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
Answers
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
Reference link will be
https://brainly.in/question/13688718
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
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