Please solve this question using heron's formula..

The perimeter of an isosceles triangle is 42cm and its base is (3/2)times each of the equal sides.Find the length of each side of the triangle,area of the triangle and height of the triangle.

2. Originally Posted by rickymylv
Please solve this question using heron's formula..

The perimeter of an isosceles triangle is 42cm and its base is (3/2)times each of the equal sides.Find the length of each side of the triangle,area of the triangle and height of the triangle.
Let x denote the length of one leg. Then the perimeter of the triangle is:

$p = 42 = x+x+\dfrac32 x~\implies~ x = 12$

That means the triangle has the side length:

a = 12, b = 12, c = 18 (c is the base)

1. Calculate the area by Heron's formula:

The half perimeter is s = 21

$A=\sqrt{21(21-12)(21-12)(21-18)}=\sqrt{5103}=27\cdot \sqrt{7} \approx 71.435...\ cm^2$

2. The height of the trinagle. Actually there are (normally) three heights in every triangle. Here 2 of the three heights are equal.

From $A=\dfrac12 \cdot side \cdot height_{side}$ you get

$h = \dfrac{2A}{side}$

Plug in the values you know to calculate the heights:

$h_{base} = \dfrac{2 \cdot 27\sqrt{7}}{18}=3 \sqrt{7}$

$h_{leg} = \dfrac{2 \cdot 27\sqrt{7}}{12}= \dfrac92 \sqrt{7}$

3. i agree with the x=12,x=12,b=18...the h goes through the middle of b
so to get h..you have to take half of b and one side of the triangle and use pytagore

12(squared)=9(squared)+h(squared)
h(squared)=12(squared)-9(squared)
h(squared)=144-81
h(squared)=63
h=3(square root)7
to get the area b*h/2
18*3(squared root)7/2
=54(squared root)7/2