Find the area (in square units) of each triangle described.

1. a = 1, b = 1, c = 1

2. a = 49, b = 33, c = 18

Guest Apr 17, 2019

#1**+4 **

Hi, thanks for posting. We will use heron's formula for this, because I assume if you posted on this forum you would have already had googled a triangle area calculator.

Heron's Formula is:

\(A=\sqrt {s(s-a)(s-b)(s-c)}\)

\(s\): The **Semiperimeter of a triangle **(Half of the perimeter)

\(a,b,c\) = The sides of the triangle.

First Question:

\(A=\sqrt {1.5(1.5-1)(1.5-1)(1.5-1)}\\A=\sqrt {1.5(.5^3)}\\A=\sqrt {0.1875}\\A \approx \boxed{0.433}\)

Second Question:

\(A=\sqrt {50(50-49)(50-33)(50-18)}\\A=\sqrt {50(544)}\\A=\sqrt {27200}\\A \approx \boxed{164.92}\)

Specify if you want a certain formula used, thanks.

Bxtterman Apr 17, 2019

#2**+2 **

I'm sorry, but I'm not quite getting something.

In the 1^{st} question you used s = half the perimeter

In the 2^{nd} question you used s = the entire perimeter (i.e., 49+33+18=100)

Could you clarify that for me, please?

?

Guest Apr 17, 2019

#7**+1 **

Hi Bxtterman,

It is absolutely not a problem.

It is actually good that little errors like this are made from time to time.

Askers should be checking our answers and they should have enough knowledge to locate errors such as this for themselves.

I often do not check my answers and this is a part of the reason.

It is really good that our guest found this error and reported his/her confusion.

Welcome to our web2.0calc forum by the way.

I have already noticed you making some great contributions.

Melody
Apr 18, 2019

#4**+3 **

1. As you can use Heron's formula for the first question, we have an equilateral triangle since a=b=c=1.

The formula to find the area of an equilateral triangle is \(\frac{\sqrt{3}}{4}s^2\)\(\), so we have \(\frac{\sqrt{3}}{4}1^2=\frac{\sqrt{3}}{4}\approx\boxed{0.433}.\)

2. Using Heron's Formula, we have: \(\sqrt{(50)(50-49)(50-33)(50-18)}=40\sqrt{17}\approx\boxed{164.92}.\)

tertre Apr 17, 2019