Find the area of a triangle with sides 3 cm, 4 cm and 5 cm

[Math Tutor]

Find the area of a triangle with sides 3 cm, 4 cm and 5 cm.

[dduclam]

Find the area of a triangle with sides 3 cm, 4 cm and 5 cm.

This triangle is a right-angled triangle,because [tex]3^2+4^2=5^2[/tex]

Area of the triangle [tex]= \frac1{2}.3.4=6[/tex]

[donaldhall]

is there a formula using only the value of the sides? i mean like of its not a right triangle? how do we solve for this?

[stefantudose]

is there a formula using only the value of the sides? i mean like of its not a right triangle? how do we solve for this?

Hi!
YES, there is. It's called the "Law of cosines" and can be aplied in each and every triangle.
Law of cosines: In the ABC triangle,
[tex]BC^{2}=AB^{2}+AC^{2}-2XABXAC[/tex]
Let's consider that our ABC triangle has the measures in the following way:
AB=3; BC=5; AC=4
Applying the Law of cosines:
[tex]5^{2}=3^{2}+4^{2}-2X3X4XcosA
cos A=0 => A=90^\circ => sinA=1[/tex]
An area formula is ABXACXsinA / 2.
So, the ABC's area is 3X4X1/2=6

Hope I helped you.

[leesajohnson]

Area of a triangle = A=hb.b/2

=> 3.4/2 = 6

[Montramugdha]

Wrongly solving ..... It's a simple problem ....

S=(a+b+c)/2
Consider a=3 b=3 and c=5
So S= (3+3+5)/2=11/2 or 5.5

We know A=√s(s-a)(s-b)(s-c)
=√ 5.5(5.5-3)(5.5-3)(5.5-5)
=√5.5 x2.5x 2.5x .5
=√17.1875
=4.146 scm

[Guest]

The easy way is to use this online calculator: https://ezcalc.me/triangle-calculator . You can calculate any three missing parameters given any three known parameters (say three sides) as well as the area and perimeter. There you can find all relevant formulas.

[HallsofIvy]

Montramugdha is using "Heron's formula" (https://en.wikipedia.org/wiki/Heron%27s_formula):
Given sides of length a, b, and c, let s be the "semi-perimeter", [tex]s= \frac{a+ b+ c}{2}[/tex].

Then the area is given by [tex]A= \sqrt{s(s-a)(s-b)(s-c)}[/tex].

In the example initially given, a= 3, b= 4, c= 5, [tex]s= \frac{3+ 4+ 5}{2}= 6[/tex] so the area is [tex]A= \sqrt{6(6-3)(6-4)(6-5)}= \sqrt{6(3)(2)(1)}= \sqrt{36}= 6[/tex] which is, of course, the same as (3)(4)/2 since this is a right triangle.

If we were given an equilateral triangle with side lengths all 4, we could find the area by dividing it into two right triangles with hypotenuse of length 4, one leg of length 2, so the other leg of length [tex]\sqrt{4^2- 2^2}= \sqrt{12}= 2\sqrt{3}[/tex]. Each right triangle has area [tex]\frac{2(2\sqrt{3}}{2}= 2\sqrt{3}[/tex] and the original equilateral triangle has area [tex]4\sqrt{3}[/tex].

By Heron's formula, [tex]s= \frac{4+ 4+ 4}{2}= 6[/tex] so the area is [tex]A= \sqrt{6(6-4)(6-4)(6-4)}= \sqrt{6(2)(2)(2)}= \sqrt{48}= \sqrt{3(16)}= 4\sqrt{3}[/tex].

[Guest]

As CK=4, then BK=4
We now have all the sides of triangle ABK which should enable us to find it's angles (by say the cosine rule to start with). That in turn will allow us to find angle AKC which in turn will help us finding AC (by the cosine rule)

[Guest]

The Answer to your question -->
And by the way if you want to read more about

Screenshot (2307).png (30.02 KiB) Viewed 828 times

\