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MathematicsClass 8Yet Things

Yet Things | Class 8 Mathematics Notes

By ConceptScroll Team · Published on 17 July 2026 · 4 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.

Investigating Patterns

This section explores interesting numerical patterns and their algebraic explanations using the identities learned earlier.

Pattern 1 examines expressions like 2(2² + 1²) = 3² + 1² and generalizes that twice the sum of squares of two numbers can be expressed as the sum of squares of their sum and difference: 2(a² + b²) = (a + b)² + (a - b)².

Pattern 2 looks at differences of squares such as 9×9 - 1×1 = 10×8 and identifies the pattern a² - b² = (a + b)(a - b).

The section encourages students to verify these patterns with various numbers and to understand the algebraic identities that explain them.

Historical notes mention mathematicians like Sridharacharya who used these identities for quick calculations.

The section also includes geometric proofs and applications of these identities to compute squares and products efficiently.

📊 Diagram: Hint: Sridharacharya (750 CE) gave an interesting method to quickly compute the squares of numbers using Identity 1C! Consider the following modified form of this identity —; Figure on page 14 showing geometric proof of (a + b)(a - b) = a² - b².

🧪 Activity: Try different pairs of numbers to verify the patterns and use the identities to compute squares and products efficiently.

🔗 Connection: Understanding these patterns prepares students to identify and correct common mistakes in algebraic expansions, as discussed in the next section.

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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