# CLASS-9PERIMETER & AREA OF QUADRILATERAL - PROBLEM & SOLUTION

Example.1) The length and breadth of a rectangular grassy plot are in the ration 7 : 4. A path 4 m wide running all around outside it has an area of 416 m². Find the dimensions of the grassy plot.

Ans.) Let the length & breadth of the grassy plot be 7a and 4a meters respectively. Then the area of the plot is = (7a X 4a) m² = (28 a²)

Now, as per the given condition, a path of 4 m wide running all around outside area and it has area of 416 As you can find from the above picture, in length there is = 4 m + 4 m = 8 m more length to be added in given length, and in breadth also there is = 4 m + 4 m = 8 m more breadth is to be added in the given breadth size.

Now, the length along with running area should be = (7a + 8) m

And, the breadth along with running area should be = (4a + 8) m

So, area of the plot including the path = (7a + 8) (4a + 8)

Now, area of the path =>  [{(7a + 8)(4a + 8)} – 28a² ]m² = 416 m²

Or,   (28a² + 32a + 56a + 64) – 28a² = 416

Or,      88 a = 416 – 64

Or,        88a = 352

Or,     a = 352/88 = 4

So, now the length of the plot is 7a = 7 X 4 = 28 m, and width of the given plot is = 4a = 4 X 4 = 16 m       (Ans.)

Example.2) A rectangular lawn 64 m by 48 m has two roads, each 5 m wide , running in the middle of it, one parallel to length and other parallel breadth. Find the cost of graveling them at \$ 2.80 per m² From the above picture, area of road ABCD = (64 X 5) m² = 320

Area of road EFGH = (48 X 5) m² = 240

Clearly, area JKLM is common to both of the above roads.

Area JKLM = (5 X 5) m² = 25

Area of the road to graveled = {(320 + 240) – 25} m² = 535

Cost of graveling of the path = \$ (2.80 X 535) = \$ 1498    (Ans.)

Example.3)  If the length and breadth of a rectangular room are each increased by 1 m, then the area of the floor increased by 21 m². if the length is increased by 1 m and breadth is decreased by 1 m, then the area is decreased by 7 m². Find the perimeter of the floor.

Ans.) Let, the original length and breadth of the room be x and y meters respectively. Then, original area of the floor = (xy) m²

When both length and breadth are increased by 1 m

New length = (x + 1) m

New breadth = (y + 1) m

New area = (x + 1)(y + 1)

New area => [{(x + 1) (y + 1)} – xy] = 21

=> xy + y + x + 1 – xy = 21

=>  y + x = 21 - 1

=>  y + x = 20 ……………………….(i)

When length is increased by 1 m and breadth is decreased by 1 m

New length = (x + 1) m

New breadth = (y - 1) m

New area = (x + 1)(y - 1)

New area => [xy - {(x + 1) (y - 1)}] = 7

=> xy – (xy + y - x – 1)  = 21

=>  x - y + 1 = 7

=>  x - y = 6 ……………………….(ii)

Adding (i) & (ii), and we get –

x + y = 20

x – y =  6

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2x = 26

Or,  x = 13

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

y + x = 20

or,     y + 13 = 20

or,        y = 7

so, the perimeter of the floor –

[2 (length + Breadth)] = 2 (13 + 7) = 2 X 20

= 40 m    (Ans.)