Step 1: Understanding the Concept:
In a right circular cone, the radius (\(r\)), the perpendicular height (\(h\)), and the slant height (\(l\)) form a right-angled triangle. The slant height (\(l\)) is the hypotenuse of this triangle.
Step 2: Key Formula or Approach:
We can use the Pythagorean theorem to relate the radius, height, and slant height of the cone:
\[ l^2 = r^2 + h^2 \]
Taking the square root gives the formula for the slant height:
\[ l = \sqrt{r^2 + h^2} \]
Step 3: Detailed Explanation:
We are given the following values:
Radius, \(r = 5\) cm
Perpendicular height, \(h = 12\) cm
Substitute these values into the formula for slant height:
\[ l = \sqrt{5^2 + 12^2} \]
\[ l = \sqrt{25 + 144} \]
\[ l = \sqrt{169} \]
\[ l = 13 \]
So, the slant height is 13 cm.
Step 4: Final Answer:
The slant height of the cone is 13 cm.
In the following figure \(\triangle\) ABC, B-D-C and BD = 7, BC = 20, then find \(\frac{A(\triangle ABD)}{A(\triangle ABC)}\). 
In the following figure \(\angle\)MNP = 90\(^\circ\), seg NQ \(\perp\) seg MP, MQ = 9, QP = 4, find NQ. 
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.