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Sagot :
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
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
______
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