We are given a semicircle with diameter \( AB \), and rectangle \( ABCD \). Let's go step-by-step to determine the dimensions and total perimeter.
The formula for the area of a semicircle with diameter \( AB \) is:
\[ \text{Area}_{\text{semi}} = \frac{1}{2} \times \pi \times \left( \frac{AB}{2} \right)^2 \]
This is given to be \( 72\pi \). Equating:
\[ \frac{1}{2} \times \pi \times \left( \frac{AB}{2} \right)^2 = 72\pi \]
Cancel \( \pi \) from both sides and simplify:
\[ \frac{1}{2} \times \left( \frac{AB^2}{4} \right) = 72 \quad \Rightarrow \quad \frac{AB^2}{8} = 72 \]
\[ AB^2 = 576 \quad \Rightarrow \quad AB = \sqrt{576} = 24 \, \text{cm} \]
The area of rectangle \( ABCD \) is given as 768 cm². Since \( AB = 24 \), and \( BC \) is the adjacent side:
\[ AB \times BC = 768 \quad \Rightarrow \quad 24 \times BC = 768 \]
\[ BC = \frac{768}{24} = 32 \, \text{cm} \]
The shape consists of three rectangle sides and one semicircular arc (along \( AB \)). The sides contributing to the perimeter are:
Therefore, the total perimeter is:
\[ \text{Perimeter} = 32 + 24 + 32 + 12\pi = 88 + 12\pi \, \text{cm} \]
\[ \boxed{\text{Perimeter} = 88 + 12\pi \, \text{cm}} \]
From one face of a solid cube of side 14 cm, the largest possible cone is carved out. Find the volume and surface area of the remaining solid.
Use $\pi = \dfrac{22}{7}, \sqrt{5} = 2.2$
The radius of a circle with centre 'P' is 10 cm. If chord AB of the circle subtends a right angle at P, find area of minor sector by using the following activity. (\(\pi = 3.14\)) 
Activity :
r = 10 cm, \(\theta\) = 90\(^\circ\), \(\pi\) = 3.14.
A(P-AXB) = \(\frac{\theta}{360} \times \boxed{\phantom{\pi r^2}}\) = \(\frac{\boxed{\phantom{90}}}{360} \times 3.14 \times 10^2\) = \(\frac{1}{4} \times \boxed{\phantom{314}}\) <br>
A(P-AXB) = \(\boxed{\phantom{78.5}}\) sq. cm.