1. Find the area of the unshaded portion of the given figure. The shaded portion are semi-circles.
Solution
\[ \begin{align*} Length &= 14 \, m \\ Breadth &= 10 \, m \\ Diameter &= 7 \, m \\ Radius &= \frac{7}{2} \, m \\ \\ \text{Area of the unshaded portion} & = \text{(Area of rectangle)} - \text{(Area of 2 semi-circle)} \\ & = (Length \times Breadth) - \text{(Area of 1 circle)} \\ & = (14 \times 10) - (\pi r^2) \\ \\ & = (140) - \left( \frac{\cancelto{11}{22}}{\cancelto{1}{7}} \times \frac{\cancelto{1}{7}}{\cancelto{1}{2}} \times \frac{7}{2} \right) \\ \\ & = (140) - \left(\frac{77}{2} \right) \\ \\ & = 140 - 38.5 \\ & = 101.5 \, m^2 \\\end{align*} \]
Answer Area of the unshaded portion \( = \color{red} 101.5 \, m^2 \)
2. Four cows are tethered at four corners of a square field of 70 m side as shown in given figure so that they can reach each other. What area of the field will remain ungrazed?
Solution
\[ \begin{align*} Side &= 70 \, m \\ Diameter &= 70 \, m \\ Radius &= 35 \, m \\ \\ \text{Area that remain ungrazed} & = \text{(Area of square)} - \text{(Area of 4 quarters of circle)} \\ & = (Side \times Side) - \text{(Area of 1 circle)} \\ & = (70 \times 70) - (\pi r^2) \\ \\ & = (4900) - \left( \frac{22}{\cancelto{1}{7}} \times \cancelto{5}{35} \times 35 \right) \\ \\ & = (4900) - (22 \times 5 \times 35) \\ & = 4900 - (110 \times 35) \\ & = 4900 - 3850 \\ & = 1050 \, m^2 \\\end{align*} \]
Answer Area that remain ungrazed \( = \color{red} 1050 \, m^2 \)