Question

The width of a rectangle is 7 ft less than the length. The area of the rectangle is 228 ft. Find the perimeter.

Answer 1

Answer:

$\huge\mathrm\blue{Solution}</p><p>$

• Let the length of rectangle be 'x' ft

Then,

• Breadth = (x - 7) ft

• Area of Rectangle = l × b = 228

• ==> x(x - 7) = 228

• ==> x^2 - 7x - 228 = 0

• ==> x^2 - 19x + 12x - 228 = 0

• ==> x(x - 19) + 12 ( x - 19 ) = 0

• ==> (x + 12) ( x - 19) = 0

• ==> (x + 12) = 0. or (x - 19) = 0

• ==> x = -12 , x = 19

• As length can't be negative.

So,

• Length (l) = x = 19 ft

Then, Breadth = 19 - 7 = 12 ft

• Perimeter of Rectangle = 2(l + b)

• ==> 2 ( 19 + 12) = 2 × 31 = 62ft

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Answer 2

$\huge\underline\mathcal\red{Given}$

• Area of rectangle = 228 ft

$\huge\underline\mathcal\red{Find\:out}$

• Find the perimeter of rectangle

$\huge\underline\mathcal\red{Solution}$

Let the length of the given rectangle be x ft then it's breadth be (x-7)ft

According to the given condition,

Area of rectangle = length × breadth

$\implies\sf x(x-7)=228$

$\implies\sf x^2-7x=228$

$\implies\sf x^2-7x-228=0$

Method : 1

To solve above equation

By splitting middle term

$\implies\sf x^2-19x+12x-228=0$

$\implies\sf x(x-19)+12(x-19)=0$

$\implies\sf (x-19)(x+12)=0$

$\sf Either\:(x-19)=0$

$\implies\sf x=19ft$

$\sf or\:(x+12)=0$

$\implies\sf x=-12ft$

Method : 2

Using quadratic formula

a = 1 b = -7 c = -228

$\implies\sf \frac{-b±\sqrt{b^2-4ac}}{2a}$

Substitute the value of a,b and c

$\implies\sf \frac{-(-7)±\sqrt{(-7)^2-4×1×(-228)}}{2×1}$

$\implies\sf \frac{7±\sqrt{49+912}}{2}$

$\implies\sf \frac{7±\sqrt{961}}{2}$

$\implies\sf \frac{7±31}{2}$

$\implies\sf \frac{7+31}{2}\:\frac{7-31}{2}$

$\implies\sf \frac{38}{2}\:\frac{-24}{2}$

$\implies\sf 19ft,-12ft$

Since x cannot negative, being a dimension, the length of rectangle is 19ft

Hence

Required length = 19ft

Required breadth = (x-7)= 19-7 = 12ft

Perimeter of rectangle :-

$\sf 2(length+breadth)$

Substitute the value of length and breadth in above formula

$\implies\sf 2(19+12)$

$\implies\sf 2×31=62m$

$\large{\boxed{\sf{Required\:perimeter\:of\:rectangle=62m}}}$

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$\large{\boxed{\mathcal{\red{Important\:point}}}}$

Area of rectangle = length×breadth

Perimeter of rectang = 2(l+b)

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