Find the dimensions of a rectangle whose perimeter is 28 m and area is 40 sq.m
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Let l,b are length and breadth of the rectangle.
Given that perimeter = 28m => 2[l+b] =28 => l+b =14.------->(1)
Area = 40 => lb =40 => b=40/l.----------->(2).
From (1) & (2),l+40/l = 14 => l2(it is l square)+40= 14l => l2-14l+40 =0.
=> l2 -10l-4l+40= 0.
=>l(l-10)-4(l-10) =0.
=>(l-10)(l-4) =0.
=> l-10 =0 (or) l-4= 0.
=> l=10 (or) l=4.
CASE (1) : IF l = 10 then b= 40/l = 40/10 = 4m.
CASE (2): IF l= 4 then b= 40/l = 40/4 = 10m.
Therefore, the length and breadth of a rectangle are 10m and 4m respectively.
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let length of the rectangle = l , breadth of the rectangle = b, according to the question , perimeter = 28m => 2(l+b) = 28 => l+b = 14 ---(1) , also area = 40 sq .m => lb = 40---(2) , from (1) l = 14 - b ---(3) , (3)in(2) => lb = 40 , (14 - b ) b = 40 , 14.b - b(square) = 40 , 14b - b.b - 40 = 0 , -b.b +14b - 40 = 0 ,multilying throughout by (-) b.b - 14b + 40 = 0 , b.b - 10b - 4b +40 = 0 , b(b-10)-4(b-10) = 0 , (b-4)(b-10) = 0 , so b could be either 10 or 4. if b is 4 then from (1) l = 14 -4= 10. or if b is 10, l = 4 , . so the dimensions of the rectangle are 10 and 4
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