Answer:
The length of the original rectangle is 73.4 mm.
The breadth of the original rectangle is 86.6 mm.
Step-by-step explanation:
Given : The perimeter of a rectangle is 320 mm. If its length increases by 10 mm and its breadth decreases by 10 mm then it its area will be 32 less.
To find : Calculate the length and breadth of the original rectangle ?
Solution :
The area of the rectangle is [tex]A=L\times B[/tex]
Let the length of the rectangle be 'x'
The breadth of the rectangle be 'y'
The area is [tex]A=xy[/tex]
Now, length increases by 10 mm i.e. L=x+10
breadth decreases by 10 mm i.e. B=y-10
The new area is [tex]A_n=(x+10)(y-10)[/tex]
According to question,
[tex]A-A_n=32[/tex]
[tex]xy-(x+10)(y-10)=32[/tex] ......(1)
The perimeter of a rectangle is 320 mm.
i.e. [tex]P=2(L+B)[/tex]
[tex]320=2(x+y)[/tex]
[tex]x+y=160[/tex]
[tex]x=160-y[/tex] .....(2)
Substitute the value of y from eqn (2) in (1),
[tex]y(160-y)-(160-y+10)(y-10)=32[/tex]
[tex]160y-y^2-(170-y)(y-10)=32[/tex]
[tex]160y-y^2-(180y-1700-y^2)=32[/tex]
[tex]160y-y^2-180y+1700+y^2=32[/tex]
[tex]1700-20y=32[/tex]
[tex]20y=1732[/tex]
[tex]y=\frac{1732}{20}[/tex]
[tex]y=86.6[/tex]
Substitute in (2),
[tex]x=160-86.6[/tex]
[tex]x=73.4[/tex]
The length of the original rectangle is 73.4 mm.
The breadth of the original rectangle is 86.6 mm.