Question

The area of ​​one rectangle is equal to the area of ​​another rectangle, which is 6m in length.
Width is less than 4m. The area of ​​the third rectangle is the same as the original rectangle, with a length of 8 m and more
Width is less than 5m; So find the length and width of the original rectangle. today is my exam so please answer fast you will not know the answer please will not answer this question​

Answer 1

$\huge\bold\red{HOLA}$

Let the length and breadth of the original rectangle be “l” & “b” respectively.

It is given that, the area of the original rectangle is equal to the area of another rectangle with length 6 m more and breadth 4 m less, so we can write the eq. as,

lb = (l+6)(b-4)

⇒ lb = lb – 4l + 6b – 24

⇒ – 4l + 6b – 24 = 0 …… (i)

Also given that, the area of the original rectangle is equal to the area of the third rectangle with length 8 m more and breadth 5 m less, so we can write the eq. as,

lb = (l+8)(b-5)

⇒ lb = lb – 5l + 8b – 40

⇒ – 5l + 8b – 40 = 0 …… (ii)

Now, multiplying eq. (i) with 5 and eq. (ii) with 4 and then subtracting bothe equations, we get

-20l + 30b -120 = 0

-20l +32b – 160 = 0

+ - +

--------------------------------

2b = 40

---------------------------------

∴ b = 20 m

Substituting b = 20 m in eq. (i), we get

– 4l + (6*20) – 24 = 0

⇒ - 4l = -96

⇒ l = 24 m