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The area of a rectangle decreases by 10cm2 if the length is decreased by 5cm and the breadth is increased by 3cm. If the length is increased by 5cm and the breadth is increased by 2cm, - Mathematics

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Question

The area of a rectangle decreases by 10cm2 if the length is decreased by 5cm and the breadth is increased by 3cm. If the length is increased by 5cm and the breadth is increased by 2cm, then the area increases by 80cm2. Find the perimeter of the rectangle.

Sum
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Solution

Here, let the original length be x cm and the breadth be y cm.

According to the given conditions,

(1) Area decreases by 10cm2 when:

  1. Length is decreased by 5cm = x − 5
  2. Breadth is increased by 3cm = y + 3

(x − 5)(y + 3) = xy − 10

x(y + 3) − 5(y + 3) = xy − 10

xy + 3x − 5y − 15 = xy − 10

xy − xy + 3x − 5y = −10 + 15

3x − 5y = 5     ...(i)

(2) Area decreases by 80cm2 when:

  1. Length is decreased by 5cm = x + 5
  2. Breadth is increased by 2cm = y + 2

(x + 5)(y + 2) = xy + 80

x(y + 2) + 5(y + 2) = xy + 80

xy + 2x + 5y + 10 = xy + 80

xy − xy + 2x + 5y = 80 − 10

2x + 5y = 70     ...(ii)

Now, adding equations (i) and (ii):

(3x − 5y) + (2x + 5y) = 5 + 70

3x − 5y + 2x + 5y = 75

3x + 2x = 75

5x = 75

x = `75/5`

∴ x = 15

Substitute x = 15 in equation (i):

3(15) − 5y = 5

45 − 5y = 5

5y = 40

y = `40/5`

∴ y = 8

Perimeter of the rectangle = 2(x + y)

= 2(15 + 8)

= 2 × 23

= 46

Hence, the perimeter of the rectangle is 46cm.

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Chapter 5: Simultaneous Linear Equations - EXERCISE 5B [Page 59]

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B Nirmala Shastry Mathematics [English] Class 9 ICSE
Chapter 5 Simultaneous Linear Equations
EXERCISE 5B | Q 20. | Page 59
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