NCERT MCQ Solutions for Class 8 Maths Ganita Prakash Chapter 6 We Distribute, Yet Things Multiply

The chapter introduces distributivity as “a property relating multiplication and addition that is captured concisely using algebra”. It is the central theme for exploring multiplication patterns and their algebraic descriptions in this chapter. 

When ‘b’ is increased by 1, the new product is a(b+1) = ab + a. This shows an increase of ‘a’ compared to the original product ‘ab’.

“a(b+c)=ab+ac”. This formula is then visualized with a diagram illustrating areas of rectangles to represent the multiplication.

When expanding (a+1)(b+1), the result is ab + b + a + 1. Comparing this to the original product ‘ab’, the increase is found to be a+b+1.

“Mathematical statements that express the equality of two algebraic expressions”. Examples include a(b+8)=ab+8a and (a + 1)(b – 1) = ab + b – a – 1.

Identity 1 is given as “(a + m)(b + n) = ab + mb + an + mn”. This identity is also visually represented by dividing a larger rectangle into four smaller rectangular areas.

The expansion is derived as (a + u)(b – v) = (a + u)b – (a + u)v = ab + ub – (av + uv) = ab + ub – av – uv. This demonstrates how integer multiplication rules apply.

The expansion is shown step-by-step: (a + b)(a + b) = (a + b)a + (a + b)b = a²+ba + ab + b². Since ba = ab, combining like terms yields a² + 2ab + b².

The “Pinch of History” section states that “The first explicit statement of the distributive property was given by Brahmagupta in his work Brahmasphuṭasiddhānta (Verse 12.55), who referred to the use of the property for multiplication as khanda-gunanam (multiplication by parts)”.

“The multiplier is broken up into two or more parts whose sum is equal to it; the multiplicand is then multiplied by each of these and the results added”. This is equivalent to (a + b)c = ac + bc.

The chapter uses an example of a square with side 65 units, splitting it into smaller regions to show its area is (60 + 5)². This visually represents the sum’s square.

Identity 1A is explicitly stated as “(a + b)² = a²+2ab + b²”. This formula is derived using the distributive property, multiplying (a + b) by (a + b).

Identity 1B is given as “(a – b)² = a² + b² – 2ab”. This identity can be derived by applying the distributive property to (a – b) × (a – b) or by substituting -b into the (a+b)² identity.

Pattern 2 introduces the relationship a²-b²=(a+b)×(a-b). Identity 1C confirms this: “(a + b) × (a – b)=a² – b²”.

Adding (a+b)² = a² + 2ab + b² and (a – b)² = a² – 2ab + b² results in 2a²+2b². This simplifies to 2(a²+b²).

The expression a² = (a + b)(a – b) + b² is a rearrangement of the identity a² – b²=(a + b)(a – b). It highlights how squaring a number can be computed using products of sums and differences.

The chapter shows that dcba × 11 = dcba × (10 + 1) = dcba × 10 + dcba. This is represented visually as adding the number to itself shifted one place to the left.

The “Math Talk” section states that “Such methods… were discussed extensively in the ancient mathematical works of Brahmagupta (628 CE), Sridharacharya (750 CE) and Bhaskaracharya (Lilavati, 1150 CE)”. Brahmagupta specifically used the term ‘ista-gunana’.

The formula derived for the number of circles at Step k is k² + 2k. For Step 15, this would be 15² + 2 × 15 = 225 + 30 = 255.

Tadang’s method calculates the area as (m + n)² – 4mn. Yusuf’s method states the area is (n – m)². The text then asks to verify that these two expressions are equivalent: (m + n)² – 4mn = (n – m)².