The perimeter of an isosceles triangle is 42 cm and its base is 3/2 times each of the equal side.
Question:

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.

Solution:

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

 

Administrator

Leave a comment

Please enter comment.
Please enter your name.