Let the dimensions of the rectangular box be $a$, $b$, and $c$. The surface area $S$ and sum of the lengths of all edges $L$ are given by:
$S = 2(ab + bc + ca) = 846$,
$L = 4(a + b + c) = 144$.
From the second equation, we get:
$a + b + c = 36$.
Now, the box is inscribed in a sphere, so the diagonal of the box is the diameter of the sphere. The diagonal of the box is:
$\sqrt{a^2 + b^2 + c^2}$.
Let $D$ be the diameter of the sphere. Thus, the radius $r$ of the sphere is:
$r = \frac{D}{2} = \frac{\sqrt{a^2 + b^2 + c^2}}{2}$.
The volume $V$ of the sphere is:
$V = \frac{4}{3}\pi r^3$
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Using the given information, we can solve for $a$, $b$, and $c$, and then find the volume of the sphere.
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.