Question

The length of a rectangle exceeds its width by 3m if the width is increased by 4m and the length is deacreased by 6m the area is decreased by 22 sq.m find the dimensions of the rectangle

Answer 1

Answer:

$8 m \times 5 m$

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 = $Length \times Breadth$

                            = $y(y+3)$

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 = $Length \times Breadth$

                                    = $(y-3)(y+4)$

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

                                    = $y^2+y-12$

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

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

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

$y-12=3y-22$

$22-12=3y-y$

$10=2y$

$y=5$

So, width = 5 m

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

Hence the dimensions of the rectangle are $8 m \times 5 m$

Answer 2

Answer:

Therefore, Area of the given rectangle = w × ( w + 3 ) m 2 . If the width is increased by 4m and the length is decreased by 6m, the area is decreased by 22 sq. m.