# In rectangles wit

Given one side is smaller than the other by 1 cm.

The two adjacent sides of the triangle are x, x + 1.

(i) perimeter p(x) = 2 × [(x) + (x + 1)]

p(x) = 4x + 2 – (1)

(ii) Area a(x) = (x) × (x + 1)

a(x) = x2 + x – (2)

(iii) by (1)

p(1) = 4 × 1 + 2 = 6

p(2) = 4 × 2 + 2 = 10

p(3) = 4 × 3 + 2 = 14

p(4) = 4 × 4 + 2 = 18

p(5) = 4 × 5 + 2 = 22

Here the difference between a number and its successor is always 4.

We can see that they are in Arithmetic Progression (AP) with a common difference of 4.

(iv) by (2)

a(1) = 12 + 1 = 2

a(2) = 22 + 2 = 6

a(3) = 32 + 3 = 12

a(4) = 42 + 4 = 20

a(5) = 52 + 5 = 30

Here the difference between a number and its previous number is increasing by 2.

a(2) – a(1) = 4

a(3) – a(2) = 6

a(4) – a(3) = 8

a(5) – a(4) = 10

Their common difference is in AP.

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