• Oct 8th 2008, 06:22 AM
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.
• Oct 8th 2008, 11:00 AM
earboth
Quote:

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:

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

$h_{leg} = \dfrac{2 \cdot 27\sqrt{7}}{12}= \dfrac92 \sqrt{7}$
• Oct 8th 2008, 03:38 PM
franckherve1
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