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:
\[ \text{Perimeter} = 32 + 24 + 32 + 12\pi = 88 + 12\pi \, \text{cm} \]
\[ \boxed{\text{Perimeter} = 88 + 12\pi \, \text{cm}} \]
On the day of her examination, Riya sharpened her pencil from both ends as shown below. 
The diameter of the cylindrical and conical part of the pencil is 4.2 mm. If the height of each conical part is 2.8 mm and the length of the entire pencil is 105.6 mm, find the total surface area of the pencil.
Two identical cones are joined as shown in the figure. If radius of base is 4 cm and slant height of the cone is 6 cm, then height of the solid is
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.