Q1 of 16 Page 147

In rectangles with one side 1 centimetre shorter than the other, take the length of the shorter side as x centimetres.

i) Taking their perimeters as p(x) centimetres, write the relation between p(x) and x as an equation.


ii) Taking their area as a(x) square centimetres, write the relation between a(x) and x as an equation.


iii) Calculate p(1), p(2), p(3), p(4), p(5). Do you see any pattern?


iv) Calculate a(1), a(2), a(3), a(4), a(5). Do you see any pattern?

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.


More from this chapter

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