Question

(iii) A rectangle has the same area as another, whose length is 6 m more and
is 4 m less. It has also the same area as the third, whose length is 8 m more and
breadth 5 m less. Find the length and breadth of the original rectangle.​

Answer 1

Let the length and breadth of the original rectangle be x and y respectively.

Let the original rectangle be A, second rectangle be B and the third rectangle be C

Area of A = xy

Length of B is 6 more than that of A, and breadth is 4 less

Hence, area of B = (x+6)(y-4)

Since the areas of A and B are equal, we can say  that

xy = (x+6)(y-4)

xy = xy+6y-4x-24

4x-6y = -24

2x-3y+12 = 0 .... (equation  [i] )

Length of C is 8 more than that of  A, whereas breadth is 5 less

Hence, area of C = (x+8)(y-5)

Since areas of A and C are equal, we can say that

xy = (x+8)(y-5)

xy = xy+8y-5x-40

5x-8y+40 = 0 .... (equation [ii] )

Solving equations [i] and [ii]

x = 24 m

y = 20 m

Hence the length of the original rectangle is 24 m and its breadth is 20 m

Areas of all the three rectangles are equal and is 480 m²

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