Finding the Area of a Parallelogram with Given Sides and Angle

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Homework Help Overview

The discussion revolves around finding the area of a parallelogram given two sides measuring 24 feet and 30 feet, with an angle of 57 degrees between them. Participants are exploring the appropriate formulas and reasoning for calculating the area without arriving at a definitive solution.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants are attempting to apply the area formula for a parallelogram and are questioning the validity of their assumptions regarding angles. There is confusion about the relationship between the angles and the sides, with some participants suggesting the use of sine and cosine functions.

Discussion Status

There is an ongoing exploration of different interpretations of the problem, with some participants offering guidance on how to approach the calculation of height and area. Multiple perspectives on the angle measures and their implications for the area calculation are being discussed.

Contextual Notes

Some participants express uncertainty about the angle measures and the properties of parallelograms, while others reference specific educational contexts, such as a regents exam question. There is mention of an expected answer, but the path to that answer remains unclear among the participants.

priscilla98
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Homework Statement



A parallelogram's two sides are 24 feet amd 30 feet. The measure of the angle between these sides is 57 degrees. Find the area of the parallelogram, to the nearest square foot.

Homework Equations



area = 1/2 (a) (b) Sin C

The Attempt at a Solution



I'm thinking if there's a 90 degree angle and then the other angle would be 33 degrees. But I'm just assuming this but when I tried it, this did not work. I'm a little troubled by this problem, thanks a lot for your responses.
 
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priscilla98 said:

Homework Statement



A parallelogram's two sides are 24 feet amd 30 feet. The measure of the angle between these sides is 57 degrees. Find the area of the parallelogram, to the nearest square foot.

Homework Equations



area = 1/2 (a) (b) Sin C
This formula isn't for a parallelogram.

The Attempt at a Solution



I'm thinking if there's a 90 degree angle and then the other angle would be 33 degrees. But I'm just assuming this but when I tried it, this did not work. I'm a little troubled by this problem, thanks a lot for your responses.
How did you come up with 33 degrees? If the angle was 90 degrees, wouldn't a parallelogram be a rectangle?
 
I came up with 33 degrees by adding 90 and 57 and i got 147 and subtracted 180 by 146and got 33 degrees. But when looking at the properties of a parallelogram, there's no right angles. There are congruent sides and angles but no right angle.
 
Okay, then the area of the parallelogram is a = b * h. But then which one would be the base? 24 feet or 30 feet?
 
Choose one to be the base, then calculate the height using the other one and the angle. It might help you to sketch the parallelogram and draw in the height so you can see how to find its length from the length of the non-base side and the angle.

You'll find it doesn't make a difference in calculating the area which side you decide is the base.
 
How do you calculate the height when you have the base given?
 
Edit: Use correct angle measure.
If you have a right triangle whose hypotenuse was 24' and the angle between the base and hypotenuse was [STRIKE]33 deg.[/STRIKE] 57 deg., can you find the altitude of this triangle? That would give you the height of your parallelogram.
 
Last edited:
priscilla98 said:

Homework Statement



A parallelogram's two sides are 24 feet amd 30 feet. The measure of the angle between these sides is 57 degrees. Find the area of the parallelogram, to the nearest square foot.

Homework Equations



area = 1/2 (a) (b) Sin C
Almost. The area of a parallelogram with side lengths a and b and angle C between them is just
ab sin(C)- no "1/2".

The Attempt at a Solution



I'm thinking if there's a 90 degree angle and then the other angle would be 33 degrees. But I'm just assuming this but when I tried it, this did not work. I'm a little troubled by this problem, thanks a lot for your responses.
You are told that the angle between the sides is 57 degrees: the area is (24)(30) cos(57). You only need one angle, why look for another?

Except in the special case of a rectangle, there is no 90 degree angle in a parallelogram. If one angle is 57 degrees, the other angle is 180- 57= 123 degrees. But cos(123)= cos(180- 57)= -sin(57) and the area formula is really an absolute value so it doesn't matter which you use.
 
Last edited by a moderator:
Okay, this is one of the questions from algebra and trigonometry regent in june. It's number 34. I'm pretty sure it doesn't have a 90 degree angle. They all ready give you the answer which is 604 but i don't understand on how to get to that answer.
 
  • #10
I have to disagree with part of HallsofIvy's answer. The area of the parallelogram is 30*24*sin(57deg), not 30*24*cos(57deg).

Assuming the long sides of the parallelogram are horizontal, and two of the angles are 57deg, the altitude is 24 sin(57deg).
 
  • #11
Thanks a lot for this everybody and especially @ Mark44
 

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