Question

(i) 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

Answer :

Step-by-step explanation:

Given 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 cm less. Find the length and breadth of the original rectangle

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

According to question rectangle with 6m is more so it will be l + 6 and breadth is 4 m less, so b – 4

Now area of rectangle is l x b. so we can write as

l x b = (l + 6)(b – 4)

       = lb + 6b – 4l – 24

   We get -4l + 6b – 24 = 0

    Or 4l – 6b + 24 = 0

Again according to question area of original rectangle is equal to area of third rectangle with length 8 m more and breadth 5 m less

Now the equation will be  

l x b = (l + 8)(b – 5)

l x b = lb + 8b – 5l – 40

       = lb – 5l + 8b – 40

We get -5l + 8b – 40 = 0

Or 5l – 8b + 40 = 0

Now we have simultaneous equations to solve,  

   4l – 6b + 24 = 0 multiply by 5

   5l – 8b + 40 = 0 multiply by 4

We get

    20l – 30b + 120 = 0

    20l – 32b + 160 = 0

             2b = 40

          So b = 20 m

Substituting b = 20 we get

   4l – 6b + 24 = 0

  4l – 6(20) + 24 = 0

  4l – 120 + 24 = 0

  4l – 96 = 0

  4l = 96

So l = 24 m

So length and breadth of rectangle will be 24 m and 20 m