Example 17 - Chapter 4 Class 9 - Exploring Algebraic Identities (Ganita Manjari I)

Transcript

Example 17 Saira has arranged a square of side 𝑥 units, 8 rectangular strips of sides 𝑥 units and width 1 unit, and 15 squares of side 1 unit to form a bigger rectangle. Find the length and breadth of the rectangle in terms of 𝑥. We can do this without drawing a figure Here, Area of square of side x units + 8 × Area of rectangle of side x units & width 1 unit + 15 × Area of square of side 1 unit = Area of bigger rectangle Putting values 𝑥 × 𝑥+8 × 𝑥 × 1+15 × 1 × 1= Area of bigger rectangle 𝑥^2+8𝑥+15= Area of bigger rectangle Area of bigger rectangle = 𝒙^𝟐+𝟖𝒙+𝟏𝟓 To find length and breadth of bigger rectangle, we need to factorise 𝒙^𝟐+𝟖𝒙+𝟏𝟓 Factorising 𝒙^𝟐+𝟖𝒙+𝟏𝟓 𝒙^𝟐+𝟖𝒙+𝟏𝟓 𝒙 × 𝒙+𝟖 × 𝒙 × 𝟏+𝟏𝟓 × 𝟏 × 𝟏= Area of bigger rectangle 𝑥^2+8𝑥+15= Area of bigger rectangle Area of bigger rectangle = 𝒙^𝟐+𝟖𝒙+𝟏𝟓 To find length and breadth of bigger rectangle, we need to factorise 𝒙^𝟐+𝟖𝒙+𝟏𝟓 Factorising 𝒙^𝟐+𝟖𝒙+𝟏𝟓 𝑥^2+8𝑥+15 Factorising by splitting the middle term = x2 + 3x + 5x + 15 = x(x + 3) + 5(x + 3) = (x + 5) (x + 3) Splitting the middle term method We need to find two numbers whose Sum = 8 Product = 1 × 15 = 15 Since sum and product both are positive, both numbers are positive = (x + 5) (x + 3) Thus, Area of bigger rectangle = 𝑥^2+8𝑥+15 = (x + 5) (x + 3) So, we can say Length of bigger rectangle = (x + 5) Breadth of bigger rectangle = (x + 3) Here is how the figure might look