Question

c) A rectangle has the same area as another, whose length is 6 m more and breadth
is 4 m less. It has also the same area as the third, whose length is 8 m more and
breadth 5 m less. Find the length and breadth of the original rectangle.

Answer 1

Length of original rectangle is 24 m and width of original rectangle is 20 m.

Step-by-step explanation:

1. Let length of original rectangle = L

        Width of original rectangle = B

        Area of first rectangle ($A_{1}$) = L× B

2. Now from given question

      length of second rectangle = L +6

      Width of second rectangle = B - 4

      Area of second rectangle ($A_{2}$) = (L + 6)× (B -4)

3. Also for third rectangle

    length of third rectangle = L + 8

    Width of third rectangle = B - 5

    Area of third rectangle($A_{3}$) = (L + 8)× (B -5)

4. Now given that

   $A_{1} = A_{2} = A_{3}$

   L× B = (L + 6)× (B -4) = (L + 8)× (B -5)   ...1)

5.  Take equation 1) can be written as

      L× B = (L + 6)× (B -4)

     LB = LB - 4L + 6B - 24

     4L - 6B = - 24      ...2)

6. Again equation 1) can be written as

    L× B = (L + 8)× (B -5)

    LB = LB - 5L + 8B - 40

    5L - 8B = - 40      ...3)

7.   On solving equation 2) and equation 3)

    We get original length (L) = 24 (m)

    and  original width (B) = 20 (m)

     

Answer 2

Answer:

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