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

14 meters. find the breat

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

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   ★ 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

   ___________________

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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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