Timeline for answer to Potentially new formula for the area of a general quadrilateral using all its sides and any one angle by Oscar Lanzi
Current License: CC BY-SA 4.0
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| when toggle format | what | by | license | comment | |
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| Jan 14 at 12:07 | vote | accept | PARTH PATEL | ||
| Jan 9 at 10:36 | comment | added | PARTH PATEL | Hi, after implementing your suggestions I have came to this formula: $\frac{1}{2}\left[ ab\sin\theta + cd\sqrt{1 - \left(\frac{c^{2}+d^{2}-a^{2}-b^{2}+2ab\cos\theta}{2cd}\right)^2 } \right]$ Here, a=AB, b=BC, c=CD, d=AD and the given angle is $ \theta $ | |
| Jan 9 at 3:06 | comment | added | PARTH PATEL | Didn't notice that at first. Thanks for your feedback! | |
| Jan 9 at 2:45 | comment | added | Oscar Lanzi | No real difference in my reckoning. When I render the cosine of angle D, it contains exactly the same terms you would use to get AC. Look closely! | |
| Jan 9 at 2:42 | history | edited | Oscar Lanzi | CC BY-SA 4.0 |
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| Jan 9 at 2:36 | comment | added | PARTH PATEL | Your method of finding area of $ \triangle ABC $ seems better. But I have a question regarding the method for $ \triangle ACD $ . Wouldn't it be better if we find AC using law of cosine in $ \triangle ABC $ and then we use the known values of AC, AD and CD to find the area of $ \triangle ACD $ using heron's formula? I feel like it would reduce the length of the formula | |
| Jan 8 at 17:43 | history | edited | hbghlyj | CC BY-SA 4.0 |
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| Jan 8 at 17:15 | history | answered | Oscar Lanzi | CC BY-SA 4.0 |