DAV Class 8 Maths Chapter 7 Worksheet 2

DAV Class 8 Maths Chapter 7 Worksheet 2

Algebraic Identities Worksheet 2


1. Find the following by using identity - II:

(i) \( (x-7)^2 \)

Solution

\[ \begin{align*} \color{green} (a-b)^2 &= \color{green} a^2 - 2ab + b^2 \\ & a = x \, , \, b = 7 \\[6pt] (x-7)^2 & = (x)^2 - 2 \times x \times 7 + (7)^2 \\[6pt] & = \color{red} x^2 - 14x + 49 \end{align*} \]

(ii) \( (5a - 4b)^2 \)

Solution

\[ \begin{align*} \color{green} (a-b)^2 &= \color{green} a^2 - 2ab + b^2 \\ & a = 5a \, , \, b = 4b \\[6pt] (5a - 4b)^2 & = (5a)^2 - 2 \times 5a \times 4b + (4b)^2 \\[6pt] & = \color{red} 25a^2 - 40ab + 16b^2 \end{align*} \]

(iii) \( \left( \dfrac{7x}{4} - \dfrac{2y}{3} \right)^2 \)

Solution

\[ \begin{align*} \color{green} (a-b)^2 &= \color{green} a^2 - 2ab + b^2 \\[6pt] & a = \frac{7x}{4} \, , \, b = \frac{2y}{3} \\[6pt] \left( \frac{7x}{4} - \frac{2y}{3} \right)^2 & = \left( \frac{7x}{4} \right)^2 - \cancel{2}^{\color{orange}1} \times \frac{7x}{{\cancel{4}_{\cancel{\color{orange}2}}}_{1}} \times \frac{{\cancel{2}^{1}}y}{3} + \left( \frac{2y}{3} \right)^2 \\[6pt] & = \color{red} \frac{49x^2}{16} - \frac{7xy}{3} + \frac{4y^2}{9} \end{align*} \]

(iv) \( (8mn - 3pq)^2 \)

Solution

\[ \begin{align*} \color{green} (a-b)^2 &= \color{green} a^2 - 2ab + b^2 \\[6pt] & a = 8mn \, , \, b = 3pq \\[6pt] (8mn - 3pq)^2 & = (8mn)^2 - 2 \times 8mn \times 3pq + (3pq)^2 \\[6pt] & = \color{red} 64m^2n^2 - 48mnpq + 9p^2q^2 \end{align*} \]

(v) \( (0.1x - 0.5y)^2 \)

Solution

\[ \begin{align*} \color{green} (a-b)^2 &= \color{green} a^2 - 2ab + b^2 \\[6pt] & a = 0.1x \, , \, b = 0.5y \\[6pt] (0.1x - 0.5y)^2 & = (0.1x)^2 - 2 \times 0.1x \times 0.5y + (0.5y)^2 \\[6pt] & = \color{red} 0.01x^2 - 0.1xy + 0.25y^2 \end{align*} \]

(vi) \( (a^2 - b^2)^2 \)

Solution

\[ \begin{align*} \color{green} (a-b)^2 &= \color{green} a^2 - 2ab + b^2 \\[6pt] & a = a^2 \, , \, b = b^2 \\[6pt] (a^2 - b^2)^2 & = (a^2)^2 - 2 \times a^2 \times b^2 + (b^2)^2 \\[6pt] & = \color{red} a^4 - 2a^2b^2 + b^4 \end{align*} \]

2. Evaluate the following by using identity - II:

(i) \( (48)^2 \)

Solution

\[ \begin{align*} \color{green} (a - b)^2 &= \color{green} a^2 - 2ab + b^2 \\[6pt] (48)^2 & = (50 -2)^2 \\[6pt] & = (50)^2 - 2 \times 50 \times 2 + (2)^2 \\[6pt] & = 2500 - 200 + 4 \\[6pt] & = 2300 + 4 \\[6pt] (48)^2 & = \color{red} 2304 \end{align*} \]

(ii) \( (9.9)^2 \)

Solution

\[ \begin{align*} (9.9)^2 & = (10 - 0.1)^2 \\[6pt] & = (10)^2 - 2 \times 10 \times 0.1 + (0.1)^2 \\[6pt] & = (10)^2 - 2 \times 1 + (0.1)^2 \\[6pt] & = 100 - 2 + 0.01 \\[6pt] & = 98 + 0.01 \\[6pt] (9.9)^2 & = \color{red} 98.01 \end{align*} \]

(iii) \( (299)^2 \)

Solution

\[ \begin{align*} (299)^2 & = (300 - 1)^2 \\[6pt] & = (300)^2 - 2 \times 300 \times 1 + (1)^2 \\[6pt] & = 90000 - 600 + 1 \\[6pt] & = 89400 + 1 \\[6pt] (299)^2 & = \color{red} 89401 \end{align*} \]

(iv) \( (98)^2 \)

Solution

\[ \begin{align*} (98)^2 & = (100 - 2)^2 \\[6pt] & = (100)^2 - 2 \times 100 \times 2 + (2)^2 \\[6pt] & = 10000 - 400 + 4 \\[6pt] & = 9600 + 4 \\[6pt] (98)^2 & = \color{red} 9604 \end{align*} \]

(v) \( (87)^2 \)

Solution

\[ \begin{align*} (87)^2 & = (90 - 3)^2 \\[6pt] & = (90)^2 - 2 \times 90 \times 3 + (3)^2 \\[6pt] & = 8100 - 540 + 9 \\[6pt] & = 7560 + 9 \\[6pt] (87)^2 & = \color{red} 7569 \end{align*} \]

(vi) \( (19.9)^2 \)

Solution

\[ \begin{align*} (19.9)^2 & = (20 - 0.1)^2 \\[6pt] & = (20)^2 - 2 \times 20 \times 0.1 + (0.1)^2 \\[6pt] & = 400 - 4 + 0.01 \\[6pt] & = 396 + 0.01 \\[6pt] (19.9)^2 & = \color{red} 396.01 \end{align*} \]