1. Find the area of a triangle whose sides are in the ratio \( \color{black} 5:12:13 \) and its perimeter is \( \color{black} 60 \, cm \).
Solution
\[ \begin{align*} \text{First side} &= 5x \\ \text{Second side} &= 12x \\ \text{Third side} &= 13x \\ \\ \text{Perimeter of triangle } & = 60 \, cm \\ 5x +12x +13x & = 60 \, cm \\ 30x & = 60 \, cm \\ x & = \frac{60}{30} \, cm \\ x & = 2 \, cm \\ \\ \text{First side} &= 5x \implies 10 \, cm\\ \text{Second side} &= 12x \implies 24 \, cm\\ \text{Third side} &= 13x \implies 26 \, cm\\ \end{align*} \]
\[ \begin{align*} \text{Consider } \triangle ABC \\ (Hypotenuse)^2 &= (AC)^2 \\ &= (26)^2 \\ &= 676 \\ \\ (Base)^2 + (Height)^2 &= (BC)^2 + (AB)^2 \\ &= (24)^2 + (10)^2 \\ &= 576 + 100 \\ &= 676 \\ (Hypotenuse)^2 & = (Base)^2 + (Height)^2 \\ \implies \triangle ABC & \text{ is a right angled triangle} \\ \\ \text{Area of triangle} &= \frac{1}{2} \times base \times height \\ \\ & = \frac{1}{\cancelto{1}{2}} \times \cancelto{12}{24} \times 10 \\ \\ & = 120 \, cm^2 \\ \end{align*} \]
Answer Area of the triangle \( = \color{red} 120 \, cm^2 \)
2. AD is the diameter of a circle of radius \( \color{black} 6 \, cm \) and \( \color{black} AB=BC=CD \). Semi-circles are drawn with AB and BD as diameters as shown in the figure. Find the perimeter and area of the shaded region.
Solution
\[ \begin{align*} OD \text{ (Radius)} &= 6 \, cm \\ AD\text{ (Diameter)} &= 12 \, cm \\ \\ AB+ BC + CD &= 12 \, cm \\ \implies AB = BC = CD &= 4 \, cm \\ \\ Radius (r_1) &= 6 \,cm \\ Radius (r_2) &= 2 \,cm \\ Radius (r_3) &= 4 \,cm \\ \\ \text{Perimeter} &= 3 \text{ semi-circles} \\ &= \pi r_1 + \pi r_2 + \pi r_3 \\ &= \pi (r_1 + r_2 + r_3) \\ &= 3.14 \times (6 + 2 + 4) \\ &= 3.14 \times 12 \\ Perimeter &= 37.68 \, cm \\ \\ \text{Area of the shaded region} &= \frac{1}{2} \pi (r_1)^2 + \frac{1}{2} \pi (r_2)^2 - \frac{1}{2} \pi (r_3)^2 \\ \\ &= \frac{1}{2} \pi [ (r_1)^2 + (r_2)^2 - (r_3)^2] \\ \\ &= \frac{1}{2} \times 3.14 \times [ (6)^2 + (2)^2 - (4)^2] \\ \\ &= \frac{1}{2} \times 3.14 \times [ 36 + 4 - 16] \\ \\ &= \frac{1}{\cancelto{1}{2}} \times 3.14 \times \cancelto{12}{24} \\ \\ &= 3.14 \times 12 \\ \\ \text{Area of the shaded region} &= 37.68 \, cm^2 \\ \\ \end{align*} \]
Answer Perimeter \( = \color{red} 37.68 \, cm \) , Area \( = \color{red} 37.68 \, cm^2 \)
3. Four equal circles are described about four corners of a square so that each touches the two of the others. Find the area of the part of the square enclosed by the circles if the side of the square is \( \color{black} 14 \, cm \).
Solution
\[ \begin{align*} \text{Diameter of each circle} &= \frac{14}{2} \, cm \\ \\ Diameter &= 7 \, cm \\ Radius &= 3.5 \, cm \\ \text{Area of 4 circles} &= 4 \times \pi r^2 \\ &= 4 \times \frac{22}{7} \times (3.5)^2 \\ \\ &= 4 \times \frac{22}{7} \times 3.5 \times 3.5 \\ \\ &= 4 \times \frac{22}{\cancelto{1}{7}} \times \cancelto{.5}{3.5} \times 3.5 \\ \\ &= 4 \times 22 \times .5 \times 3.5 \\ &= 14 \times 11 \\ &= 154 \, cm^2 \\ \end{align*} \]
Answer Area enclosed by 4 circles \( = \color{red} 154 \, cm^2 \)