Math, asked by yamsmaya0293, 3 months ago

If the length of a certain rectangle is decreased by 4 cm and width is increased by 3 cm, a square with the same area as the original rectangle would result. Find the perimeter of the original rectangle.​

Answers

Answered by Anonymous
61

QUESTION

If the length of a certain rectangle is decreased by 4 cm and width is increased by 3 cm, a square with the same area as the original rectangle would result. Find the perimeter of the original rectangle.

FIND

The perimeter of original rectangle =?

Answer

50 cm^2

Explanation

  • let the length be X cm.
  • and breadth be Y cm

★According to Question★

 \bf x \:  - 4 = y + 3 \ \\  \:  \bf  \ x - y = 7........(1)eq

And

 \bf x.y + 3x - 4y - 12 =  x .y \\  \:  \:  \:  \:  \bf or \\  \:  \bf 3x - 4y = 12......(2)eq

From equation (i) and (ii)

 \large \bf   \frac{x}{ - 12 + 28}  =  \frac{y}{21 - 12}  =  \frac{ - 1}{ - 4 + 3}

oR

  \large \bf   \frac{x}{16}  =  \frac{y}{9}  = 1

● so x=16cm

●and y= 9 cm

★Therefore the perimeter of original rectangle

= 2.(16+9)= 50cm

Attachments:
Answered by BrainlyConqueror0901
49

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

\green{\tt{\therefore{Perimeter\:of\:original\:triangle=50\:cm}}}

\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}

\green{\underline{\bold{Given :}}} \\  \tt:  \implies Length \: of \: rectangle \: decreased \: by \: 4 \: cm \\  \\ \tt:  \implies Width\: of \: rectangle \: increased \: by \: 3\: cm \\  \\ \red{\underline{\bold{To \: Find :}}} \\  \tt:  \implies Perimeter \: of \:original \:rectangle =M

• According to given question :

 \green{\star } \tt \: Let \: length \: be \: 'l' \\  \\\green{\star } \tt \: \: Breadth \: be \:  'b' \\  \\\green{\star } \tt \: New \: length = l - 4\\  \\  \green{\star } \tt \: New \:  breadth = b  +3\\  \\  \bold{As \: we \: know \: that} \\  \tt:  \implies  New \: length \:  = New \: breadth  \:  \:  \:  \: \\  \\  \tt:  \implies l - 4 = b +3 \\  \\ \tt:  \implies l - b = 7 -  -  -  -  - (1) \\  \\  \bold{As \: we \: know \: that} \\  \tt:  \implies Area \: of \: new \: rectangle = Area \: of \: original\:rectangle \\  \\ \tt:  \implies (l - 4)(b  +  3) = l \times b \\  \\ \tt:  \implies lb - 4b + 3l - 12 =lb \\  \\ \tt:  \implies 3l - 4b = 12 -  -   -   -  - (2) \\  \\  \text{Solving \: (1) \: and \: (2)}

\tt:  \implies b = 9 \:  \:  \: and \:  \:  \:l = 16 \\  \\  \bold{As \: we \: know \: that} \\  \tt:  \implies Perimeter \: of \: original \: rectangle =2(l + b) \\  \\ \tt:  \implies Perimeter \: of \: original \: rectangle =2(16 + 9) \\  \\ \tt:  \implies Perimeter \: of \: orginal \: rectangle =2 \times 25 \\  \\  \green{\tt:   \implies Perimeter \: of \: original\: rectangle =50 \: cm}

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