I want a wire of length 86 CM in bent in the form of a rectangle such that its length is 7 cm more than its breadth find the length and breadth of the rectangle so formed
Answers
Answered by
0
Let the breadth be x
Let the length of wire = 86 = perimeter
Thus length = 7 + x
Thus perimeter =2l + 2b
86 = 14 + 2x + 2x
86 -14 = 4x
72 / 4 = x
X = 18
Breadth = 18
Length = 7 + x
= 7 + 18 = 25
Let the length of wire = 86 = perimeter
Thus length = 7 + x
Thus perimeter =2l + 2b
86 = 14 + 2x + 2x
86 -14 = 4x
72 / 4 = x
X = 18
Breadth = 18
Length = 7 + x
= 7 + 18 = 25
Answered by
1
Let's assume the breadth as x cm
So, the length as (x + 7) cm
We know,
Length of wire = Perimeter of rectangle
= 86 cm
∵ Perimeter of Rectangle
= 2(length + breadth)
According to Question,
Perimeter of Rectangle = 86 cm
2(x + 7 + x) = 86
⇒ 2x + 7 = 86 / 2
⇒ 2x + 7 = 43
⇒ 2x = 43 - 7
⇒ 2x = 36
⇒ x = 36 / 2
⇒ x = 18
Required Measurements -
Breadth = x = 18 cm
Length = x + 7 = 18 + 7 = 25 cm
Hence, the length and breadth are 25 cm and 18 cm respectively
So, the length as (x + 7) cm
We know,
Length of wire = Perimeter of rectangle
= 86 cm
∵ Perimeter of Rectangle
= 2(length + breadth)
According to Question,
Perimeter of Rectangle = 86 cm
2(x + 7 + x) = 86
⇒ 2x + 7 = 86 / 2
⇒ 2x + 7 = 43
⇒ 2x = 43 - 7
⇒ 2x = 36
⇒ x = 36 / 2
⇒ x = 18
Required Measurements -
Breadth = x = 18 cm
Length = x + 7 = 18 + 7 = 25 cm
Hence, the length and breadth are 25 cm and 18 cm respectively
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