For (a + b)(a – b), remove corner square b² from square a²; rearrange the remaining L-shape into rectangle (a + b) by (a – b). For (a + b + c)², divide a square into 9 blocks yielding a² + b² + c² + 2ab + 2bc + 2ca.
Identities in algebra can sometimes be shown as area relationships. The figure shown corresponds to the identity (a + b)² = a² + 2ab + b². Do you see how? Draw figures corresponding to the identities (a + b)(a – b) = a² – b² and (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca.
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In Fig. 6.41, side length (a + b) forms a square split into regions a², ab, ab and b², summing to a² + 2ab + b².
For (a + b)(a – b) = a² – b²: Start with a square of side a (area a²). Cut out a corner square of side b (area b²). The remaining L-shaped region splits into two rectangles of dimensions (a – b) by a and (a – b) by b, which combine into one rectangle of sides (a + b) and (a – b).
For (a + b + c)²: Partition a square of side (a + b + c) into 9 sub-rectangles: three squares a², b², c² and six rectangular regions giving 2ab, 2bc, 2ca.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-6/