Question

the diagonal of a rectangular field is 60meters more than the shorter side .If the longer side is 30meters more than the shorter side, find the sides of the field. ​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

$\huge\sf\color{lime}{Figure}$

$\setlength{\unitlength}{1.5cm}\begin{picture}(8,2)\thicklines\put(7.7,3){\large{A}}\put(9.6,1.7){\sf{\large{(n + 60)}}}\put(7.7,1){\large{B}}\put(9.1,0.7){\sf{\large{(n + 30)}}}\put(11.1,1){\large{C}}\put(8,1){\line(1,0){3}}\put(8,1){\line(0,2){2}}\put(11,1){\line(0,3){2}}\put(8,3){\line(3,0){3}}\put(11.1,2){\sf{\large{n}}}\put(8,1){\line(3,2){3}}\put(11.1,3){\large{D}}\put(10.8,1){\line(0,2){0.2}}\put(10.8,1.2){\line(2,0){0.2}}\end{picture}$

$\huge\sf\pink{Answer}$

☞ Length = 120 m

☞ Breadth = 90 m

☞ Diagonal = 150 m

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$\huge\sf\blue{Given}$

✭ The diagonal of a rectangular field is 60 m more than the shorter side

✭ Long side is 30 m more than the shorter side

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$\huge\sf\gray{To\:Find}$

➳ The sides of the field?

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$\huge\sf\purple{Steps}$

Let Shorter Side (Breadth) of Rectangular Field be n metre.

Diagonal of Field = (n + 60) metre

Length of Field = (n + 30) metre

Using Pythagoras theorem in ∆BCD

$:\implies\sf (Hypotenuse)^2=(Perpendicular)^2+(Base)^2\\\\:\implies\sf (n+60)^2=(n)^2+(n+30)^2\\\\:\implies\sf (n)^2 +(60)^2 + (2.n.60) = n^2 + (n)^2 + (30)^2 + (2.n.30)\\ \\ :\implies\sf n^2 + 3600 + 120n = 2n^2 + 900 + 60n\\\\:\implies\sf 0 = 2n^2 - n^2 + 60n - 120n + 900 - 3600\\\\:\implies\sf n^2 - 60n - 2700 = 0\\\\:\implies\sf n^2 - (90 - 30)n - 2700 = 0\\\\:\implies\sf n^2 - 90n + 30n - 2700 = 0\\\\:\implies\sf n(n - 90) + 30(n - 90) = 0\\\\:\implies\sf (n - 90)(n + 30) = 0\\\\:\implies\green{\sf n = 90} \quad \sf or \quad \red{n = -\:30}</p><p>$

➝ Ignoring Negative , value of n = 90

$\rule{140}{1}$

$\bullet\:\:\textsf{Length = (n + 30) = \textbf{120 m}}\\\bullet\:\:\textsf{Breadth = n = \textbf{90 m}}\\\bullet\:\:\textsf{Diagonal = (n + 60) = \textbf{150 m}}$

$\rule{130}3$