Answers

2014-10-08T20:26:52+05:30
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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2014-10-09T12:32:27+05:30
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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