NCERT Solutions for Class 8 Maths Ganita Prakash Chapter 6 — Figure it Out (Pages 154–156)#
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. 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
$$= 40^2 + 2 \times 40 \times 6 + 6^2$$
$$= 1600 + 480 + 36$$
$$= 2116$$
(ii) 397 × 403
$$= 400^2 - 3^2$$
$$= 160000 - 9$$
$$= 159991$$
(iii) 91²
$$= 100^2 - 2 \times 100 \times 9 + 9^2$$
$$= 10000 - 1800 + 81$$
$$= 8281$$
(iv) 43 × 45
43 and 45 are 1 less and 1 more than 44:
$$43 \times 45 = (44 - 1)(44 + 1)$$$$= 44^2 - 1^2$$
$$= 1936 - 1$$
$$= 1935$$
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)² (vi) (7p) × (3r) × (p + 2) #
(i) (p – 1)(p + 11)
$$= p^2 + 10p - 11$$
(ii) (3a – 9b)(3a + 9b)
Using Identity 1C:
$$(3a - 9b)(3a + 9b) = (3a)^2 - (9b)^2$$$$= 9a^2 - 81b^2$$
(iii) –(2y + 5)(3y + 4)
$$= -(6y^2 + 23y + 20)$$
$$= -6y^2 - 23y - 20$$
(iv) (6x + 5y)²
Using Identity 1A:
$$(6x + 5y)^2 = (6x)^2 + 2 \times 6x \times 5y + (5y)^2$$$$= 36x^2 + 60xy + 25y^2$$
(v) (2x – 1/2)²
Using Identity 1B:
$$\left(2x - \dfrac{1}{2}\right)^2 = (2x)^2 - 2 \times 2x \times \dfrac{1}{2} + \left(\dfrac{1}{2}\right)^2$$$$= 4x^2 - 2x + \dfrac{1}{4}$$
(vi) (7p) × (3r) × (p + 2)
$$= 21p^2r + 42pr$$
3. For each statement identify the appropriate algebraic expression(s). (i) Two more than a square number. 2 + s, (s + 2)², s² + 2, s² + 4, 2s², 2²s (ii) The sum of the squares of two consecutive numbers m² + n², (m + n)², m² + 1, m² + (m + 1)², m² + (m – 1)², (m + (m + 1))², (2m)² + (2m + 1)² #
(i) Two more than a square number
A square number is $s^2$, and two more than it is $s^2 + 2$.
The others do not fit: $2 + s$ adds 2 to $s$ itself, $(s + 2)^2$ is the square of $s + 2$, $s^2 + 4$ is four more, and $2s^2$ and $2^2 s$ are products.
(ii) The sum of the squares of two consecutive numbers
Two consecutive numbers can be written as $m$ and $m + 1$, or as $m - 1$ and $m$. So both $m^2 + (m + 1)^2$ and $m^2 + (m - 1)^2$ are correct.
The others do not fit:
- $m^2 + n^2$: $m$ and $n$ need not be consecutive.
- $(m + n)^2$ and $(m + (m + 1))^2$: these are squares of a sum, not sums of squares.
- $m^2 + 1$: this is $m^2 + 1^2$, which is consecutive only when $m$ is 0 or 2.
- $(2m)^2 + (2m + 1)^2$: this covers only pairs whose smaller number is even (like 4 and 5), not all pairs.
4. Consider any 2 by 2 square of numbers in a calendar, as shown in the figure. Find products of numbers lying along each diagonal — 4 × 12 = 48, 5 × 11 = 55. Do this for the other 2 by 2 squares. What do you observe about the diagonal products? Explain why this happens. Hint: Label the numbers in each 2 by 2 square as a, (a + 1), (a + 7), (a + 8). #
Solution
Some more 2 by 2 squares from the February calendar:
| 2 by 2 square | Diagonal products | Difference |
|---|---|---|
| 4, 5 / 11, 12 | $4 \times 12 = 48$, $5 \times 11 = 55$ | 7 |
| 3, 4 / 10, 11 | $3 \times 11 = 33$, $4 \times 10 = 40$ | 7 |
| 13, 14 / 20, 21 | $13 \times 21 = 273$, $14 \times 20 = 280$ | 7 |
Observation: the two diagonal products always differ by 7.
Why: in a calendar, the number below is 7 more, since a week has 7 days. So any 2 by 2 square is
| $a$ | $a + 1$ |
|---|---|
| $a + 7$ | $a + 8$ |
and the two diagonal products are:
$$a(a + 8) = a^2 + 8a$$$$(a + 1)(a + 7) = a^2 + 7a + a + 7$$
$$= a^2 + 8a + 7$$
The second is always 7 more than the first.
5. Verify which of the following statements are true. (i) (k + 1) (k + 2) – (k + 3) is always 2. (ii) (2q + 1) (2q – 3) is a multiple of 4. (iii) Squares of even numbers are multiples of 4, and squares of odd numbers are 1 more than multiples of 8. (iv) (6n + 2)² – (4n + 3)² is 5 less than a square number. #
(i) (k + 1)(k + 2) – (k + 3) is always 2.
False.
$$(k + 1)(k + 2) - (k + 3) = k^2 + 3k + 2 - k - 3$$$$= k^2 + 2k - 1$$
This depends on $k$. For example, $k = 1$ gives $2$, but $k = 2$ gives $7$.
(ii) (2q + 1)(2q – 3) is a multiple of 4.
False.
$$(2q + 1)(2q - 3) = 4q^2 - 6q + 2q - 3$$$$= 4q^2 - 4q - 3$$
$$= 4(q^2 - q - 1) + 1$$
It is always 1 more than a multiple of 4. In fact, it is a product of two odd numbers, so it is odd. For example, $q = 2$ gives $5 \times 1 = 5$.
(iii) Squares of even and odd numbers
True.
An even number is $2n$:
$$(2n)^2 = 4n^2$$which is a multiple of 4.
An odd number is $2n + 1$:
$$(2n + 1)^2 = 4n^2 + 4n + 1$$$$= 4n(n + 1) + 1$$
One of $n$ and $n + 1$ is even, so $n(n + 1)$ is even and $4n(n + 1)$ is a multiple of 8. So the square is 1 more than a multiple of 8 (for example, $5^2 = 25 = 24 + 1$).
(iv) (6n + 2)² – (4n + 3)² is 5 less than a square number.
False (in general).
$$(6n + 2)^2 - (4n + 3)^2 = (36n^2 + 24n + 4) - (16n^2 + 24n + 9)$$$$= 20n^2 - 5$$
This is 5 less than $20n^2$, which is not a square number for $n = 1, 2, 3, \ldots$ (for example, $20$ and $80$ are not squares).
6. A number leaves a remainder of 3 when divided by 7, and another number leaves a remainder of 5 when divided by 7. What is the remainder when their sum, difference, and product are divided by 7? #
Solution
Let the numbers be $7a + 3$ and $7b + 5$.
Sum:
$$(7a + 3) + (7b + 5) = 7a + 7b + 8$$$$= 7(a + b + 1) + 1$$
Remainder 1.
Difference (the second number minus the first):
$$(7b + 5) - (7a + 3) = 7(b - a) + 2$$Remainder 2. (If the first number is the larger one, the difference is $7(a - b) - 2 = 7(a - b - 1) + 5$, remainder 5.)
Product:
$$(7a + 3)(7b + 5) = 49ab + 35a + 21b + 15$$$$= 7(7ab + 5a + 3b + 2) + 1$$
Remainder 1.
7. Choose three consecutive numbers, square the middle one, and subtract the product of the other two. Repeat the same with other sets of numbers. What pattern do you notice? How do we write this as an algebraic equation? Expand both sides of the equation to check that it is a true identity. #
Solution
Examples:
- 4, 5, 6: $5^2 - 4 \times 6 = 25 - 24 = 1$
- 9, 10, 11: $10^2 - 9 \times 11 = 100 - 99 = 1$
- 19, 20, 21: $20^2 - 19 \times 21 = 400 - 399 = 1$
Pattern: the answer is always 1.
As an equation: with the middle number $n$, the numbers are $n - 1$, $n$ and $n + 1$:
$$n^2 - (n - 1)(n + 1) = 1$$Check: using Identity 1C, $(n - 1)(n + 1) = n^2 - 1$, so
$$n^2 - (n - 1)(n + 1) = n^2 - (n^2 - 1)$$$$= 1$$
So it is a true identity.
8. What is the algebraic expression describing the following steps—add any two numbers. Multiply this by half of the sum of the two numbers? Prove that this result will be half of the square of the sum of the two numbers. #
Solution
Let the two numbers be $a$ and $b$. Their sum is $a + b$, and half of the sum is $\dfrac{1}{2}(a + b)$. The steps give:
$$(a + b) \times \dfrac{1}{2}(a + b)$$Proof:
$$(a + b) \times \dfrac{1}{2}(a + b) = \dfrac{1}{2}(a + b)(a + b)$$$$= \dfrac{1}{2}(a + b)^2$$
which is half of the square of the sum. For example, with 3 and 5: $8 \times 4 = 32$, and $\dfrac{1}{2} \times 8^2 = 32$.
9. Which is larger? Find out without fully computing the product. (i) 14 × 26 or 16 × 24 (ii) 25 × 75 or 26 × 74 #
(i) 14 × 26 or 16 × 24
Both pairs add up to 40, so both are centred on 20. By Identity 1C:
$14 \times 26 = (20 - 6)(20 + 6) = 400 - 36$
$16 \times 24 = (20 - 4)(20 + 4) = 400 - 16$
∴ 16 × 24 is larger (by 20).
(ii) 25 × 75 or 26 × 74
Both pairs add up to 100, so both are centred on 50:
$25 \times 75 = (50 - 25)(50 + 25) = 2500 - 625$
$26 \times 74 = (50 - 24)(50 + 24) = 2500 - 576$
∴ 26 × 74 is larger (by 49).
In general, of two pairs with the same sum, the pair whose numbers are closer together has the larger product.
10. A tiny park is coming up in Dhauli. The plan is shown in the figure. The two square plots, each of area g² sq. ft., will have a green cover. All the remaining area is a walking path w ft. wide that needs to be tiled. Write an expression for the area that needs to be tiled. #
Solution
Each plot is a square of side $g$ ft. In the plan, the path is $w$ ft wide along the outer edges and $2w$ ft wide between the two plots.
Size of the park:
Length $= w + g + 2w + g + w$
Length $= 2g + 4w$
Breadth $= w + g + w$
Breadth $= g + 2w$
Area of the park:
$$(2g + 4w)(g + 2w) = 2g^2 + 4gw + 4gw + 8w^2$$$$= 2g^2 + 8gw + 8w^2$$
Area to be tiled = area of the park − area of the two plots:
$$2g^2 + 8gw + 8w^2 - 2g^2 = 8gw + 8w^2$$$$= 8w(g + w)$$
∴ The area to be tiled is $8w(g + w)$ sq. ft., that is $8gw + 8w^2$ sq. ft.
11. For each pattern shown below, (i) Draw the next figure in the sequence. (ii) How many basic units are there in Step 10? (iii) Write an expression to describe the number of basic units in Step y. #
Pattern (a) — the yellow staircase
How it grows: Step $y$ is a block of $y$ rows of $y + 2$ squares, with an arm of $y + 2$ squares going up from the right end and another going down from the left end.
| Step | Block | Arms | Total |
|---|---|---|---|
| 1 | $1 \times 3 = 3$ | $3 + 3$ | 9 |
| 2 | $2 \times 4 = 8$ | $4 + 4$ | 16 |
| 3 | $3 \times 5 = 15$ | $5 + 5$ | 25 |
(i) Step 4: a $4 \times 6$ block with two arms of 6:
(iii) Step $y$:
$$y(y + 2) + 2(y + 2) = (y + 2)(y + 2)$$$$= (y + 2)^2$$
(ii) Step 10: $(10 + 2)^2 = 12^2 = $ 144 units.
Note: in the textbook picture, the arms of Step 3 are drawn with 6 squares each. Following Steps 1 and 2, each arm should be as long as the block is wide (5), which gives 25. NCERT’s answer also uses $(y + 2)^2$.
Pattern (b) — the blue squares
How it grows: Step $y$ is a square of $(y + 1) \times (y + 1)$ units with one extra row of $y$ units below it.
| Step | Square | Extra row | Total |
|---|---|---|---|
| 1 | $2^2 = 4$ | 1 | 5 |
| 2 | $3^2 = 9$ | 2 | 11 |
| 3 | $4^2 = 16$ | 3 | 19 |
(i) Step 4: 5 rows of 5 and a row of 4:
(iii) Step $y$: $(y + 1)^2 + y$, that is $y^2 + 3y + 1$.
(ii) Step 10: $11^2 + 10 = 121 + 10 = $ 131 units.