### Algebraic Expressions Class 8th Mathematics AP Board Solution

##### Question 1.Verify the identity (a + b)2 ≡ a2 + 2ab + b2 geometrically by takinga = 2 units, b = 4 unitsAnswer: (a + b)2Ξa2 + 2ab + b2Draw a square with the side a + b i.e.,2 + 4L.H.S of the whole square = (2 + 4)2 = (6)2 = 36R.H.S = Area of the square with 2 units + Area of the square with 4 units +Area of the 2,4 units + Area of the square with 4 ,2 units = 22 + 42 + 2×4 + 2×4 = 4 + 16 + 8 + 8 = 36L.H.S = R.H.SHence,the identity is verified.Question 2.Verify the identity (a + b)2 ≡ a2 + 2ab + b2 geometrically by takinga = 3 units, b = 1 unitAnswer: (a + b)2Ξa2 + 2ab + b2Draw a square with the side a + b i.e.,3 + 1L.H.S of the whole square = (3 + 1)2 = (4)2 = 16R.H.S = Area of the square with 3 units + Area of the square with 1 unit +Area of the 3,1 unit + Area of the square with 1 ,3 units = 32 + 12 + 3×1 + 1×3 = 9 + 1 + 3 + 3 = 16L.H.S = R.H.SHence, the identity is verified.Question 3.Verify the identity (a + b)2 ≡ a2 + 2ab + b2 geometrically by takinga = 5 units, b = 2 unitAnswer: (a + b)2Ξa2 + 2ab + b2Draw a square with the side a + b i.e.,5 + 2L.H.S of the whole square = (5 + 2)2 = (7)2 = 49R.H.S = Area of the square with 5 units + Area of the square with 2 units +Area of the 5,2 units + Area of the square with 2 ,5 units = 52 + 22 + 5×2 + 2×5 = 49L.H.S = R.H.SHence,the identity is verified.Question 4.Verify the identity (a – b)2 ≡ a2 - 2ab + b2geometrically by takinga = 3 units, b = 1 unitAnswer: (a – b)2 ≡ a2 - 2ab + b2Consider a square with side a.i.e.a = 3The square is divided into 4 regions.It consists of 2 squares with sides a-b and b respectively and 2 rectangles with length and breadth as ‘a-b’ and ‘b’ respectively.Here a = 3 and b = 1.Therefore the 2 squares consist of sides ‘3-1’ and ‘1’ respectively and 2 rectangles with length and breadth as ‘3-1’ and ‘1’ respectively.Now area of figure I = Area of whole square with side ‘a’ i.e.3 units-Area of figure II-Area of figure III –Area of figure IVL.H.S of area of figure I = (3-1)(3-1) = 2(2) = 4 unitsR.H.S = Area of whole square with side 3 units-Area of figure II with 1,(3-1)units-Area of figure III with 1,(3-1) units –Area of figure IV with 1,1 unit = 32-(1×(3-1))-(1×(3-1))-(1×1) = 4 unitsL.H.S = R.H.SHence,the identity is verified.Question 5.Verify the identity (a – b)2 ≡ a2 - 2ab + b2geometrically by takinga = 5 units, b = 2 unitsAnswer: (a – b)2 ≡ a2 - 2ab + b2Consider a square with side a.i.e.a = 5The square is divided into 4 regions.It consists of 2 squares with sides a-b and b respectively and 2 rectangles with length and breadth as ‘a-b’ and ‘b’ respectively.Here a = 5 and b = 2.Therefore the 2 squares consist of sides ‘5-2’ and ‘2’ respectively and 2 rectangles with length and breadth as ‘5-2’ and ‘2’ respectively.Now area of figure I = Area of whole square with side ‘a’ i.e.5 units-Area of figure II-Area of figure III –Area of figure IVL.H.S of area of figure I = (5-2)(5-2) = 3(3) = 9 unitsR.H.S = Area of whole square with side 5 units-Area of figure II with 2,(5-2)units-Area of figure III with 2,(5-2) units –Area of figure IV with 2,2 units = 25-(2×3)-( 2×3)-22 = 25-6-6-4 = 9 unitsL.H.S = R.H.SHence,the identity is verified.Question 6.Verify the identity (a + b) (a – b) ≡ a2 – b2 geometrically by takinga = 3 units, b = 2 unitsAnswer:Remove square from this whose side is ‘b’.(b<a) Consider a square with side ‘a’.  We get the above figure by removing the square with side ‘b’.It consists of 2 partsI and II.So a2-b2 = Area of figure I + Area of figure II = a(a-b) + b(a-b) = (a-b)(a + b)Thus a2-b2 = (a-b)(a + b).Here a = 3 units and b = 2 unitsSo,L.H.S = a2-b2 = 32-22 = 5 unitsR.H.S = (a-b)(a + b) = (3-2)(3 + 2) = 1(5) = 5 unitsTherefore L.H.S = R.H.SHence,the identity is verified.Question 7.Verify the identity (a + b) (a – b) ≡ a2 – b2 geometrically by takinga = 2 units, b = 1 unitAnswer:Refer the above figure.Here a2-b2 = Area of figure I + Area of figure II = a(a-b) + b(a-b) = (a-b)(a + b)Thus a2-b2 = (a-b)(a + b).Here a = 2 units and b = 1 unitsSo, L.H.S = a2-b2 = 22-12 = 3 unitsR.H.S = (a-b)(a + b) = (2-1)(2 + 1) = 3 unitsTherefore L.H.S = R.H.SHence,the identity is verified.

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