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MathematicsClass 8Yet Things

Yet Things | Class 8 Mathematics Notes

By ConceptScroll Team · Published on 17 July 2026 · 5 min read

Yet Things | Class 8 Mathematics Notes

Yet Things – this guide gives you a concise, exam-ready overview of Yet Things from Class 8 Mathematics, written by ConceptScroll editors and reviewed against the latest NCERT textbook.

Special Cases of the Distributive Property

This section explores special algebraic identities derived from the distributive property, focusing on squares of sums and differences and the product of sum and difference of two numbers.

Starting with geometric visualization, the area of a square with side length (a + b) is decomposed into smaller parts: a square of side a, a square of side b, and two rectangles each with sides a and b. This leads to the algebraic identity (a + b)² = a² + 2ab + b².

Similarly, the square of a difference (a - b)² is expanded using the distributive property and is found to be a² - 2ab + b². The section also shows how this can be derived from the square of a sum by substituting b with -b.

Another important identity is the product of sum and difference: (a + b)(a - b) = a² - b². This identity is verified algebraically and geometrically.

The section includes examples of expanding expressions using these identities and encourages students to use these identities to compute squares of numbers efficiently, such as 104² or 37².

Patterns involving sums and differences of squares are also investigated, showing how algebraic identities explain observed numerical patterns.

📊 Diagram: What if we write 65² as (30 + 35)² or (52 + 13)²? Draw the figures and check the area that you get.; Using the distributive property, (a + b)² can be expanded as; If you have difficulty remembering or using the general rule, you can just apply the distributive property to multiply and get the desired result.; We can get the area of the square of sidelength 55 by taking the area of the square of sidelength 60 and removing the areas of the two rectangles of sidelengths 60 and 5, i.e., 60² - (60 × 5) - ...

🧪 Activity: Use the identities to find squares of numbers like 104², 37², 99², and 58² by decomposing them into sums or differences of numbers whose squares are easy to compute.

🔗 Connection: This section's identities are foundational for understanding algebraic expansions and patterns, which are further explored in the next section on investigating patterns.

Frequently asked questions

By expanding the expressions, verify that all three expressions are equivalent. If x = 8 and y = 3, find the area of the shaded region.

The three expressions given are: Vaishnavi's method: x(x + 2y) - 3xy, Aditya's method: x(x - y), and the third expression (not explicitly stated here but implied to be equivalent). Expanding Vaishnavi's expression: x(x + 2y) - 3xy = x^2 + 2xy - 3xy = x^2 - xy. Expanding Aditya's expression: x(x - y) = x^2 - xy. Both expressions are equal. Substituting x = 8 and y = 3: Area = 8^2 - 8*3 = 64 - 24 = 40. Therefore, the area of the shaded region is 40 square units.

Write an expression for the area of the dashed region in the figure below. Use more than one method to arrive at the answer. Substitute p = 6, r = 3.5, and s = 9, and calculate the area.

Method 1: Consider the larger rectangle with dimensions p and (r + s + r) = p × (2r + s). The dashed region is the area of this rectangle minus the two smaller rectangles of dimensions p × r each. Area = p(2r + s) - 2pr = p(2r + s - 2r) = p × s. Method 2: Alternatively, sum the areas of the two rectangles of dimensions p × r and the rectangle p × s, then subtract the two p × r rectangles to get the dashed area. Substituting p = 6, r = 3.5, s = 9: Area = 6 × 9 = 54 square units.

1. Compute these products using the suggested identity. (i) 462 using Identity 1A for (a + b)^2 (ii) 397 × 403 using Identity 1C for (a + b)(a – b) (iii) 912 using Identity 1B for (a – b)^2 (iv) 43 × 45 using Identity 1C for (a + b)(a – b)

(i) 462 = (400 + 60 + 2)^2 can be simplified using (a + b)^2 = a^2 + 2ab + b^2. Let a = 460, b = 2 (or better to split as 40 + 20 + 2, but here the identity is for two terms, so better to consider 40 + 22 or 400 + 62). Since 462 is a number, likely the question means 46^2. Assuming 46^2: 46^2 = (40 + 6)^2 = 40^2 + 2×40×6 + 6^2 = 1600 + 480 + 36 = 2116. (ii) 397 × 403 = (400 - 3)(400 + 3) = 400^2 - 3^2 = 160000 - 9 = 159991. (iii) 912 = (900 + 12)^2 or (a - b)^2? Assuming 91^2: 91^2 = (90 + 1)^2

2. Use either a suitable identity or the distributive property to find each of the following products. (i) (p – 1)(p + 11) (ii) (3a – 9b)(3a + 9b) (iii) –(2y + 5)(3y + 4) (iv) (6x + 5y)^2 (v) (2x – 1)^2 (vi) (7p) × (3r) × (p + 2)

(i) (p – 1)(p + 11) = p^2 + 11p – p – 11 = p^2 + 10p – 11. (ii) (3a – 9b)(3a + 9b) = (3a)^2 – (9b)^2 = 9a^2 – 81b^2. (iii) –(2y + 5)(3y + 4) = –[2y×3y + 2y×4 + 5×3y + 5×4] = –[6y^2 + 8y + 15y + 20] = –(6y^2 + 23y + 20) = –6y^2 – 23y – 20. (iv) (6x + 5y)^2 = (6x)^2 + 2×6x×5y + (5y)^2 = 36x^2 + 60xy + 25y^2. (v) (2x – 1)^2 = (2x)^2 – 2×2x×1 + 1^2 = 4x^2 – 4x + 1. (vi) (7p) × (3r) × (p + 2) = 21pr(p + 2) = 21prp + 42pr = 21p^2r + 42pr.

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