Find the blue shaded area | Geometry Problem | Important Geometry and Algebra Skills Explained

Find the blue shaded area | Geometry Problem | Important Geometry and Algebra Skills Explained
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[@じーちゃんねる-v4n]

Think in terms of x-y coordinates with Q as the origin. A(-3, 6√3) F(-6, 3√3) Straight line AQ: y=-2√3x① Straight line FP: (1/2√3)(x+6)+3√3=x/( 2√3)+4√3② From ①②, the y coordinate of P is y=(48/13)√3 ∴△ABP=(1/2)6(6-48/13)√3=(90/13)√ 3

[@davidellis1929]

Using coordinate geometry with E at (0,0) and Q at (0,3), because AE=6*sqrt(3) as in the video, A is at (0,6*sqrt(3)) and F is at (-3,3*sqrt(3)). We can determine the equations of AQ and its perpendicular FP, yielding the coordinates of P. Then the area of the blue triangle is half of AB times the distance from P to AB.

[@jimlocke9320]

Using an HP 32S scientific calculator, I find the area of ΔABP as follows. The distance between parallel sides of a regular hexagon is side length times √3. In this case, the side length is 6 and the parallel sides are 6√3 apart.

[@sarathbabunalam5931]

Excellent!!

[@vierinkivi]

Piirretään pystysuora pisteestä F, jaketaan janaa AQ ylös, kohtaavat pisteessä G. FG = 9*sqrt3 FP nousee ( 3/(6sqrt3) )/1 GQ alenee (( 6sqrt3)/3)/1 Yhteensä (13/6) sqrt3 Kohtaavat etäisyydellä 54/13 linjasta FG ja 15/13 pisteestä A oikealle Kolmion alaksi saadaan 1/2 ((15/13)/3)6*sqrt3*6 = 90/13*sqrt3

[@giorgoschanis8919]

Cos(a-b)=cosa•cosb+sina•sinb

[@jonathancapps1103]

11:55 It's been a long time since I've been in a math class. I followed everything up to this point. But I don't follow cos(120⁰-θ)=cos120⁰cosθ+sin120⁰sinθ. I'm guessing that's a known identity? Does it have a name I can google to see its derivation?

[@satyamsharma9570]

can you please make a video on solution of these equation √x + y=4 , x+√y =15

[@ΓΕΩΡΓΙΟΣΛΕΚΚΑΣ-μ9μ]

Καλησπέρα σας. Πολύπλοκη άσκηση που απαιτεί ΚΑΙ χρήση τριγωνομετρίας και calculator. (την έλυσα με διαφορετικό τρόπο από τον δικό σας, είναι όμως μακρύς και δεν μπορώ να τον παραθέσω). Απαιτούνται υπολογιστής ή πίνακες τριγωνομετρικών αριθμών. Δεν μπόρεσα να την λύσω αποκλειστικά με χρήση Γεωμετρίας. Αν κάποιος την έλυσε ΑΠΟΚΛΕΙΣΤΙΚΑ με Ευκλείδεια Γεωμετρία, κι αν του είναι εύκολο ας παραθέσει την λύση του. Ευχαριστώ.

[@RamKumar-zi6dc]

Is view

[@marioalb9726]

Angle EAQ : tan α = 3/ (12 cos30°) α = 16,102° Angle FAQ: β = α +30° = 46,102° Cos β = AP / AF AP = AF cosβ = 6 cosβ AP = 4,2603 cm Height of triangle: h = AP cos α h = 4,2603 cos α h = 3,997 cm Area = b. h / 2 Area = 6 . 3,997 / 2 Area= 11,99 cm² ( Solved √ )

[@rabotaakk-nw9nm]

NO CALCULATOR!

[@marioalb9726]

​​@@rabotaakk-nw9nm No calculator?? A message through time, 50 years ago !!! How did you do this !!! In what year do you live?? unbelievable!! If you think so, your car mast be made of wood, blood traction, made without any machine or calculator !!!