Question

The diagonal of a rectangle is 28m more than is breadth and the length is 14m more than its breadth. Find the length and breadth of the rectangle.
( Quadratic Equations )​

Answer 1

$\sf\large\underline\purple{Let:-}$

$\sf{\implies Rectangle\:_{(breadth)}=x}$

$\sf{\implies Rectangle\:_{(length)}=14+x}$

$\sf{\implies Rectangle\:_{(diagonal)}=28+x}$

$\sf\large\underline\purple{To\: Find:-}$

$\sf{\implies Rectangle\:_{(Length\:and\: breadth)}=?}$

$\sf\large\underline\purple{Solution:-}$

• To calculate the dimensions of rectangle ,at first we have to assume the breadth of rectangle be x. After that we have to assume the length and diagonal of rectangle as per the given clue in the question. Then applying Pythagoras theorem to calculate length and breadth of rectangle:-]

$\sf\large\underline\purple{Formula\: used:-}$

$\sf{\implies Diagonal^2=Perpendicular^2+Base^2}$

$\tt\large\underline{Here, D=28+x\:\:,P=14+x\:\:,B=x:-}$

$\tt{\implies (28+x)^2=(14+x)^2+(x)^2}$

$\tt{\implies 784+56x+x^2=196+28x+x^2+x^2}$

$\tt{\implies 2x^2-x^2+28x-56x+196-784=0}$

$\tt{\implies x^2-28x-588=0}$

• Here splitting the middle term:-]

$\tt{\implies x^2+14x-42x-588=0}$

$\tt{\implies x(x+14)-42(x+14)=0}$

$\tt{\implies (x+14)(x-42)=0}$

$\tt{\implies \therefore\:x=42,\:\:-14}$

• Here the dimensions of rectangle can't be in negative so, we have to take positive value hence, x=42:-]

$\sf\large{Hence,}$

$\sf{\implies Rectangle\:_{(breadth)}=42m}$

$\sf{\implies Rectangle\:_{(length)}=14+x=56m}$

$\sf{\implies Rectangle\:_{(diagonal)}=28+x=70m}$

Answer 2

Given :

• The diagonal of a rectangle is 28m more than its breadth.
• The length of the rectangle is 14m more than its breadth.

To Find :

• Length and breadth of the rectangle.

Solution :

Let,

• The breadth of the rectangle be (x)

then, According to the question :

• The diagonal of the rectangle will be (x + 28)
• The length of the rectangle will be (x + 14)

$\underline \blue{{\boxed{ \bf{ \red{required \: figure}}}}}$

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Using Pythagoras theorem :

༒ Hypotenuse² = Perpendicular² + Base²

⪼ (x + 28)² = (x + 14)² + x²

⪼ x² + 784 + 56x = x² + 196 + 28x + x²

⪼ 784 - 196 = 2x² - x² + 28x - 56x

⪼ 588 = x² - 28x

⪼ x² - 28x - 588 = 0

༒ By middle term split method

⪼ x² - (42 - 14)x - 588 = 0

⪼ x² - 42x + 14x - 588 = 0

⪼ x(x - 42) + 14(x - 42) = 0

⪼ (x - 42) (x + 14) = 0

⪼ x = 42, x = -14

• As we know that, the breadth of the rectangle can't be negative, So (x = 42) will be taken.

Hence,

• Breadth of the rectangle = 42m.
• Length of the rectangle = (42 + 14) = 56m.
• Diagonal of the rectangle = (42 + 28) = 70m.