The perimeter of an isosceles triangle is 42 cm and its base is 3/2 times each of the equal side. Find the length of each of the triangle, area of the triangle and the height of the triangle.
Let 'x' be the length of two equal sides,
Therefore the base =1/2 × x
Let the sides a, b, c of a triangle be 1/2 × x, x and x respectively
So, the perimeter = 2s = a + b + c
42 = a + b + c
42 = 3/2 × x + x + x
Therefore, x = 12 cm
So, the respective sides are
a = 12 cm
b = 12 cm
c = 18 cm
Now, semi perimeter
$s=\frac{a+b+c}{2}$
$=\frac{12+12+18}{2}$
= 21 cm
By using Heron's Formula
The area of a triangle $=\sqrt{s \times(s-a) \times(s-b) \times(s-c)}$
$=\sqrt{21 \times(21-12) \times(21-12) \times(21-18)}$
$=71.42 \mathrm{~cm}^{2}$
Thus, the area of a triangle is $71.42 \mathrm{~cm}^{2}$
The altitude will be smallest provided the side corresponding to this altitude is longest.
The longest side = 18 cm
Area of the triangle = 12 × h × 18
$12 \times h \times 18=71.42 \mathrm{~cm}^{2}$
h = 7.93 cm
Hence the length of the smallest altitude is 7.93 cm