A logo was designed with four orange petals made from semicircles inscribed in a circle of diameter $14\sqrt{2}$ units. The orange part is the total area of the four petals. Find its area. (Take $\pi = \tfrac{22}{7}$). 
Step 1: Understand the figure.
- A circle of diameter $14\sqrt{2}$ is drawn.
- Radius $R = \dfrac{14\sqrt{2}}{2} = 7\sqrt{2}$.
- Inside, four semicircles are drawn with diameters along the sides of a square inscribed in the circle.
- The petals are the overlaps of these semicircles at the center.
Step 2: Side of the square.
If the circle radius = $R$, the side of an inscribed square = $R\sqrt{2} = 7\sqrt{2}\times \sqrt{2} = 14$. So, each semicircle has diameter $14$, radius $7$.
Step 3: Area of one petal (lens formed by overlap of two semicircles).
Each petal is the intersection of two semicircles of radius 7.
Formula for area of lens from two equal circles: \[ A_{\text{lens}} = 2r^2\cos^{-1}\left(\frac{d}{2r}\right) - \frac{d}{2}\sqrt{4r^2 - d^2} \] where $r = 7$, $d = 7$.
Step 4: Substitute values.
\[ A_{\text{lens}} = 2(49)\cos^{-1}\left(\frac{7}{14}\right) - \frac{7}{2}\sqrt{196 - 49} \] \[ = 98\cos^{-1}\left(\tfrac{1}{2}\right) - \tfrac{7}{2}\sqrt{147} \] \[ = 98. \frac{\pi}{3} - \tfrac{7}{2}. 7\sqrt{3} \] \[ = \frac{98\pi}{3} - \frac{49\sqrt{3}}{2} \] Step 5: Total orange area.
There are 4 petals: \[ A_{\text{orange}} = 4\left(\frac{98\pi}{3} - \frac{49\sqrt{3}}{2}\right) \] \[ = \frac{392\pi}{3} - 98\sqrt{3} \] Substitute $\pi = \tfrac{22}{7}$: \[ A_{\text{orange}} = \frac{392}{3}. \frac{22}{7} - 98. 1.732 \] \[ = \frac{392\times 22}{21} - 169.736 \approx 411.81 \] Final Answer: \[ \boxed{412 \, \text{units}^2 \, \text{(approx)}} \]
In the adjoining figure, $\triangle CAB$ is a right triangle, right angled at A and $AD \perp BC$. Prove that $\triangle ADB \sim \triangle CDA$. Further, if $BC = 10$ cm and $CD = 2$ cm, find the length of AD. 
In the diagram, the lines QR and ST are parallel to each other. The shortest distance between these two lines is half the shortest distance between the point P and the line QR. What is the ratio of the area of the triangle PST to the area of the trapezium SQRT?
Note: The figure shown is representative

\( AB \) is a diameter of the circle. Compare:
Quantity A: The length of \( AB \)
Quantity B: The average (arithmetic mean) of the lengths of \( AC \) and \( AD \). 
O is the center of the circle above. 



