Question 2

We will use algebraic identities to find these products without direct multiplication.

Step 1 — Calculate $(41)^2$

Let's write 41 as a sum. We can write $41 = 40 + 1$. We use the identity $(a + b)^2 = a^2 + 2ab + b^2$.

$(41)^2 = (40 + 1)^2$

$= (40)^2 + 2 \times 40 \times 1 + (1)^2$

$= 1600 + 80 + 1$

$\boxed{1681}$

Step 2 — Calculate $(27)^2$

Let's write 27 as a difference. We can write $27 = 30 - 3$. We use the identity $(a - b)^2 = a^2 - 2ab + b^2$.

$(27)^2 = (30 - 3)^2$

$= (30)^2 - 2 \times 30 \times 3 + (3)^2$

$= 900 - 180 + 9$

$\boxed{729}$

Step 3 — Calculate $(23 \times 17)$

Let's express the numbers as sums and differences. We can write $23 = 20 + 3$ and $17 = 20 - 3$. We use the identity $(a + b)(a - b) = a^2 - b^2$.

$23 \times 17 = (20 + 3)(20 - 3)$

$= (20)^2 - (3)^2$

$= 400 - 9$

$\boxed{391}$

Step 4 — Calculate $(135)^2$

Let's write 135 as a sum. We can write $135 = 100 + 35$. We use the identity $(a + b)^2 = a^2 + 2ab + b^2$.

$(135)^2 = (100 + 35)^2$

$= (100)^2 + 2 \times 100 \times 35 + (35)^2$

$= 10000 + 7000 + 1225$

$\boxed{18225}$

Step 5 — Calculate $(97)^2$

Let's write 97 as a difference. We can write $97 = 100 - 3$. We use the identity $(a - b)^2 = a^2 - 2ab + b^2$.

$(97)^2 = (100 - 3)^2$

$= (100)^2 - 2 \times 100 \times 3 + (3)^2$

$= 10000 - 600 + 9$

$\boxed{9409}$

Step 6 — Calculate $(18 \times 29)$

Let's express the numbers using a common term. We can write $18 = 20 - 2$ and $29 = 20 + 9$. We use the identity $(x + a)(x + b) = x^2 + (a + b)x + ab$.

$18 \times 29 = (20 - 2)(20 + 9)$

$= (20)^2 + (-2 + 9) \times 20 + (-2) \times 9$

$= 400 + (7) \times 20 - 18$

$= 400 + 140 - 18$

$\boxed{522}$

Step 7 — Calculate $(34 \times 43)$

Let's express the numbers using a common term. We can write $34 = 38 - 4$ and $43 = 38 + 5$. We use the identity $(x + a)(x + b) = x^2 + (a + b)x + ab$.

$34 \times 43 = (38 - 4)(38 + 5)$

$= (38)^2 + (-4 + 5) \times 38 + (-4) \times 5$

$= 1444 + (1) \times 38 - 20$

$= 1444 + 38 - 20$

$\boxed{1462}$

Step 8 — Calculate $(205)^2$

Let's write 205 as a sum. We can write $205 = 200 + 5$. We use the identity $(a + b)^2 = a^2 + 2ab + b^2$.

$(205)^2 = (200 + 5)^2$

$= (200)^2 + 2 \times 200 \times 5 + (5)^2$

$= 40000 + 2000 + 25$

$\boxed{42025}$

Answer

(i) 1681 (ii) 729 (iii) 391 (iv) 18225 (v) 9409 (vi) 522 (vii) 1462 (viii) 42025