6-3 Area of a Triangle

Hey guys! So we officially only have wednesday, thursday, friday, another week after that and then four days after that till christmas break!!!! That's only 12 days- we can do it!...

Anyway lets learn about the Area of a Triangle...

Objective:
Given the measures of two sides and the included angle, find the area of the triangle.

Property: Area of a Triangle:

As we already know the area of a triangle = (1/2)(base)(height)

We can alter this formula to equal:

sin A= h/c

h= c sin A which can then be replaced in the original A= (1/2)(b)(h) to equal:

Area = (1/2)bc sin A

The area of a triangle equals half the product of two of its sides and the sine of the included angle.

Example Problem 1:

Find the area of triangle ABC given:

a=12 in.

b=10 in.

c= 62 degrees We can plug this into the formula Area = (1/2)bc sin A to get

A= (1/2)(10)(12) (sin62) = 52.9768 in. squared

Property: Hero's Formula:
(Hero)

In triangle ABC, the area is given by
Area=



where s is the semiperimeter (half the perimeter) found by taking 1/2 (a+b+c)

Example Problem 2:

Use Hero's Formula to calculate the area of this triangle.

a= 4

b=11

c= 8

The perimeter (4+11+8) = 23 so s= (1/2)(23)= 11.5

A=

For Extra Help go to:

http://regentsprep.org/regents/mathb/5E1/areatriglesson.htm or http://astro.temple.edu/~dhill001/trianglearea/trianglearea.html

Personalization: I had a ballet performance of the
Nutcracker on saturday and sunday and here's a picture from Snow which comes right before the Land of Sweets!

Reminder: Arash you are next!