The problem consists of two parts: (a) A window is in the shape of a semi-circle with radius 70 cm. We need to find the length of the diameter. (b) We need to calculate the perimeter of the window, using $\pi = \frac{22}{7}$. Then, another problem is provided: (a) Express $4 \times 8^3 + 3 \times 8^2 + 5 \times 8$ as a number in base 8. (b) Evaluate $431_5 - 244_5$, giving your answer in base 10.

1. Problem Description

The problem consists of two parts:

(a) A window is in the shape of a semi-circle with radius 70 cm. We need to find the length of the diameter.

(b) We need to calculate the perimeter of the window, using $\pi = \frac{22}{7}$.

Then, another problem is provided:

(a) Express $4 \times 8^3 + 3 \times 8^2 + 5 \times 8$ as a number in base

8. (b) Evaluate $431_5 - 244_5$, giving your answer in base

1

0.

2. Solution Steps

Problem 4:

(a) The diameter of a circle is twice its radius. Therefore, the diameter is $2 \times 70 = 140$ cm.

(b) The perimeter of the semi-circle is half the circumference of the full circle plus the diameter. The circumference of a full circle is $2 \pi r$. Thus, half the circumference is $\pi r$.

The perimeter of the semi-circle is $\pi r + 2r = r(\pi + 2)$.

Substituting $r = 70$ and $\pi = \frac{22}{7}$, we get:

Perimeter $= 70 (\frac{22}{7} + 2) = 70 (\frac{22}{7} + \frac{14}{7}) = 70 (\frac{36}{7}) = 10 \times 36 = 360$ cm.

Problem 5:

(a) $4 \times 8^3 + 3 \times 8^2 + 5 \times 8 = 4 \times 8^3 + 3 \times 8^2 + 5 \times 8^1 + 0 \times 8^0$

This is already in base

8. So the number is $4350_8$.

(b) Convert both numbers to base

1

0. $431_5 = 4 \times 5^2 + 3 \times 5^1 + 1 \times 5^0 = 4 \times 25 + 3 \times 5 + 1 = 100 + 15 + 1 = 116$

$244_5 = 2 \times 5^2 + 4 \times 5^1 + 4 \times 5^0 = 2 \times 25 + 4 \times 5 + 4 = 50 + 20 + 4 = 74$

Therefore, $431_5 - 244_5 = 116 - 74 = 42$.

3. Final Answer

Problem 4:

(a) 140 cm

(b) 360 cm

Problem 5:

(a) $4350_8$

(b) 42