Arectangular field is 100 meters wide and 140meters long. give the length and width of another rectangular field that has the same perimeter but a smaller area.

The length of rectangle 2  = 80 m

The width of rectangle 2  = 160 m

Step-by-step explanation:

The width of rectangle 1  = 100 m

The length of rectangle 1  = 140 m

Now the PERIMETER OF A RECTANGLE = 2 (L+B)

So, here the Perimeter of R 1 =  2 ( 100 + 140)  = 2 x 240  = 480 m

Also, AREA OF THE RECTANGLE = LENGTH x WIDTH

So, here the Area of R 1 =  100 x 140  = 14,000  sq m  

Now, in the Rectangle  2:

Let us assume the width of rectangle 2  = x

Assume the length of rectangle 2  = y  

Now,Perimeter of rectangle 2= Perimeter of rectangle 1

⇒ 2 (x + y)  = 480 m ......... (1)

Also, Area of rectangle 2   <  Area of rectangle 1

⇒ x y  < 14,000  .....  (2)

Now, solving both  the equations, we get

(x +y ) = 240  ⇒  y = 240 - x

x y < 14,000

Substituting y = 240 - x in (2), we get:

x (240-x) < 14,000

or,

Solving the above equation, we get

x < 140

Now, if we take x  = 80, then y = 160

So, for x = 80, y = 160,

Perimeter = 2 ( 80 + 160) = 240 m

Area = x y  = 80 x 160  = 12,800 < 14,000

Hence, one possible pair of solution for (x,y) =(80,160)

the answer is 2/54

step-by-step explanation: