Questions & Answers: "Perimeter and Area"

The four circles are arranged such that their centers form a square. The side length of this square is 2 times the radius of a circle. Given radius (r) = 3.5 cm. Side of the square formed by centers = 2 * 3.5 = 7 cm. Area of this square = side^2 = 7^2 = 49 sq.cm. The shaded portion is the area of this square minus the area of four quadrants (one from each circle) that lie within this square. Area of four quadrants = 4 * (1/4) * π * r^2 = π * r^2. Area of four quadrants = (22/7) * (3.5)^2 = (22/7) * (7/2)^2 = (22/7) * (49/4) = (22 * 7) / 4 = 11 * 7 / 2 = 77 / 2 = 38.5 sq.cm. Area of shaded portion = Area of square - Area of four quadrants = 49 - 38.5 = 10.5 sq.cm.

The sides of the triangular plot are $50$m, $30$m, and $40$m. Let $a=30$, $b=40$, $c=50$. We check if it's a right-angled triangle: $a^2 + b^2 = 30^2 + 40^2 = 900 + 1600 = 2500$. Also, $c^2 = 50^2 = 2500$. Since $a^2 + b^2 = c^2$, it is a right-angled triangle. The area of a right-angled triangle is $\frac{1}{2} \times \text{base} \times \text{height}$. So, Area $= \frac{1}{2} \times 30 \times 40 = \frac{1}{2} \times 1200 = 600$ sq.m. To convert to 'Are', we use the conversion $1$ Are $= 100$ sq.m. So, $600$ sq.m. $= \frac{600}{100}$ Are $= 6$ Are.

The area of a circle $A$ is in direct variation with the square of its radius $r$, which means $A = k r^2$, where $k$ is the constant of proportionality (in this case, $k=\pi$). Given: $A_1 = 616$ sq.cm. when $r_1 = 14$ cm. We can find the constant $k$ (which is $\pi$): $616 = k (14)^2$ $616 = k (196)$ $k = \frac{616}{196} = \frac{22}{7}$. So, $k=\pi$. Now, we need to find the area $A_2$ when $r_2 = 42$ cm. $A_2 = k r_2^2$ $A_2 = \frac{22}{7} (42)^2$ $A_2 = \frac{22}{7} (42 \times 42)$ $A_2 = 22 \times (\frac{42}{7}) \times 42$ $A_2 = 22 \times 6 \times 42$ $A_2 = 132 \times 42$ $A_2 = 5544$ sq.cm. Alternatively, since $A = k r^2$, we have $\frac{A_1}{r_1^2} = \frac{A_2}{r_2^2}$. We notice that $r_2 = 42 = 3 \times 14 = 3 r_1$. So, $A_2 = k r_2^2 = k (3r_1)^2 = k (9r_1^2) = 9 (k r_1^2) = 9 A_1$. $A_2 = 9 \times 616 = 5544$ sq.cm. Therefore, the area of the circle whose radius is $42$ cm is $5544$ sq.cm.

The garden is semi-circular with a radius $r = 420$ m. The perimeter of a semi-circular garden consists of the length of the semi-circular arc and the length of the diameter. Length of the semi-circular arc $= \frac{1}{2} (2\pi r) = \pi r$. Length of the diameter $= 2r$. Total perimeter for one round of fencing $= \pi r + 2r = r(\pi + 2)$. Using the approximation $\pi = \frac{22}{7}$: Perimeter $= 420 \left(\frac{22}{7} + 2\right)$ Perimeter $= 420 \left(\frac{22}{7} + \frac{14}{7}\right)$ Perimeter $= 420 \left(\frac{22+14}{7}\right)$ Perimeter $= 420 \left(\frac{36}{7}\right)$ Perimeter $= (60 \times 7) \times \frac{36}{7}$ Perimeter $= 60 \times 36 = 2160$ m. The garden is to be fenced with five rounds of wire. Total length of wire needed $= 5 \times \text{Perimeter for one round}$ Total length of wire needed $= 5 \times 2160 \text{ m} = 10800 \text{ m}$. The cost of the wire is Rs. $60$ per metre. Total cost of the wire $= \text{Total length of wire} \times \text{Cost per metre}$ Total cost $= 10800 \text{ m} \times \text{Rs. } 60/\text{m}$ Total cost $= \text{Rs. } 648000$. Therefore, the cost of the wire needed for fencing is Rs. $6,48,000$.