12. The length of a rectangle is greater than twice is breadth by 1 cm. fits
breadth is doubled and length is decreased by 4

By Remi

12. The length of a rectangle is greater than twice is breadth by 1 cm. fits
breadth is doubled and length is decreased by 4 cm, its area remains same
Find the dimension of the original rectangle​

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Remi

1 thought on “12. The length of a rectangle is greater than twice is breadth by 1 cm. fits<br />breadth is doubled and length is decreased by 4”

  1. [tex]\large\bf{✯ \: understanding \: the \: concept✯}[/tex]

    The length of the rectangle is greater than twice its breadth by 1 cm.

    and if the breadth is doubled and the length decreased by 4 cm , its area remains the same

    [tex]\large\fbox\red{To\:find:}[/tex]

    Here we need to find the area of the original rectangle .

    [tex]\sf\pink{Lets\:do\:it\:mate!}[/tex]

    let the breadth of the rectangle be b

    and the length be l

    Therefore length (l) = 2b+1

    Now,

    if the breadth(b) is doubled then,

    breadth ( b) = 2x

    and length (l) = l 4

    then the area remains same .

    [tex]\sf{(l \times b) = (l \times b)} \\ [/tex]

    [tex]\sf{⇒(2b – 1) \times b = (2b + 1 – 4)(2b)}[/tex]

    [tex]\sf{⇒2b {}^{2} + b = 4b {}^{2} – 6b}[/tex]

    [tex]\sf{⇒2b {}^{2} = 7b} \\ \sf{⇒b {}^{2} = \frac{7b}{2} } \\ \sf{⇒b {}^{2} = 3.5b}[/tex]

    [tex]\sf{⇒b = \frac{3.5b}{b} } \\ \sf{⇒b \: = 3.5 \: }[/tex]

    Therefore,

    breadth = 3.5 cm

    Now lets find the length

    [tex]\sf{⇒l = 2b + 1} \\ \sf{⇒l =2 \times 3.5 + 1} \\ \sf{⇒l = 8}[/tex]

    Answer:

    The breadth of the rectangle is 3.5 cm and the length of the rectangle is 8 cm .

    Hope it helps uh dear mate

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