Question

The length of a rectangle exceeds its width by 3M if the width increase by 4 metre and the length is decreased by 6 metre the area decreased by 22 to square metre find the dimensions of the rectangle

Answer 1

Answer:

Width = 5m

Length = 8m

Step-by-step explanation:

Define x:

Let the width be x m

The length is (x + 3) m

Find the area:

Area = length x width

Area = x(x + 3) m²

Find the new dimensions:

width = (x + 4) m

length = (x + 3) - 6 = (x - 3) m

Find the area:

Area = length x width

Area = (x + 4)(x - 3)

Solve x:

The difference in area is 22 m²

x(x +3) - (x + 4)(x - 3) = 22

x² + 3x - x² + 3x - 4x + 12 = 22

2x + 12 = 22

2x = 10

x = 5

Find the dimension:

Width = x  = 5 m

Length = x + 3 = 5 + 3 = 8 m

Answer: The rectangle is 5m by 8m

Answer 2

$\tt 8 m\times 5m 8m×5m$

Step-by-step explanation:

Let the breadth of the rectangle be y

We are also given that The length of a rectangle exceeds its width by 3m

So, length = y+3

Area of rectangle = Length \times BreadthLength×Breadth

= y(y+3)y(y+3)

Now we are given that the width is increased by 4 m and the length is decreased by 6 m

So,he length of the rectangle is y+3-6=y-3

So, the breadth of the rectangle is y+4

So, Area of rectangle = Length \times BreadthLength×Breadth

= (y-3)(y+4)(y−3)(y+4)

= y^2+4y-3y-12y

2

+4y−3y−12

= y^2+y-12y

2

+y−12

Now we are given that the area is decreased by 22 sq.m

So,]y^2+y-12=y(y+3)-22]y

2 +y−12=y(y+3)−22

]y^2+y-12=y^2+3y-22]y

2

+y−12=y

2

+3y−22

y-12=3y-22y−12=3y−22

22-12=3y-y22−12=3y−y

10=2y10=2y

y=5y=5

So, width = 5 m

Length = y+3=5+3=8 m

Hence the dimensions of the rectangle are

$\tt 8 m \times 5m \;8m×5m$