Question

the area of a rectangle. is 100 m if its length decreased by 2 m and breadth increase 3 m the area increased 44 m square find length and breadth​

Answer 1

$\large\boxed{\rm{\pink{Correct\: Question:}}}$

The perimeter of a rectangle. is 100 m.If its length decreased by 2 m and breadth increase 3 m ,the area increased 44 m².Find length and breadth of the rectangle.

$\large\boxed{\rm{\blue{Given:}}}$

Perimeter of rectangle=100m

Lenght if decreased by 2m then breadth increase by 3m

And the area also increased by 44m²

$\large\boxed{\rm{\orange{To\:find:}}}$

Length

Breadth

$\large\boxed{\rm{\red{Solution:}}}$

Let us take the length of rectangle as x.

Thus, Perimeter of rectangle $\rm{=100m}$

We will first find the Breadth.

As we know Perimeter of rectangle=2(L + B)

Putting the values:

$\rm{100=2(x+Breadth)}$

$\rm{x+Breadth=50}$

$\rm{Breadth=(50-x)m}$

Now,Area of the given rectangle

= Length × Breadth

$\rm{=x(50-x)m}^{2}$

New Length $\rm{=(x-2)m}$

New Breadth $\rm{=(50-x+3)=(53-x)m}$

∴Thr area of the rectangle $\rm{=(x-2)(53-x)m}^{2}$

According to the Question,

Area of new rectangle - Area of given rectangle =44

$\rm{(x-2)(53-x)-x(50-x)=44}$

$\rm{53x-x^2-106+2x-50x+x^2=44}$

$\rm{5x-106=44}$

$\rm{5x=44+106}$

$\rm{5x=150}$

$\sf{x=}\cancel\dfrac{150}{5}{=30m}$

Therefore,we got the length of the given rectangle as $\rm{30m}$

And ,the breadth $\rm{50-30=20m}$

__________________________

$\large\boxed{\rm{\green{Verification:}}}$

Area of rectangle $\sf{=(30}\times{20)m}^{2}$

$\sf{=600m}^{2}$

New Length $\sf{=(30-2)m=28m}$

New Breadth $\sf{=(20+3)m=23m}$

•Area of new rectangle $\sf{=(27}\times{23)m}^{2}{=644m}^{2}$

Area of new rectangle - Area of given rectangle

$\sf{=(644-600)m}^{2}$

$\sf{=44m}^{2}$(that is same as given)

Hence,our Solution is Verified.

Answer 2

$\large\boxed{\rm{\pink{Correct\: Question:}}}$

The perimeter of a rectangle. is 100 m.If its length decreased by 2 m and breadth increase 3 m ,the area increased 44 m².Find length and breadth of the rectangle.

$\large\boxed{\rm{\blue{Given:}}}$

• Perimeter of rectangle=100m

• Lenght if decreased by 2m then breadth increase by 3m

• And the area also increased by 44m²

$\large\boxed{\rm{\orange{To\:find:}}}$

• Length

• Breadth

$\large\boxed{\rm{\red{Solution:}}}$

Let us take the length of rectangle as x.

Thus, Perimeter of rectangle$\rm{=100m}$

We will first find the Breadth.

As we know Perimeter of rectangle=2(L + B)

Putting the values:

$\rm{100=2(x+Breadth)}$

$\rm{x+Breadth=50}$

$\rm{Breadth=(50-x)m}$

Now,Area of the given rectangle

= Length × Breadth

$\rm{=x(50-x)m}^{2}$

New Length$\rm{=(x-2)m}$

New Breadth $\rm{=(50-x+3)=(53-x)m}$

∴Thr area of the rectangle $\rm{=(x-2)(53-x)m}$

According to the Question,

Area of new rectangle - Area of given rectangle =44

$\rm{(x-2)(53-x)-x(50-x)=44}$

$\rm{53x-x^2-106+2x-50x+x^2=44}$

$\rm{5x-106=44}$

$\rm{5x=44+106}$

$\rm{5x=150}$

$\sf{x=}\cancel\dfrac{150}{5}{=30m}$

Therefore,we got the length of the given rectangle as $\rm{30m}$

And ,the breadth$\rm{50-30=20m}$

__________________________

$\large\boxed{\rm{\green{★Verification★}}}$

Area of rectangle $\sf{=(30}\times{20)m}^{2}$

$\sf{=600m}^{2}$

New Length $\sf{=(30-2)m=28m}$

New Breadth $\sf{=(20+3)m=23m}$

•Area of new rectangle $\sf{=(27}\times{23)m}^{2}{=644m}^{2}$

Area of new rectangle - Area of given rectangle

$\sf{=(644-600)m}^{2}$

$\sf{=44m}^{2}$

(that is same as given)

Hence,our Solution is Verified.

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