Rectángulo RPQU DE 20*40 m; V es la proyección ortogonal de U sobre RS; UV corta a RT en Y; RT y SU se cortan en H y Z es su proyección ortogonal sobre RS. QR=√(40²+20²)=20√5---> La razón de semejanza entre RVU y RPQ es s=40/20√5=2√5/5---> UV=40s=16√5--->RV=8√5---> VS=20√5-8√5=12√5---> Respecto a RS, la pendiente de RT es 1--->RV=VY y RZ=ZH---> Respecto a RS, la pendiente de SU es VU/VS=4/3---> 1*RZ=(4/3)(RS-RZ)---> RZ=80√5/7=ZH--> Área sombreada =RSH-RVY =(1/2)[(RS*HZ)-(RV*VY)] =[(RS*HZ)-(RV²)]/2 =(800*5/7)-160 =411,42857...m². Gracias y saludos.
@MathandEngineering
Ай бұрын
Wow, an excellent solution, The thing I like the most about it, is that you didn't do any approximations, thank you so much for the amazing solution.
@xualain3129
Ай бұрын
Thank you very much for the marking points V, Y and H. Using them I got an alternative solution 2880/7 as yours. QR=RS=TS=sqrt(20^2+40^2)=20*sqrt(5) Let
@MathandEngineering
Ай бұрын
Yet another perfect solution, by our great Mr Xu Alain, thank you for this selfless hard work, if I am opportuned 1 day I'll come visit you, you really are an expert in trigonometry, I'll like to learn more from you, if you notice in my videos, I have adopted you method of calculating sin(a) from cos(a), or vice verse using the identity sin²(x) + cos²(x) = 1, Honestly I knew the identity, but it never occurred to me to use it that way, untill I saw it in one of your solutions and I thought it was the best. Thanks
@xualain3129
Ай бұрын
@@MathandEngineeringThank you for your compliments. But to be honest with you, I like to resort to trigonometry for the reason that very often I lack the instinct to construct proper lines to solve geometry problems. My best regards.
@jeffthompson2967
28 күн бұрын
I Cheated, and used Autodesk Inventor to Lay-It-Out, and had the Software calculate the Area for me. "Santiagoarosam430" is Spot On; Autodesk Inventor calculated 411.428571429 m^2 Just My Opinion; The Video Answer is WAAAY OFF. Thanks for showing such a great solution "Santiagoarosam430" !!!!.
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