l = 4x,b = 3{x^2}$
Area of the rectangle= $4x \times 3{x^2}$
$
   = 4 \times {x^1} \times 3 \times {x^2} \\
   = 4 \times 3 \times {x^1} \times {x^2} \\
   = 12 \times {x^{1 + 2}} \\
  \left( {\because {a^m} \times {a^n} = {a^{m + n}}} \right) \\
   = 12 \times {x^3} \\
   = 12{x^3} \\
$
Therefore, the area of the rectangle with sides $4x,3{x^2}$ is $12{x^3}$ square units.
$\left( v \right)3mn,4np$
Here we are given with 3mn, 4np as the length and breadth of the rectangle respectively.
Area of the rectangle= $l \times b$ , where $l = 3mn,b = 4np$
Area of the rectangle = $3mn \times 4np$
‘3mn’ can be written as $3 \times {m^1} \times {n^1} \times {p^0}$
‘4np’ can be written as $4 \times {m^0} \times {n^1} \times {p^1}$
$
  3mn \times 4np = 3 \times {m^1} \times {n^1} \times {p^0} \times 4 \times {m^0} \times {n^1} \times {p^1} \\
   = 3 \times 4 \times {m^1} \times {m^0} \times {n^1} \times {n^1} \times {p^0} \times {p^1} \\
   = 12 \times {m^{1 + 0}} \times {n^{1 + 1}} \times {p^{0 + 1}} \\
  \left( {\because {a^m} \times {a^n} = {a^{m + n}}} \right) \\
   = 12 \times {m^1} \times {n^2} \times {p^1} \\
   = 12m{n^2}p \\
$
Therefore, the area of the rectangle with sides $3mn,4np$ is $12m{n^2}p$ square units.

Note: When the units of the measurements are not mentioned, just consider them as units like x units, y units and after calculating the area square units must be mentioned after the area value. If volume is asked in the question, then mention cube units.