The side of an equilateral triangle is 10 cm. Find its altitude.

Correct option is D

Given:

side of an equilateral triangle = 10 cm 

Altitude of equilateral triangle = ?

Concept Used:

In an equilateral triangle, all three sides are equal in length, and the altitude divides the triangle into two right triangles. The altitude is perpendicular to the base and bisects the base into two equal parts.

Pythagorean theorem:   $$H^2=P ^2+B^2$$​​

Solution:

Let the equilateral triangle be ABC, with side length AB = BC = CA = 10 cm. Drop a perpendicular from vertex A to the base BC, which will intersect BC at the midpoint,  M. This perpendicular line is the altitude.

The altitude divides the equilateral triangle into two congruent right triangles. In each right triangle:

The hypotenuse is the side of the equilateral triangle, which is 10 cm.

One leg is half the length of the base, so $$BM = MC = \frac{10}{2} = 5 \, \text{cm}$$ 

The other leg is the altitude (h), which we need to find.

In the right triangle $△ABM$, the sides satisfy the Pythagorean theorem:

​$$AB^2 = AM^2 + BM^2$$​​

​Substituting the known values:

​$$10^2 = h^2 + 5^2$$ 

$$100 = h^2 + 25$$ 

$$h^2 = 100 - 25 = 75$$ 

​$$h = \sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}$$ 

​Thus, the altitude of the equilateral triangle is $$5\sqrt{3} \, \text{cm}$$.​