Question

If one side of a square is increased by 5metres and the other side is reduced by 5 metres , a rectangle is formed whose perimeter is 30 m . Find the side of the original square .​

Answer 1

In a square, all the sides are equal.

Let the side of the square be x.

Therefore, the sides of the rectangle are (x+5) and (x - 5) .

The perimeter of the rectangle is

2[(l+b)

=2[( x+5) + (x - 5)]

=2x + 10 + 2x - 10

= 4x

Now,

4x = 30

= x = 30/4

= x = 7.5

Answer 2

✴ Required Answer:

✏GiveN:

• One side of a square increased by 5
• Other side decreased by 5
• A rectangle was formed.
• Perimeter of the rectangle = 30 m

✏To FinD:

• Side of original square....?

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✴ How to solve?

Here, we need to maximise and minimise the sides of the square to form a rectangle. We need to know,

• Square has 4 equal sides.
• Rectangle has opposite sides equal.
• Perimeter of rectangle = 2(Length + breadth)

By keeping these points in mind, let's solve the question.

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✴ Solution:

Let the side of the square be x

It was changed to a rectangle with l and b,

Given,

• New length = x + 5
• New breadth = x - 5
• Perimeter = 30 m

By using formula for perimeter of rectangle,

$\large{ \rm{ \longrightarrow \: 2(l + b) = 30 \: m}} \\ \\ \large{ \rm{ \longrightarrow \: 2(x + \cancel{5} + x - \cancel{ 5}) = 30 \: m}} \\ \\ \large{ \rm{ \longrightarrow \:2(2x) = 30 \: m}} \\ \\ \large{ \rm{ \longrightarrow \: 4x = 30 \: m}} \\ \\ \large{ \rm{ \longrightarrow \: x = \boxed{ \red{ \rm{7.5 \: m}}}}}$

✏ We had our side of square = x,

So, x = 7.5 m

$\large{ \therefore{ \underline{ \underline{ \purple{ \rm{Hence \: solved \: \dag}}}}}}$

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