Question

the diagonal of a rectangular field is 60 m more than its bredth .if the length is 30 m more than its bredth find the dimensions of the field​

Answer 1

Step-by-step explanation:

Given :-

The diagonal of a rectangular field is 60 m more than its bredth and the length is 30 m more than its breadth.

To find:-

Find the dimensions of the field ?

Solution:-

Let the breadth of the rectangle be X m

Length of the rectangle = (X+30) m

We know that

Diagonal of the rectangle = √(l^2+b^2) units

=> d = √[(X^2+(X+30)^2] m

=> d =√[X^2+X^2+60X+900]

=> d = √(2X^2+60X+900) m

According to the given problem

Diagonal of the rectangle = (X+60) m

=> √(2X^2+60X+900) = X+60

On squaring both sides

=> [√(2X^2+60X+900)]^2= (X+60)^2

=> 2X^2+60X+900 = X^2+120X+3600

=> 2X^2+60X+900-X^2-120X-3600 = 0

=>X^2-60X-2700 = 0

=> X^2-90X+30X-2700 = 0

=> X(X-90)+30(X-90) = 0

=> (X-90)(X+30) = 0

=> X-90 = 0 or X+30 = 0

=> X= 90 or X=-30

X cannot be negative since it is a length of the breadth

X= 90 m

Breadth = 90 m

Length = 90+30 = 120 m

Diagonal = 90+60 = 150 m

Answer:-

The length of the rectangle = 120 m

The breadth of the rectangle = 90 m

The diagonal of the rectangle = 150 m

Used formulae :-

• Diagonal of the rectangle = √(l^2+b^2) units
• l = length of the rectangle
• b=breadth of the rectangle