length of a rectangle is 20 more then 3 times of it breadth if perimeter of rectangle is 60m then find a linear equation this
Answers
Answer:
- 3x - y + 20 = 0 , x + y - 30 = 0
Given that:
- Length of rectangle is 20 more than 3 times of breadth.
- Perimeter of rectangle is 60m.
To Find:
- Linear equation representing this situation.
Solution:
Let the breadth of the rectangle be x and the length be y.
In the given question, the relation between the length and breadth (i.e., x and y) is given as,
=> Length = 3 × Breadth + 20
=> y = 3x + 20
=> 3x - y + 20 = 0
No matter what variables you give to the dimensions of the rectangle, it will always follow the following general relation:
=> Length = 3 × Breadth + 20
Relating the length and breadth to the perimeter of the rectangle, we have
=> 2 (Length + Breadth) = 60
=> 2(x + y) = 60
=> x + y = 30
=> x + y - 30 = 0
Question:
- Length of a rectangle is 20 more then 3 times of it's breadth if perimeter of rectangle is 60 m. Find a linear equation this
Answer:
- Required eqⁿ is 3n - m + 20 = 0.
Step-by-step explanation:
Given :-
- Length of a rectangle is 20 more then 3 times of it's breadth if perimeter of rectangle is 60 m.
To Find :-
- Linear equation for this?
Solution :-
- Let length of rectangle be m and breadth of rectangle be n.
According to the given question;
- Length of a rectangle is 20 more then 3 times of it's breadth if perimeter of rectangle is 60 m.
Therefore;
➙ Length = 3(Breadth) + 20
➙ m = 3n + 20
➙ 0 = 3n + 20 - m
We can write it as;
➙ 0 = 3n - m + 20
➙ 3n - m + 20 = 0
So, required equation is 3n - m + 20 = 0.
Learn More :-
- Perimeter of any figure is calculated by sum of its all sides.
- Perimeter of square = 4 × side
- Area of square = (side)²
- Perimeter of equilateral ∆ = 3 × side
- Area of equilateral ∆ = √3/4 (side)²
- Perimeter of rhombus = 4 × side
- Area of rhombus = ½ × d₁ × d₂
- Perimeter of circle = 2πr
- Area of circle = πr²
- Perimeter of rectangle = 2(l + b)
- Area of rectangle = l × b
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