SOLUTION: Farmer Jessie has a field shaped as quadrilateral ABCD. She measures three of the sides AB = 50 meters, BC = 65 meters, and CD = 80 meters. She also determines that angle ABC = 130

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Question 1209009: Farmer Jessie has a field shaped as quadrilateral ABCD. She measures three of the sides AB = 50 meters, BC = 65 meters, and CD = 80 meters. She also determines that angle ABC = 130 degrees and BCD = 52 degrees.
(a) What is the distance between points A and C?
(b) Show one way to find the area of triangle ABC.
(c) Show a different way to find the area of triangle ABC.
(d) What is the area of quadrilateral ABCD?
Round each result to 3 decimal places.
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(53979) (Show Source):
You can put this solution on YOUR website!
.
Farmer Jessie has a field shaped as quadrilateral ABCD. She measures three of the sides
AB = 50 meters, BC = 65 meters, and CD = 80 meters. She also determines
that angle ABC = 130 degrees and BCD = 52 degrees.
(a) What is the distance between points A and C?
(b) Show one way to find the area of triangle ABC.
(c) Show a different way to find the area of triangle ABC.
(d) What is the area of quadrilateral ABCD?
Round each result to 3 decimal places.
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(a) Use the cosine law formula for triangle ABC AC = = = 104.418 meters (rounded). ANSWER (b) Use the formula for the area of a triangle = = = = 1244.822 m^2 (rounded). ANSWER

(a) and (b) are solved.

Answer by math_tutor2020(3842) (Show Source):
You can put this solution on YOUR website!

Part (a)
Here is a diagram if your teacher/textbook has not provided it.

Let's draw in diagonal AC so we split the quadrilateral into triangles ABC and ACD.

x = length of segment AC

Focus on triangle ABC only.
Either cover up the other triangle, or draw triangle ABC off to the side.
We use the Law of Cosines to find x.
b^2 = a^2 + c^2 - 2*a*c*cos(B)
x^2 = 65^2 + 50^2 - 2*65*50*cos(130)
x^2 = 10903.11946296
x = sqrt(10903.11946296)
x = 104.41800354
This value is approximate.
Unless otherwise stated, each decimal value mentioned from here on out will also be approximate.
The distance from A to C is roughly 104.418 meters when rounding to 3 decimal places.
Please make sure that your calculator is set to degrees mode.

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Part (b)

Triangle ABC has these side lengths
AB = 50 (given)
BC = 65 (given)
AC = 104.41800354 (approximate; see part (a) )

We can use Heron's triangle area formula to get the following
s = semi perimeter = (a+b+c)/2 = (65+104.41800354+50)/2 = 109.70900177
area = sqrt(s*(s-a)*(s-b)*(s-c))
area = sqrt(109.70900177*(109.70900177-65)*(109.70900177-104.41800354)*(109.70900177-50))
area = 1244.822220001852
Triangle ABC has area of approximately 1244.822 square meters when rounding to 3 decimal places.

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Part (c)

The previous part relies on calculating the length of AC.

Luckily there's a much faster way to find the area of triangle ABC without needing AC at all.
I'll use the SAS triangle area formula. SAS stands for "side angle side".
area = 0.5*side1*side2*sin( included angle )
area = 0.5*AB*BC*sin(angle ABC )
area = 0.5*50*65*sin(130)
area = 1244.822220068339
When rounding to 3 decimal places, we get 1244.822 square meters once again.

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Part (d)

Let y = measure of angle ACB
The adjacent angle ACD is 52-y. Angles ACB and ACD add to 52 degrees (measure of angle BCD).

Since we found x = AC = 104.41800354 approximately, we can update the diagram to this

Focus on triangle ABC.
Use the Law of Cosines to determine angle y.
c^2 = a^2+b^2 - 2*a*b*cos(C)
50^2 = 65^2+104.41800354^2 - 2*65*104.41800354*cos(y)
I'll skip a few steps and leave the arithmetic for the student to do.
You should arrive at y = 21.51940163 degrees approximately.

Alternatively, you can use the Law of Sines
sin(B)/b = sin(C)/c
sin(130)/104.41800354 = sin(y)/50
I'll let the student handle the scratch work to solve for angle y.
You should arrive at the previously mentioned y value. Or very close to it.
This is the approximate measure of angle ACB.

Adjacent to this is angle ACD, which has measure of 52-y = 52-21.51940163 = 30.48059837 degrees approximately.
Now we can use the SAS area formula to find the area of triangle ACD.
area = 0.5*side1*side2*sin( included angle )
area = 0.5*AC*CD*sin(angle ACD )
area = 0.5*104.41800354*80*sin(30.48059837)
area = 2118.62695314
Like with all of the other decimal values mentioned, this value is approximate.

Then we add the two triangle areas to get the area of the quadrilateral.
Refer to either part (b) or part (c) to get the area of triangle ABC.
area(ABCD) = area(ABC) + area(ACD)
area(ABCD) = 1244.82222000 + 2118.62695314
area(ABCD) = 3363.44917314
The area of quadrilateral ABCD is approximately 3363.449 square meters.

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Answers Summary
(a) 104.418 meters
(b) 1244.822 square meters
(c) 1244.822 square meters
(d) 3363.449 square meters
Each decimal value is approximate.