Question

The area of a rectangle is 192 sq cm and its perimeter is 56 cm . Find the dimensions of the rectangle .

Answer 1

The area of a rectangle = l x b = 192 cm(sq) 
= l = 192/b 
The perimeter = 2l + 2b = 56 
= 2*(192/b) + 2b = 56 
= (384/b) +2(b^2) =384 
rearranging&dividing by 2 = b^2 - 28b + 192 =0 
= b^2 - 16b - 12b + 192 = 0 
= b(b - 16) - 12(b - 16) = 0 
= (b - 16) (b -12) =0 
so b-16=0 i.e b= 16 
or 
b-12 = 0 i.e b = 12 
if l = 12 then b = 16 or vice versa 

area = l * b = 12 * 16 = 192 
peri = (2 *12)+(2 * 16) = 24 +52 = 56

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Answer 2

given area of rectangle = 192 sq cm
            i.e., l×b = 192 ------------------ 1
given its perimeter is 56 cm
               2(l+b) = 56
                l+b = 28
                b = 28 - l -----------------------------2
substitute 2 in 1
then we get  l(28-l) = 192
                  l² - 28l + 192 = 0
                 l² -16l - 12l +192 = 0
                 l(l-16) -12(l-16) = 0
                 l-12 = 0 or l-16 = 0
then l= 12 or 16
by substituting them in 1 we get
                  12 × b = 192 or  16 × b = 192
   then b = 16 or 12
hence if l=12 then b=16
                   or
if l=16 then b=12