Can you find area of the Green shaded region? | Two semicircles | (Easy explanation) |

[@amarendrasingh7327]

Impressive 😍😍 question very nice 👍👍🙂🙂

[@PreMath]

Thanks for watching ❤️🌹

[@TheSMNM]

Shorter method is difference between areas of sector ABD with 30deg and smaller semicircle

[@wackojacko3962]

I love those 30⁰ 60⁰ 90⁰ special triangles! And that love is right up there for the love for I love I have for you and your PreMath channel! 🙂

[@PreMath]

Yes! Thank you! ❤️🌹

[@johnbrennan3372]

|DB|= 4root3 +2root5 by using the rt triangles QCB and QCD. In the triangle DPB we know the sides are length 4,6 and 4root3 + 2root5. Then use cosine rule to find measure of angle dpb.That gives measure of angle dpa. So then find area of triangle DPB and sector ADP.Add the two areas and subtract area of smaller semicircle.

[@ExploredMaths]

Excellent 👍

[@adept7474]

R = 6, r = 4. ∠QCP = ∠ABD = 30° (QC= 2QP). S(sector ADP) = 1/3S(big semicircle) = 6π. S(▲DBP) = S(▲ADP) (DP - median) = R²√3/4 = 9√3. S(green) = 6π + 9√3 - 8π = 9√3 - 2π.

[@ybodoN]

AP = PB = PD = R = 6. AQ = QC = r = 4. QB = 2R − r = 8. Since the hypotenuse QB is twice QC, then △QCB is a 30° - 60° - 90° special right triangle. As soon as we know that ∠ABD = 30°, we also know that ∠APD = 60° (the angle at the center of a circle is twice the angle at the circumference). Therefore, △APD is equilateral and its area is ¼ R²√3 = 9√3. Since △APD and △PBD have the same height and AP = PB ⇒ area △PBD also is 9√3. The area of the circular sector APD = ⅙ R² π = 6π. The area of the small semicircle is 8π. So the green shaded region is 6π + 9√3 − 8π = 9√3 − 2π.

[@johnbrennan3372]

Very nice method.Thanks for sharing

[@phungpham1725]

Very impressive! Thank you!

[@PreMath]

Thank you too! ❤️

[@robertbourke7935]

Got it out. Many thanks

[@PreMath]

Thank you too ❤️

[@txt.myhome7979]

You can connect C and P and C and Q and then do it

[@PreMath]

Thanks

[@manuelantoniobahamondesa.3252]

Excelente!!

[@PreMath]

Thanks ❤️🌹

[@Ibrahimfamilyvlog2097l]

Very nice 👍👍❤❤

[@PreMath]

Many many thanks 🌹

[@JAMESYUN-e3t]

Really excellent PT on the excellent math question. Reasoning power is being strengthened thanks to these math questions.😅

[@PreMath]

Great to hear! Thanks ❤️🌹

[@mohabatkhanmalak1161]

Intense 'O' level geometry, it got crowded a bit but very, very interesting. Enjoyed watching to the solution. I didn't get it, as I quickly glossed over it with a cup of coffee in hand, but will work through it for practice. Thank you teacher!💫

[@PreMath]

Glad you enjoyed it! ❤️🌹

[@jimlocke9320]

Green area = big semicircle - small semicircle - sector BPD + ∆BEP + ∆DEP = 18π - 8π - 12π + (1/2)(3)(3√3) + (1/2)(3)(3√3) = 9√2 - 2π I am using dimensions and angles found in the video. Once we find

[@giuseppemalaguti435]

Con la geometria analitica calcolo P,intersezione big circle e retta..in sintesi Agr=9√3+6π-8π=9√3-2π

[@PreMath]

Super! Thank you ❤️

[@randiwijaya9609]

👍🙏

[@PreMath]

Thanks dear ❤️🌹

[@JSSTyger]

9.305

[@PreMath]

Great ❤️

[@mega_mango]

Велком ту примат

[@kasarlamahesh]

who knows all this properties and all these maths look like bullshit