The base of an isosceles triangle is 8 cm and one of its equal sides is 5 cm. The height of the vertex opposite to the base from the base is:

Correct option is B

Given:
Base of isosceles triangle = 8 cm
One of the equal sides = 5 cm
Formula Used:
Area of triangle = $$\frac{1}2$$​ × base × Height
Area of triangle (Heron's Formula) = √ s(s - a)(s - b)(s - c)
s = (a + b + c) ÷ 2 Sides of the triangle = a, b, c
Solution:
s = (8 + 5 + 5) ÷ 2 = 9
$$\frac{1}2$$​ × 8 × Height = √ 9(9 - 8)(9 - 5)(9 - 5)
=> 4 × Height = 3 × 4
=> Height = 3 cm

Alternate Solution:
Given:
The base of the isosceles triangle AB = 8cm.
The two equal sides AC = BC = 5 cm.
triangle into two right triangles by drawing a height from
the vertex C to the midpoint D of the base AB. Since D is the midpoint, AD = DB = $$\frac{8}{2}$$ = 4 cm.
Right triangle △ADC
AC = 5 cm , AD = 4 cm
height from C to D be h.
Using the Pythagorean theorem in △ADC:
AC² = AD²+h²
5²= 4²+ h²
5² = 4² + h²
25 = 16 + h²
h^2​ = 25 - 16
h² = 9
h = √ 9
= 3 cm