Math, asked by shrutipataliya121, 7 months ago

The length of the greater sides of a rectangle is exceeds the length of the smaller sides by 5cm.If the area of the rectangle is 84sq.cm.Find the dimensions​

Answers

Answered by ExᴏᴛɪᴄExᴘʟᴏʀᴇƦ
10

\displaystyle\large\underline{\sf\red{Given}}

✭ Length of the longer side of a rectangle exceeds the smaller side by 5

✭ Area of the rectangle is 84 cm²

\displaystyle\large\underline{\sf\blue{To \ Find}}

◈ The dimensions of the rectangle?

\displaystyle\large\underline{\sf\gray{Solution}}

Area of a rectangle is given by,

\displaystyle\sf \underline{\boxed{\sf Area_{Rectangle} = Length \times Breadth}}

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\underline{\bigstar\:\textsf{According to the given Question :}}

So here let the breadth of the rectangle [as per the question, the shorter side] will be x. And also the length [here the longer side] will be (x+5)

So now substituting these values in the formula,

\displaystyle\sf Area_{Rectangle} = Length \times Breadth

  • Area = 84 cm²
  • Length = (x+5)
  • Breadth = x

\displaystyle\sf 84 = (x+5)\times x

\displaystyle\sf 84 = x^2+5x

\displaystyle\sf x^2+5x-84 = 0

\displaystyle\sf x^2+12x-7x-84 = 0

\displaystyle\sf x(x+12)-7(x+12)

\displaystyle\sf \purple{(x-7)(x+12)}

On Equating them with zero,

»» \displaystyle\sf x-7 = 0

»» \displaystyle\sf x = 7

Similarly,

»» \displaystyle\sf x+12 = 0

»» \displaystyle\sf x = -12

On ignoring negative sign,

›› \displaystyle\sf \orange{x = 7}

So then the length of the triangle will be,

\displaystyle\sf x+5

\displaystyle\sf 7+5

\displaystyle\sf \pink{Length = 12 \ cm}

\displaystyle\sf \therefore\:\underline{\sf Length = 12 \ cm, Breadth = 7 \ cm}

\displaystyle\sf \star\: Diagram\:\star

\setlength{\unitlength}{1.5cm}\begin{picture}(8,2)\linethickness{0.4mm}\put(7.7,3){\large\sf{A}}\put(7.8,2){\sf{\large{x}}}\put(7.7,1){\large\sf{B}}\put(9.3,0.7){\sf{\large{x+5}}}\put(11.1,1){\large\sf{C}}\put(11.1,3){\large\sf{D}}\put(8,1){\line(1,0){3}}\put(8,1){\line(0,2){2}}\put(11,1){\line(0,3){2}}\put(8,3){\line(3,0){3}}\end{picture}

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