How many different combinations of 3 numbers for a 4 number code artemis?
1 answer:
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Answer:
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)