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# Help

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4 What is the area of the triangle to the nearest tenth?

Apr 18, 2020

#1
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Use law of cosines to find the 3rd side length

then use Heron's Rule to calcualte the area of a triangle given three side lengths.

Or drop a vertical from the apex......use law of sines to find height  (2.44)      then area = 1/2  b x h   = 1/2   * 5.6  * 2.44    yd2

Apr 18, 2020
edited by Guest  Apr 18, 2020
#2
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I just learned this but I'll give it a shot:) I hope I can help!

You can solve for the other length of the triangle by using the pythagorean theorem, a^2 + b^2 = c^2. Substitute your values.

So, 2.6^2 + 5.6^2 = c^2

6.76 + 31.36 = c^2

38.12 = c^2

Sqrt 38.12 = c

c = 6.174139616

Now you must find the semi-perimeter by using (a+b+c)/2. Substitute your values.

(2.6+5.6+6.2)/2 = s

14.4 / 2 = s

s = 7.2

Now use your semi-perimeter to solve for the area using the formula, a = [sqrt (s (s - a) (s - b) (s - c) ) ]. Substitute your values where s is your semi-perimeter.

a = sqrt [ 7.2 (7.2 - 2.6) (7.2 - 5.6) (7.2 - 6.2) ]

a = sqrt [ 7.2 (4.6) (1.6) (1) ]

a = sqrt [ 52.992 ]

a = 7.3

If you have any questions, then let me know! I'm glad to help!

Apr 18, 2020
#3
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Pretty good try....but....Pythagorean Theorem only applies in RIGHT triangles..... Guest Apr 18, 2020
#4
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Guest is correct -- but there is an easier way ...

Using the Law of Cosines:

[You can't use the Pythagorean Theorem because it isn't a right triangle.]

Use the Law of Cosines to find the third side of the triangle:  c2  =  a2 + b2 - 2·a·b·cos(C)

--->   c2  =  (2.6)2 + (5.6)2 - 2·(2.6)·(5.6)·cos(70)

--->   c2  =  28.160

--->    c  =  5.3

Now, use Heron's formula:

s  =  (2.6 + 5.6 + 5.3) / 2  =  6.75

Area  =  sqrt( 6.75 · (6.75 - 2.6) · (6.75 - 5.6) · (6.75 - 5.3) )  =  6.83

But, there is an easier way:  use this formula for area:  Area  =  ½ · a · b · sin(C)

Area  =  ½ · (2.6) · (5.6) · sin(70)

Area  =  6.84

The difference between the two answers is due to a rounding error in my using Heron's formula.

Apr 18, 2020