# CLASS-9SIMULTANEOUS LINEAR EQUATIONS - PROBLEM ON AREAS

PROBLEM ON AREAS

Example.1) If the length and breadth of a room are increased by 2 m each, its area is increased by 22 m². If the length is increased by 2 m and breadth decreased by 2 m, the area is decreased by 6 m². Find the area of the room.

Ans.) Let the length and breadth of the room is x meters and y meters respectively. Then the area of the room = xy

Case.1) As per the given situation, when length = (x + 2) m, and breadth = (y + 2) m

In this case, area of the room =>  [(x + 2) (y + 2)] m²

As per the given condition =>  (x + 2) (y + 2) – xy = 22

=>   xy + 2x + 2y + 4 – xy = 22

=>   2 (x + y) =  22 – 4

=>    x + y =  9 …………………(i)

Case.2)  As per the given condition, when length = (x + 2) m, and breadth = (y – 2) m

In this case, area of the room =>  [(x + 2) (y – 2)]

As per the given condition => xy - [(x + 2) (y – 2)] = 6

=> xy – (xy + 2y – 2x – 4) = 6

=> xy – xy – 2y + 2x + 4 = 6

=>  2x – 2y =  6 – 4

=>  2 (x – y)  =  2

=>   x – y = 1  ………………………(ii)

Now, we will add (i) and (ii), and we get –

x + y = 9

x – y = 1

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2x = 10

=>          x = 5

Now, we will substitute the value of x in (i), and we get –

x + y = 9

=>       5 + y = 9

=>         y = 9 – 5

=>         y =  4

Hence the area of the room is xy = (5 X 4) = 20 m²           (Ans.)

Example.2) If the length and breadth of a room are decreased by 3 m each, its area is decreased by 24 m². If the length is increased by 4 m and breadth decreased by 1 m, the area is increased by 9 m². Find the area of the room.

Ans.) Let the length and breadth of the room is x meters and y meters respectively. Then the area of the room = xy

Case.1) As per the given situation, when length = (x – 3) m, and breadth = (y – 3) m

In this case, area of the room => [(x – 3) (y – 3)]

As per the given condition =>  xy – [(x – 3) (y – 3)] =  24

=>  xy – (xy – 3y – 3x + 9) = 24

=>  xy – xy + 3y + 3x – 9 = 24

=>    3 (x + y)  =  24 + 9

=>    x + y =  11 …………………(i)

Case.2) As per the given condition, when length = (x + 4) m, and breadth = (y – 1) m

In this case, area of the room =>  [(x + 4) (y – 1)]

As per the given condition =>  [(x + 4) (y – 1)] – xy = 10

=>  (xy + 4y – x – 4) – xy =  10

=>   xy + 4y – x – 4 – xy = 10

=>     4y – x = 10 + 4

=>     4y – x  =  14 ………………………(ii)

Now, we will add (i) and (ii), and we get –

x + y =  11

– x + 4y = 14

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5y =  25

=>          y = 5

Now, we will substitute the value of y in (i), and we get –

x + y  = 11

=>       x + 5  = 11

=>        x  = 11 – 5

=>         x = 6

Hence the area of the room is xy = (5 X 6) = 30 m²           (Ans.)