A rectangle is 15 meters long and its perimeter is the same as that of square whose side is

14 meters. find the breat

A rectangle is 15 meters long and its perimeter is the same as that of square whose side is

14 meters. find the breath of rectangle.​

2 thoughts on “A rectangle is 15 meters long and its perimeter is the same as that of square whose side is <br /><br /> 14 meters. find the breat”

  1. Given :

    • Length of the Rectangle = 15m
    • Perimeter of Rectangle is same as Perimeter of Square.
    • Side of Square = 14m

    To Find :

    • Breadth of the Rectangle

    Solution :

    It is given that :

    ⟶⠀Square’s peri. = Rectangle’s peri.

    ⟶⠀4 × side = 2(Length + Breadth)

    ⟶⠀4 × 14 = 2(15 + Breadth)

    ⟶⠀56 = 2(15 + Breadth)

    ⟶⠀56/2 = 15 + Breadth

    ⟶⠀28 = 15 + Breadth

    ⟶⠀28 – 15 = Breadth

    ⟶⠀13m = Breadth

    So, Breadth of the Rectangle is 13m

    ___________________

    Additional Info :

    Formulas Related to Rectangle:

    • Perimeter of Rectangle = 2( l + b)
    • Area = Length × Breadth
    • Length = Area / Breadth
    • Breadth = Area / Length
    • Diagonal = √(l)² + (b)²

    Formulas Related to Square :

    • Perimeter = 4 × Side
    • Area = Side × Side
    • Side of Square = √Area
    • Diagonal = √2 × Side
    • Area = ½ × (Diagonal)²
    • Area = (Perimeter ÷ 4)²

    ___________________

  2. Given:

    • A rectangle is 15 meters long and its perimeter is the same as that of square whose side is 14 meters

    To Find:

    • The breadth of the rectangle

    Solution:

    ➤ Here we’re given with the side of the square and the length of the rectangle. It is said that the perimeter of the square is equal to the perimeter of the rectangle and we’ve asked to find the breadth of the rectangle.

    Now,

    • Let’s firstly find the perimeter of the square

    As we know that,

    [tex] \: \: \: \: \: \: \: \: \: \: \dag \bigg( \bf \: perimeter _{(square)} = 4 \times side \bigg)[/tex]

    Where,

    • Side of the square is 14m

    [tex]{ \underline{ \bf{ \bigstar \: Substituting \: the \: values : }}}[/tex]

    [tex]{ : \implies} \sf \: Perimeter _{(square)} = 4 \times side \: \: \\ \\ \\ { : \implies} \sf \: Perimeter _{(square)} = 4 \times 14cm \\ \\ \\ { : \implies} \sf \: Perimeter _{(square)} = { \blue{ \boxed{ \frak{56cm}} \star}} \: \: \: [/tex]

    • Henceforth the perimeter of the square is 56cm

    We know that,

    • Perimeter of the square is equal to the perimeter of the rectangle

    So,

    • Perimeter of the rectangle is 56cm

    Now,

    • Let’s find the breadth of the rectangle

    As we know that,

    [tex] \: \: \: \: \: \: \: \: \: \: \dag \bigg( \bf \: perimeter _{(rectangle)} = 2(lenght + breadth) \bigg)[/tex]

    Where,

    • Perimeter = 56m
    • Length = 15cm

    [tex]{ \underline{ \bf{ \bigstar \: Substituting \: the \: values : }}}[/tex]

    [tex]{ : \implies} \sf \: Perimeter _{(rectangle)} = 2(lenght + breadth) \\ \\ \\ { : \implies} \sf 56cm = 2(15 + b) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ { : \implies} \sf 56cm = 30 + 2b \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ { : \implies} \sf 2b = 26cm \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ { : \implies} \sf b = \frac{26}{2} cm \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ { : \implies} \sf { \purple{\underline{ \boxed{\frak{b = 13cm}}}\star}}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: [/tex]

    • Henceforth the breadth of the rectangle is 13cm

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