This problem requires finding the area and altitude of an equilateral triangle with a given side length.
The formula for the area of an equilateral triangle with side length '$a$' is:
$ \text{Area} = \frac{\sqrt{3}}{4} a^2 $
Given the side length '$a = 2$ cm':
$ \text{Area} = \frac{\sqrt{3}}{4} (2 \text{ cm})^2 $
$ \text{Area} = \frac{\sqrt{3}}{4} \times 4 \text{ cm}^2 $
$ \text{Area} = \sqrt{3} \text{ cm}^2 $
The formula for the altitude (height) '$h$' of an equilateral triangle with side length '$a$' is:
$ h = \frac{\sqrt{3}}{2} a $
Given the side length '$a = 2$ cm':
$ h = \frac{\sqrt{3}}{2} (2 \text{ cm}) $
$ h = \sqrt{3} \text{ cm} $
The calculated area is $\sqrt{3} \text{ cm}^2$ and the altitude is $\sqrt{3} \text{ cm}$.
ABCDEF is a regular hexagon. Side of the hexagon is 36 cm. What is the area of the triangle AOB ?
If ∆ABC ~ ∆DEF, and BC = 4 cm, EF = 5 cm and the area of triangle ABC = 80 cm 2, then the area of the triangle DEF is:
If Δ ABC is right angled at B, AB = 12 cm and ∠CAB = 60°, determine the length of BC.
If ΔABC and ΔDEF are congruent triangles, then which of the following is FALSE?
D and E are points on the sides AB and AC, respectively, of ΔABC such that DE is parallel to BC and AD ∶ DB = 7 ∶ 9. If CD and BE intersect each other at F. then find the ratio of areas of ΔDEF and ΔCBF.