NCERT Solutions for Class 8 Maths Ganita Prakash Chapter 6 — Figure it Out (Page 149)#
The identities used here:
- 1A: $(a + b)^2 = a^2 + 2ab + b^2$
- 1B: $(a - b)^2 = a^2 - 2ab + b^2$
- 1C: $(a + b)(a - b) = a^2 - b^2$
1. Which is greater: (a – b)² or (b – a)²? Justify your answer. #
Solution
We expand both using Identity 1B:
$(a - b)^2 = a^2 - 2ab + b^2$
$(b - a)^2 = b^2 - 2ba + a^2$
These are the same expression, since $ab = ba$.
Another way: $b - a = -(a - b)$, and squaring a number and its negative gives the same result.
∴ Neither is greater; they are equal. For example, $(7 - 3)^2 = 16$ and $(3 - 7)^2 = (-4)^2 = 16$.
2. Express 100 as the difference of two squares. #
Solution
By Identity 1C, $a^2 - b^2 = (a + b)(a - b)$. So we need two numbers whose product is 100, and then:
- $a + b$ = the larger factor and $a - b$ = the smaller factor.
- Both factors must be even (or both odd) so that $a$ and $b$ are whole numbers.
$100 = 50 \times 2$, with both factors even:
$$a + b = 50 \quad \cdots (1)$$$$a - b = 2 \quad \cdots (2)$$
Adding equation (1) and equation (2):
$$\begin{array}{rcl} a + b &=& 50 \\ a - b &=& 2 \\ \hline 2a &=& 52 \end{array}$$$$\Rightarrow a = 26$$
Substituting $a = 26$ in equation (1):
$$a + b = 50$$$$26 + b = 50$$
$$\Rightarrow b = 24$$
Check: $26^2 - 24^2 = 676 - 576 = 100$.
∴ $100 = $ $26^2 - 24^2$. (Using $100 = 10 \times 10$ gives the simpler $100 = 10^2 - 0^2$.)
3. Find 406², 72², 145², 1097², and 124² using the identities you have learnt so far. #
406²
Using Identity 1A with $406 = 400 + 6$:
$$406^2 = (400 + 6)^2$$$$= 400^2 + 2 \times 400 \times 6 + 6^2$$
$$= 160000 + 4800 + 36$$
$$= 164836$$
72²
Using Identity 1A with $72 = 70 + 2$:
$$72^2 = (70 + 2)^2$$$$= 4900 + 280 + 4$$
$$= 5184$$
145²
Using Identity 1A with $145 = 140 + 5$:
$$145^2 = (140 + 5)^2$$$$= 19600 + 1400 + 25$$
$$= 21025$$
1097²
Using Identity 1B with $1097 = 1100 - 3$:
$$1097^2 = (1100 - 3)^2$$$$= 1100^2 - 2 \times 1100 \times 3 + 3^2$$
$$= 1210000 - 6600 + 9$$
$$= 1203409$$
124²
Using Identity 1A with $124 = 120 + 4$:
$$124^2 = (120 + 4)^2$$$$= 14400 + 960 + 16$$
$$= 15376$$
4. Do Patterns 1 and 2 hold only for counting numbers? Do they hold for negative integers as well? What about fractions? Justify your answer. #
Solution
The two patterns are the identities:
- Pattern 1: $2(a^2 + b^2) = (a + b)^2 + (a - b)^2$
- Pattern 2: $a^2 - b^2 = (a + b)(a - b)$
Both were proved by expanding with the distributive property, which is true for all numbers. So the patterns hold for counting numbers, negative integers and fractions alike.
Negative integers, $a = -3$, $b = 5$:
$2((-3)^2 + 5^2) = 2(9 + 25) = 68$
$(-3 + 5)^2 + (-3 - 5)^2 = 4 + 64 = 68$ ✓
Fractions, $a = \dfrac{1}{2}$, $b = \dfrac{1}{3}$:
$a^2 - b^2 = \dfrac{1}{4} - \dfrac{1}{9} = \dfrac{5}{36}$
$(a + b)(a - b) = \dfrac{5}{6} \times \dfrac{1}{6} = \dfrac{5}{36}$ ✓