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If the sides of rectangle are increased by 6% find the percentage increase in the diagonals.


1. 8%
2. 6%
3. 12%
4. 9%

1 Answer

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Correct Answer - Option 2 : 6%

Formula used:

Diagonal of a rectangle = \(\sqrt{l^2 + b^2}\)

Calculation:

Let the length and breath be 'l' and 'b'

After increase of 6%, 

Length = l + 6l/100

⇒ 106l/100 = 53l/50

After increase of 6%,

Breath  = b = 53b/50

Diagonal of a rectangle = \(\sqrt{l^2 + b^2}\)

After increase 6% diagnals = \(\sqrt{(53l/50)^2 + (53b/50)^2}\)

⇒ \(\sqrt{(53/50)^2(l^2 + b^2)}\) = \((53/50)^2\sqrt{l^2 + b^2}\)

The length of diagonals originally = \(\sqrt{l^2 + b^2}\)

Percentage increase in diagonals =( \(\sqrt{(53/50)^2(l^2 + b^2)}\)/\(\sqrt{l^2 + b^2}\))× 100

⇒ 53/50((\(\sqrt{l^2+ b^2}-\sqrt{l^2 - b^2}\))/\(\sqrt{l^2 + b^2}\)) ×100

⇒Required percentage change = [(53/50) - 1] × 100 = 6%

∴ Required percentage change is 6%

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