Question

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

Answer 1

Answer:

$\begin{gathered}{\large{\textsf{\textbf{\underline{\underline{Given :}}}}}}\end{gathered}$

• ➳ The diagonal of a rectangular field is 60 metres more than the shorter side.
• ➳ The longer side of Rectangular field is 30 metres more than the shorter.

$\begin{gathered}\end{gathered}$

$\begin{gathered}{\large{\textsf{\textbf{\underline{\underline{To Find :}}}}}}\end{gathered}$

• ➳ Sides of Rectangular field

$\begin{gathered}\end{gathered}$

$\begin{gathered}{\large{\textsf{\textbf{\underline{\underline{Solution :}}}}}}\end{gathered}$

$\bigstar{\underline{ \underline{\pmb{\frak{\red{Let \: the : }}}}}}$

• ➺ The shorter side of rectangle = x metres
• ➺ The diagonal of rectangle = (x + 60) metres
• ➺ The longer side of rectangle = (x + 30) metres

$\rule{200}2$

$\bigstar{\underline{ \underline{\pmb{\frak{\red{According \: to \: pythagoras\: theorem: }}}}}}$

${: \implies{\sf{ {(Diagonal)}^{2} = ({Longer \: Side})^{2} + (Small \: Side )^{2}}}}$

• Substituting the values

${: \implies{\sf{ {(x + 60)}^{2} = (x + 30)^{2} + (x )^{2}}}}$

${: \implies{\sf{ {( {x}^{2} + 120x + 3600)}= {x}^{2} + {x}^{2} + 60x + 900 }}}$

${: \implies{\sf{ { {x}^{2} + 60x - 120x + 900 - 3600 = 0}}}}$

${: \implies{\sf{ { {x}^{2} - 60x - 2700 = 0}}}}$

${: \implies{\sf{ { {x}^{2} - 90x + 30x - 2700 = 0}}}}$

${: \implies{\sf{ { {x( x- 90)+ 30 - (x - 90) = 0}}}}}$

${: \implies{\sf{ { {( x - 90)+ (x + 30) = 0}}}}}$

${: \implies{\sf{x = 90, -30}}}$

• We ignore –30. Since length cannot be in negative.

$\rule{200}2$

$\bigstar{\underline{ \underline{\pmb{\frak{\red{Hence : }}}}}}$

• ➳ Shorter side = x = 90 metres
• ➳ Longer side = x + 30 = 90 + 30 = 120 metres

$\begin{gathered}\end{gathered}$

$\begin{gathered}{\large{\textsf{\textbf{\underline{\underline{Answer :}}}}}}\end{gathered}$

• ➽ The shorter side of Rectangular field is 90 m.
• ➽ The longer side of Rectangular field is 120 meter.

$\begin{gathered}\end{gathered}$

$\begin{gathered}{\large{\textsf{\textbf{\underline{\underline{Learn More :}}}}}}\end{gathered}$

$\;\sf{\leadsto\;\;Area\;\;of\;\;Square\;=\;(Side)^{2}}$

$\;\sf{\leadsto\;\;Area\;\;of\;\;Rectangle\;=\;Length\;\times\;Breadth}$

$\;\sf{\leadsto\;\;Area\;\;of\;\;Circle\;=\;\pi r^{2}}$

$\;\sf{\leadsto\;\;Area\;\;of\;\;Triangle\;=\;\dfrac{1}{2}\;\times\;Base\;\times\;Height}$

$\;\sf{\leadsto\;\;Area\;\;of\;\;Parallelogram\;=\;Base\;\times\;Height}$

$\;\sf{\leadsto\;\;Perimeter\;\;of\;\;Square\;=\;4\;\times\;(Side)}$

$\;\sf{\leadsto\;\;Perimeter\;\;of\;\;Rectangle\;=\;2\;\times\;(Length\;+\;Breadth)}$

$\;\sf{\leadsto\;\;Perimeter\;\;of\;\;Circle\;=\;2\pi r}$

Answer 2

Given :-

• The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side.

To Find :-

• The sides of the field ?

Solution :-

• Let shorter side be x m
• diagonal = ( x + 60) m
• longer side = (x + 30) m

We know that,

• In right ∆ ABC

• Using formula,

• (AC)² = (AB)² + (BC)²

• Putting all values in formula,

➻ (x + 60)² = (x)² + (x + 30)²

➻ (x)² + 2 × x × 60 + (60)² = x² + (x)² + 2 × x × 30 + (30)²

➻ x² + 120x + 3600 = x² + x² + 60x + 900

➻ x² + 60x + 900 - 120x - 3600 = 0

➻ x² - 60x - 2700 = 0

➻ x² - 90x - 30x - 2700 = 0

➻ x(x - 9) + 30(x - 90) = 0

➻ (x - 90) ( x + 30) = 0

Now,

➻ (x - 90) = 0

➻ x = 0 + 90

➻ x = 90

Then,

➻ (x + 30) = 0

➻ x = 0 - 30

➻ x = -30

• Hence, the side of the field cannot be negative so, the length of the shorter side will be 90m. and length of longer side will be (x + 30) = (90 + 30) = 120 m.

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