Statement I is true because the perimeter of any triangle is indeed greater than the sum of the lengths of its medians. This is a well-established property in geometry, and it holds for all types of triangles.
Statement II is also true. It is known as a triangle inequality for a point inside the triangle: if D is any point on side BC, then the sum of the distances from A to D, B to D, and C to D is always greater than the length of any side of the triangle. This inequality holds true for all triangles and is one of the fundamental results in geometry
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.