Question

The sum of the length and the breadth of a rectangle is 240 cm. If the length is
decreased by 20% and the breadth is increased by 10% the perimeter remains the
same. Find the dimensions of the rectangle.

please tell me the answer fast and no unnecessary messages please​ it's too urgent please

Answer 1

$\huge{\blue{\bold{\underline{\underline{Answer :}}}}}$

$\:\:$

$\large{\green{\underline \bold{\tt{Given :-}}}}$

$\:\:$

• Sum of the length and the breadth of a rectangle is 240 cm

$\:\:$

• When the length is decreased by 20% and the breadth is increased by 10% the perimeter remains the same

$\:\:$

$\large{\red{\underline \bold{\tt{To \: Find :-}}}}$

$\:\:$

• Dimensions of the rectangle

$\:\:$

$\large{\orange{\underline{\tt{Solution :-}}}}$

$\:\:$

• Let the length of rectangle be "x"

• Let the breadth of rectangle be "y"

$\:\:$

$\purple{\underline \bold{According \: to \: the \ question :}}$

$\:\:$

〚 Sum of the length and the breadth is 240 cm 〛

$\:\:$

So,

$\:\:$

➜ x + y = 240

$\:\:$

➜ x = 240 - y ------ (1)

$\:\:$

〚 Also given that , when the length is decreased by 20% and the breadth is increased by 10% the perimeter remains the same 〛

$\:\:$

$\underline{\bold{\texttt{New length :}}}$

$\:\:$

Length is decreased by 20%

$\:\:$

➜ $\sf x- \dfrac { 20 } { 100 } \times x$

$\:\:$

➜ $\sf x - \dfrac { 1 } { 5 } \times x$

$\:\:$

➜ $\sf \dfrac { 5x - x } { 5 }$

$\:\:$

➜ $\sf \dfrac { 4x } { 5 }$

$\:\:$

$\underline{\bold{\texttt{New breadth :}}}$

$\:\:$

Breadth is increased by 10%

$\:\:$

➜ $\sf y + \dfrac { 10 } { 100 } \times y$

$\:\:$

➜ $\sf y + \dfrac { 1 } { 10 } \times y$

$\:\:$

➜ $\sf \dfrac { 10y + y} { 10 }$

$\:\:$

➜ $\sf \dfrac { 11y } { 10 }$

$\:\:$

$\underline{\bold{\texttt{Perimeter of rectangle :}}}$

$\:\:$

➠ 2(length + breadth)

$\:\:$

$\underline{\bold{\texttt{Perimeter with original dimensions :}}}$

$\:\:$

➜ 2(length + breadth)

$\:\:$

➜ 2(240)

$\:\:$

➜ 480 cm

$\:\:$

$\underline{\bold{\texttt{Perimeter with new dimensions :}}}$

$\:\:$

➜ 2(length + breadth)

$\:\:$

➜ $\sf 2(\dfrac { 4x } { 5 } + \dfrac { 11y } { 10 })$

$\:\:$

➜ $\sf 2(\dfrac {8x + 11y } { 10 })$

$\:\:$

➜ $\sf \dfrac { 8x + 11y } { 5 }$ cm ----- (2)

$\:\:$

$\underline{\bold{\texttt{Putting the value of "x" from (1) to (2) }}}$

$\:\:$

➜ $\sf \dfrac { 8(240 - y) + 11y } { 5 }$

$\:\:$

➜ $\sf \dfrac { 1920 - 8y + 11y } { 5 }$

$\:\:$

➜ $\sf \dfrac { 1920 + 3y } { 5 }$ cm

$\:\:$

Given that perimeter remains unchanged

$\:\:$

So,

$\:\:$

➜ $\sf \dfrac { 1920 + 3y } { 5 } = 480$

$\:\:$

➜ $\sf 1920 + 3y = 480 \times 5$

$\:\:$

➜ $\sf 1920 + 3y = 2400$

$\:\:$

➜ $\sf 3y = 2400 - 1920$

$\:\:$

➜ $\sf 3y = 480$

$\:\:$

➜ $\sf y = \dfrac { 480 } { 3 }$

$\:\:$

➨ $\sf y = 160$

$\:\:$

• Hence breadth of rectangle is 160 cm

$\:\:$

$\underline{\bold{\texttt{Putting y = 160 in (1) }}}$

$\:\:$

➜ x = 240 - y

$\:\:$

➜ x = 240 - 160

$\:\:$

➨ x = 80

$\:\:$

• Hence length of rectangle is 80 cm

$\:\:$

Therefore the length and breadth of rectangle is 80 cm & 160 cm respectively

$\:\:$

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Additional information

$\:\:$

Properties of rectangle

• Area of rectangle = length × breadth

• The opposite sides are parallel and equal to each other.

• Each interior angle is equal to 90 degrees.

• The sum of all the interior angles is equal to 360 degrees.

• The diagonals bisect each other.

• Both the diagonals have the same length.

Answer 2

$\huge{\blue{\bold{\underline{\underline{Answer :}}}}}$

$\:\:$

$\large{\green{\underline \bold{\tt{Given :-}}}}$

$\:\:$

• Sum of the length and the breadth of a rectangle is 240 cm

$\:\:$

• When the length is decreased by 20% and the breadth is increased by 10% the perimeter remains the same

$\:\:$

$\large{\red{\underline \bold{\tt{To \: Find :-}}}}$

$\:\:$

• Dimensions of the rectangle

$\:\:$

$\large{\orange{\underline{\tt{Solution :-}}}}$

$\:\:$

Let the length of rectangle be "x"

Let the breadth of rectangle be "y"

$\:\:$

$\purple{\underline \bold{According \: to \: the \ question :}}$

$\:\:$

〚 Sum of the length and the breadth is 240 cm 〛

$\:\:$

So,

$\:\:$

➜ x + y = 240

$\:\:$

➜ x = 240 - y ------ (1)

$\:\:$

〚 Also given that , when the length is decreased by 20% and the breadth is increased by 10% the perimeter remains the same 〛

$\:\:$

$\underline{\bold{\texttt{New length :}}}$

$\:\:$

Length is decreased by 20%

$\:\:$

➜ $\sf x- \dfrac { 20 } { 100 } \times x$

$\:\:$

➜ $\sf x - \dfrac { 1 } { 5 } \times x$

$\:\:$

➜ $\sf \dfrac { 5x - x } { 5 }$

$\:\:$

➜ $\sf \dfrac { 4x } { 5 }$

$\:\:$

$\underline{\bold{\texttt{New breadth :}}}$

$\:\:$

Breadth is increased by 10%

$\:\:$

➜ $\sf y + \dfrac { 10 } { 100 } \times y$

$\:\:$

➜ $\sf y + \dfrac { 1 } { 10 } \times y$

$\:\:$

➜ $\sf \dfrac { 10y + y} { 10 }$

$\:\:$

➜ $\sf \dfrac { 11y } { 10 }$

$\:\:$

$\underline{\bold{\texttt{Perimeter of rectangle :}}}$

$\:\:$

➠ 2(length + breadth)

$\:\:$

$\underline{\bold{\texttt{Perimeter with original dimensions :}}}$

$\:\:$

➜ 2(length + breadth)

$\:\:$

➜ 2(240)

$\:\:$

➜ 480 cm

$\:\:$

$\underline{\bold{\texttt{Perimeter with new dimensions :}}}$

$\:\:$

➜ 2(length + breadth)

$\:\:$

➜ $\sf 2(\dfrac { 4x } { 5 } + \dfrac { 11y } { 10 })$

$\:\:$

➜ $\sf 2(\dfrac {8x + 11y } { 10 })$

$\:\:$

➜ $\sf \dfrac { 8x + 11y } { 5 }$ cm ----- (2)

$\:\:$

$\underline{\bold{\texttt{Putting the value of "x" from (1) to (2) }}}$

$\:\:$

➜ $\sf \dfrac { 8(240 - y) + 11y } { 5 }$

$\:\:$

➜ $\sf \dfrac { 1920 - 8y + 11y } { 5 }$

$\:\:$

➜ $\sf \dfrac { 1920 + 3y } { 5 }$ cm

$\:\:$

Given that perimeter remains unchanged

$\:\:$

So,

$\:\:$

➜ $\sf \dfrac { 1920 + 3y } { 5 } = 480$

$\:\:$

➜ $\sf 1920 + 3y = 480 \times 5$

$\:\:$

➜ $\sf 1920 + 3y = 2400$

$\:\:$

➜ $\sf 3y = 2400 - 1920$

$\:\:$

➜ $\sf 3y = 480$

$\:\:$

➜ $\sf y = \dfrac { 480 } { 3 }$

$\:\:$

➨ $\sf y = 160$

$\:\:$

Hence breadth of rectangle is 160 cm

$\:\:$

$\underline{\bold{\texttt{Putting y = 160 in (1) }}}$

$\:\:$

➜ x = 240 - y

$\:\:$

➜ x = 240 - 160

$\:\:$

➨ x = 80

$\:\:$

Hence length of rectangle is 80 cm

$\:\:$

• Therefore the length and breadth of rectangle is 80 cm & 160 cm respectively

$\:\:$

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