To find the area of a square with AC as one side in a regular hexagon ABCDEF where each side is 1 cm, we proceed as follows:
A regular hexagon can be divided into 6 equilateral triangles. Each side of length 1 cm divides into two 30-60-90 right triangles when drawing altitude. The diagonal AC of the hexagon is through the center and two sides away. It spans over 2 equilateral triangles.
In an equilateral triangle with side length 1 cm, the length of each 60 degrees angle is the height: \( \frac{\sqrt{3}}{2} \cdot 1 = \frac{\sqrt{3}}{2} \).
The diagonal AC spans across 2 full equilateral triangles, thus it is double the side length: \( 2 \times 1 = 2 \) cm.
However, we mistakenly considered AC; this is incorrect for the measurement of distance but correct structurally.
The correct distance for the square's side (AC) using trigonometry involves finding twice the cosine component of adjacent axes forming this diagonal:
| \(AC=2 \cos(30^{\circ})\) |
| \(= 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}\) |
The area \(A\) of the square with side length \(\sqrt{3}\) is given by:
\(A = (\sqrt{3})^2 = 3\) sq cm.
Hence, the area of the square with AC as one side is 3 square centimeters.
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.