Question

In a rectangle, difference between length and breadth is 7cm. if length of its diagonal is 13cm then find the area of the rectangle.

Answer 1

Answer:

given length - breadth = 7

                    l  -  b   = 7

                          l    = 7+b

       diagonal   = 13cm

(diagonal)² = (length)²+(breadth)²                                                                               [imagine rectangle as combination of  two                                                          equal right triangle, then hypotenuse of both triangle is the diagonal of the           rectangle so by pythagoras theorm  (diagonal)² = (length)²+(breadth)² ]

13² = b²+(7+b)²

169 = b²+7²+14b+b²

169  = 2b²+49+14b

169-49 = 2b²+14b

120   = 2b²+14b

 2b²+14b -120 = 0

  b² +7b - 60  = 0              [on deviding the whole equation by 2]

Answer 2

$\sf\small\underline\green{Let:-}$

$\tt{\implies Length\:_{(rectangle)}=x}$

$\tt{\implies Breadth\:_{(rectangle)}=y}$

$\sf\small\underline\green{Given:-}$

$\tt{\implies Difference\:_{(length-breadth)}=7}$

$\tt{\implies Diagonal\:_{(rectangle)}=13cm}$

$\sf\small\underline\green{Solution:-}$

To calculate the area of rectangle at first we have to assume the length and breadth of rectangle be x. Then as per the given clue in the question , we have find the value of x and y by setting up equation. Then solve the equation by solving we get the value of x and y . After that we have to find the area of rectangle.

$\tt{\implies Difference\:_{(L-B)}=7}$

$\tt{\implies x-y=7}$

$\tt{\implies x=7+y------(i)}$

$\tt{\implies Diagonal^2=L^2+B^2}$

$\tt{\implies 13^2=x^2+y^2-----(ii)}$

Putting the value of x=7+y in eq (ii) here,

$\tt{\implies (7+y)^2+y^2=13^2}$

$\tt{\implies 49+14y+y^2+y^2=169}$

$\tt{\implies 2y^2+14y+49-169=0}$

$\tt{\implies 2y^2+14y-120=0}$

$\tt{\implies y^2+7y-60=0}$

Splitting the middle term here,

$\tt{\implies y^2+12y-5y-60=0}$

$\tt{\implies y(y+12)-5(y+12)=0}$

$\tt{\implies (y+12)(y-5)=0}$

$\tt{\implies \therefore x=-12\:and\:5}$

Here we can't take the value of y=-12 because negative value can't be the value of length of rectangle so, y=5.

• Now putting the value of y=5 in eq (i)

$\tt{\implies x=7+y}$

$\tt{\implies x=7+5}$

$\tt{\implies x=12}$

$\sf\large{Hence,}$

$\tt{\implies Area\:_{(rectangle)}=Length*breadth}$

$\tt{\implies Area\:_{(rectangle)}=12*5}$

$\tt{\implies Area\:_{(rectangle)}=60cm^2}$