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May 10th, 2018, 09:41 AM  #1  
Member Joined: Jan 2018 From: Belgrade Posts: 55 Thanks: 2  Calculate the area of a quadrilateral
While thinking of strategies ... Quote:
My reason was that because it does not matter where on arc $\displaystyle BC$ is $\displaystyle M$, the area must be the same wherever it is positioned. I'm concerned that this kind of reasoning isn't mathematically rigorous enough. Isn't it? On the other hand, the solution could be anticipated. When you move $\displaystyle M$ toward, for example, point $\displaystyle C$, our quadrilateral $\displaystyle APQD$ becomes triangle $\displaystyle ACD$, with area of $\displaystyle \frac{a^{2}}{2}$. So, is there a more general solution? Last edited by skipjack; May 11th, 2018 at 06:53 AM.  
May 11th, 2018, 04:03 AM  #2 
Member Joined: Jan 2018 From: Belgrade Posts: 55 Thanks: 2 
NOTE: I just realized that the word quadrilateral instead of quadruple should stay in the above topic title and in the post itself. Sorry for this inconvenience.
Last edited by lua; May 11th, 2018 at 04:06 AM. 
May 11th, 2018, 07:00 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,107 Thanks: 1907 
I've corrected the wording. As triangles ADQ, DPA are similar, QA/QD = AD/AP. Hence QA·AP = AD·QD, and so triangles QAP and CDQ have equal area. It follows that quadrilateral APQD's area = triangle ADC's area = a²/2. 
May 11th, 2018, 09:03 AM  #4 
Member Joined: Jan 2018 From: Belgrade Posts: 55 Thanks: 2 
I don't know how, but I don't see that from $\displaystyle QA·AP = AD·QD$ comes out that triangles $\displaystyle QAP$ and $\displaystyle CDQ$ have equal area. Can you explain that in more detail? It is easier for me to think this way. From $\displaystyle \frac{AD}{AQ}=\frac{DP}{AD}$ we have: $\displaystyle a^{2}=AD^{2}=AQ \cdot DP=2 \cdot P_{APQD}$ (the surface of a quadrilateral with normal diagonals is equal to half of their product). So, $\displaystyle P_{APQD}=\frac{a^{2}}{2}$. Last edited by lua; May 11th, 2018 at 09:52 AM. 
May 11th, 2018, 09:19 AM  #5 
Global Moderator Joined: Dec 2006 Posts: 20,107 Thanks: 1907 
Do you see why angles QAP and CDQ are equal? As a consequence, this area formula will give the same value for the two triangles. You can avoid explicit use of trigonometry by constructing an appropriate altitude for each triangle, though doing that would use rightangled triangles in a way that's obviously trigonometric in nature.
Last edited by skipjack; May 11th, 2018 at 09:47 AM. 
May 11th, 2018, 09:27 AM  #6 
Member Joined: Jan 2018 From: Belgrade Posts: 55 Thanks: 2 
Yes, they are inscribed angles over arc $\displaystyle CM$. Thanks. I'm trying to avoid using trigonometry because this is problem for middle school (13 years of age) students. P.S. Link is bad. Erase 'v' from the beginning of the URL. Last edited by lua; May 11th, 2018 at 09:30 AM. 
May 11th, 2018, 10:27 AM  #7 
Global Moderator Joined: Dec 2006 Posts: 20,107 Thanks: 1907 
I've corrected that typo, thanks. Your suggested method of using the fact that the area of the quadrilateral APQD is AQ·DP/2 because AQ and PD are perpendicular to each other is also valid, and is shorter. That area result is a special case of this standard trigonometrical formula, where $\sin(\theta) = \sin(90^\circ\!) = 1$, but can easily be proved without explicit use of trigonometry. By the way, I haven't come across your notation for the area of a quadrilateral before. It isn't a standard notation as far as I know. 
May 11th, 2018, 10:46 AM  #8 
Member Joined: Jan 2018 From: Belgrade Posts: 55 Thanks: 2 
Oh, I see. It is standard notation (among others) for the area where I am from (Belgrade is the capital of Republic of Serbia. In Serbian, area is said površina, and that first letter p  capitalized  is very often used as a notation for area.). Often, name of the figure is in the subscript of P. Last edited by skipjack; May 11th, 2018 at 05:20 PM. 

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area, calculate, quadrilateral, quadruple, surface 
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