Complete the following table by observing the given figures: 
| \hline Points | Figure 1 (Myopia) | Figure 2 (Hypermetropia) |
| \hline (a) Name of the defect | Myopia (Nearsightedness) | Hypermetropia (Farsightedness) |
| \hline (b) Position of the image | In front of the retina | Behind the retina |
| \hline (c) Lens used to correct the defect | Concave lens | Convex lens |
| \hline |
The colour of the solution observed after about 1 hour of placing iron nails in copper sulphate solution is:
In the following figure \(\triangle\) ABC, B-D-C and BD = 7, BC = 20, then find \(\frac{A(\triangle ABD)}{A(\triangle ABC)}\). 
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.