A square is altered so that one dimension is increased by 5 meters and the other dimension is increased by 3 meters. If the area of the resulting rectangle is 99 square meters, find the area of the original square
Answers
Step-by-step explanation:
Given:-
A square is altered so that one dimension is increased by 5 meters and the other dimension is increased by 3 meters. If the area of the resulting rectangle is 99 square meters.
To find:-
Find the area of the original square ?
Solution:-
Let the length of the side of the square be X m
If square is altered
Length of the rectangle = X is Increased by 5 m
=> (X+5) m
Breadth of the rectangle =X is increased by 3 m
=> (X+3) m
Area of a rectangle = lb sq.units
=> (X+5)(X+3) sq.m
=> X(X+3)+5(X+3)
=> X^2+3X+5X+15
=>X^2+8X+15 sq.m
According to the given problem
Area of the rectangle = 99 m^2
=> X^2+8X+15 sq.m = 99 sq.m
=> X^2+8X+15 = 99
=> X^2+8X+15-99 = 0
=>X^2+8X-84 = 0
=> X^2+14X-6X-84 = 0
=> X(X+14) -6(X+14) = 0
=> (X+14)(X-6) = 0
=> X+14 = 0 or X-6 = 0
=>X = -14 or X = 6
But X can not be negative since it is a length of the side
X≠-14 m
Therefore X = 6 m
Length of the side of the square = 6 m
Area of a square = Side×Side sq.units
=>Area of the given square = 6×6 sq.m
=> 36 sq.m
Answer:-
Area of the original square is 36 m^2 for the given problem
Used formulae:-
- Area of a rectangle = lb sq.units
- Area of a square = Side×Side sq.units
- Area of a square = a^2 sq.units
- a = length of the side of a square
- l=length
- b=breadth