The length of a rectangle is $3$ times its breadth, if the length is decreased by $3$ cm and the breadth increased by $5$ cm the area of the rectangle is increased by $57$ $c m^{2}$ The perimeter of the rectangle is

18

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48

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24

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Solution

The correct option is D $48$ cm
Let the breadth of the rectangle be $x$ cm.
Then, length = $3$ $x$ cm
New breadth = $(x + 5)$ cm
New length = $(3 x - 3)$ cm
Then, $(x + 5) (3 x - 3) - 3 x \times x = 57$
$\Rightarrow 3 x^{2} + 12 x - 15 - 3 x^{2} = 57$
$\Rightarrow 12 x = 57 + 15 = 72 \Rightarrow x = 6$
$∴$ Breadth = $6$ cm, Length = $18$ cm
Perimeter = $2 (6 c m + 18 c m) = 2 \times 24 c m = 48$ cm

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The length of a rectangle is 3 times its breadth, if the length is decreased by 3 cm and the breadth increased by 5 cm the area of the rectangle is increased by 57 cm2 The perimeter of the rectangle is

The correct option is D 48 cmLet the breadth of the rectangle be x cm.Then, length = 3x cmNew breadth =(x+5) cmNew length =(3x−3)cmThen, (x+5)(3x−3)−3x×x=57⇒3x2+12x−15−3x2=57⇒12x=57+15=72⇒x=6∴ Breadth = 6 cm, Length = 18 cmPerimeter =2(6cm+18cm)=2×24cm=48 cm

The length of a rectangle is $3$ times its breadth, if the length is decreased by $3$ cm and the breadth increased by $5$ cm the area of the rectangle is increased by $57$ $c m^{2}$ The perimeter of the rectangle is