Question

a rectangle has the same area as another,whose length is 6m more and breadth is 4m less.it has also the same area as the third,whose length is 8m more and breadth 5m less.find the length and breadth of the original rectangle

Answer 1

Answer:

hey mate here is ur answer

Step-by-step explanation:

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Answer 2

Answer:

Length = 24m

Breadth = 20m

Step-by-step explanation:

$\boxed{\text{Area = Length x Breadth}}$

Let the length of the first rectangle be x

Let breadth of the second rectangle by y

∴ Area₁ = xy

Given:

Area₁ = Area₂

Therefore,

Area₂ = xy

Also given,

Length₂ = 6 + x

Breadth₂ = y - 4

Therefore,

Area₂ = (6 + x)(y - 4)

But, we know Area₂ = xy

Therefore,

xy = (6 + x)(y - 4)

∴ xy = 6 (y - 4) + x (y - 4)

∴ xy = 6y - 24 + xy - 4x

Given:

Area₂ = Area₃

Therefore,

Area₃ = xy

Also given,

Length₃ = 8 + x

Breadth₃ = y - 5

Therefore,

Area₃ = (8 + x)(y - 5)

But, we know Area₃ = xy

Therefore,

xy = (8 + x)(y - 5)

∴ xy = 8 (y - 5) + x(y - 5)

∴ xy = 8y - 40 + xy - 5x

Comparing Area₁ and Area₂:

xy = 6y - 4x - 24 + xy

∴ xy - xy = 6y - 4x - 24

∴ 6y - 4x = 24

Comparing Area₂ and Area₃:

6y - 4x - 24 = 8y - 40 - 5x

∴ 6y - 4x - 24 + xy = 8y - 40 + xy - 5x

∴ -4x + 5x - 24 + 40 = 8y - 6y + xy - xy

∴ x + 16 = 2y

Now, the two equations are:

$\boxed{\text{x + 16 = 2y}}$ - Eq. 1

and

$\boxed{\text{6y - 4x = 24}}$ - Eq. 2

From equation 1:

$y = \frac{x}{2} + 8$

If we substitute this value of y in Eq. 2:

$6(\frac{x}{2} + 8) - 4x= 24$

$3x + 48 - 4x = 24$

$\boxed{\text{Answer: x = 24m}}$

Therefore,

Length = 24m

and,

$y = \frac{x}{2} + 8$

$y = \frac{24}{2} + 8\\\\\\\boxed{\text{Answer: y = 20m}}$

Breadth = 20m

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