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The length of a rectangle is 3 more than its width. If its perimeter is 54 m. Find its length & breadth and area and area.​

Respuesta :

Step-by-step explanation:

Given:

  • Length of the Rectangle is 3 more than its width.
  • Perimeter of rectangle is 54 m

To Find:

  • Length and Breadth
  • Area of the rectangle

Solution:

We are given length of a rectangle is 3 more than its width

Let's assume:

  • Breadth of the rectangle = x
  • Length of the rectangle = x + 3

We know that,

[tex]\dashrightarrow \sf \: \: Perimeter_{(rectangle)} = 2(L + B) [/tex]

[tex]\dashrightarrow \sf \: \: 54 = 2(x + x + 3) [/tex]

[tex]\dashrightarrow \sf \: \: 54 = 2(2x + 3)[/tex]

[tex]\dashrightarrow \sf \: \: \dfrac{54}{2} = 2x + 3[/tex]

[tex]\dashrightarrow \sf \: \: 27 = 2x + 3[/tex]

[tex]\dashrightarrow \sf \: \: 27 - 3 = 2x [/tex]

[tex]\dashrightarrow \sf \: \: 24 = 2x [/tex]

[tex]\dashrightarrow \sf \: \: \dfrac{24}{2} = x [/tex]

[tex]\dashrightarrow \: \: {\underline{\boxed{\pink{\pmb{\mathfrak{12 = x}}}}}} [/tex]

Hence,

  • Breadth of rectangle = x = 12 m
  • Length of rectangle = 3 + x = 15 m

Now, Finding its area:

[tex]\dashrightarrow \: \: { \sf{ Area_{(rectangle)} = L \times B }} \: \\ [/tex]

[tex]\dashrightarrow \sf \: \: Area = 12 \times 15 [/tex]

[tex]\dashrightarrow \sf \: \: {\underline{\boxed{\pink{\pmb{\mathfrak{Area = 180 \: {m}^{2}}}}}}} [/tex]

Hence,

  • Length and breadth of rectangle is 12 m and 15 m

  • Area of the rectangle is 180 m²
blackshades

Answer:

Width = 12 m

Length = 15 m

Area = 180 m²

Step-by-step explanation:

Let, width = x

Length = x + 3

Perimeter = 54 m

Perimeter = 2 (l + w)

54 = 2 (x + 3 + x)

54/2 = 2x + 3

27 - 3 = 2x

24 = 2x

24/2 = x

12 m = x (width)

Length = x + 3 = 12 + 3 = 15 m

Area

= l × w

= 12 × 15

= 180 m²

______

Hope it helps ⚜

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