Can you find the area of the Green shaded region? | (Quarter circle in a square) |

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Learn how to find the area of the Green shaded region. Quarter circle is inscribed in a square. Important Geometry and Algebra skills are also explained: area of a triangle formula; area of a sector formula; Pythagorean theorem. Step-by-step tutorial by PreMath.com
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Комментарии : 56

@ProfessorDBehrman 10 месяцев назад

Nice problem! Thanks!

@PreMath 10 месяцев назад

Happy to help! Thanks a lot ❤️🌹

@georgebliss964 10 месяцев назад

Angle PBC = 15 degrees. Area of triangle PBC = 0.5 x 2 x root 2 x sin 15 degrees. Root 2 x sin 15. Triangle QBC same area, therefore sum of both = 2 x root 2 x sin 15. Root 8 x sin 15 = 0.732. Area of 30 degree sector = Pi x 4 /12 = Pi /3 = 1.0472. Green area = 1.0472 - 0.732 = 0.315.

@marioalb9726 10 месяцев назад

sin α = ½R / R = 1/2 α = 30° Side of green region (s +1) = R cos α = 2.√3/2 s = 0,73205 cm = √3-1 Area of angular sector: A= ½αR² = ½.30°.2² A = 1,0472 cm² = π/3 Area of 2 triangles: A = 2 . (½ b.h) A = s. R/2 = (√3-1).1 A = 0,73205 cm² = √3-1 Area of green region: A = A₁ - A₂ A = 1,0472 - 0,73205 A = 0,31515 cm² ( Solved √ )

@幕天席地-w9c 10 месяцев назад

Alternatively, let M and N be the midpoint of AH and HG, it's easy to calculate the area of AHG, AMP. AMP=QNG, so MPQNH=AHG-2*AMP, PCQ=MCNH - MPQNH

@Okkk517 10 месяцев назад

How do you calculate the area of AMP ?!

@hyung-yulcho841 10 месяцев назад

The area of 1/4 Circle that includes green area is 1/4(2X2 )X 3.14.units = 3.14 units and there are three 1 by 1 squares, which is 3 square unis altogether. Therefore, 3.14units minus three units are approximately 0,14 units.

@parthtomar6987 10 месяцев назад

Nice solution sir 🎉🎉 First comment

@PreMath 10 месяцев назад

Thanks dear ❤️

@phungpham1725 10 месяцев назад

By observation (the picture makes me think of the unit circle), we can see the BPG is an equilateral triangle so PD= sqrt3, so PC=sqrt3-1 and the sum of the area of the 2 congruent triangles BCP and BCQ = sqrt3-1 Area of the green region= Area of the 30 degree sector - area of the sum of BCP and BCQ = pi x 4 /12 - (sqrt3-1)= pi/3 - sqrt3+1

@quigonkenny 6 месяцев назад

Draw BP, BG, and BC. Let M be the midpoint of BG and N be the midpoint of BA. As BP and BG are radii of the quarter circle, BP = BG = 2. As BP = 2 and BM = 1, ∆PMB is a 30-60-90 special right triangle, and PM = √3. By symmetry, QN is also √3. By observation, PC and QC are √3-1. Triangle ∆CBP: A = bh/2 = (√3-1)1/2 = (√3-1)/2 By symmetry, ∆CBQ = (√3-1)/2 as well, so quadrilateral CPBQ = √3-1. As ∠PBM = 60°, ∠ABP = ∠GBQ = 30°, so ∠PBQ = 90-30-30 = 30°. Sector PBQ: A = (θ/360)πr² = (30/360)π2² = 4π/12 = π/3 Green area: A = Sector PBQ - CPBQ A = π/3 - (√3-1) = 1 + π/3 - √3 ≈ 0.315

@giuseppemalaguti435 10 месяцев назад

Procedo con geometria e integrali ..π/3-(√3-1)

@PreMath 10 месяцев назад

Thanks ❤️🌹

@aljawad 6 дней назад

I resorted to using integral calculus to reach the same result.

@elmer6123 10 месяцев назад

π = area of 1/4 circle; π/3 + √3/2 = area in 1/4 circle above center, and the same area in 1/4 circle to the right of center. Thus, 2(π/3 + √3/2) - 1 = area in 1/4 circle excluding green area, and green area is π - [(2π/3 + √3) - 1] = 0.31515...

@DB-lg5sq 9 месяцев назад

شكرا لكم ليكن Xهو المجهول و s مساحة القطاع الدائري PBQ وb مساحة المثلث PBQ وC مساحة المثلث PCQ إذن X=s-b+c X=pi/3 +(1- جذر 3)

@misterenter-iz7rz 10 месяцев назад

x=pi/2-2arcsin(1/c2)=pi/2-2×pi/6=pi/2-pi/3=pi/6, area=(1/2)2^2 (pi/6)-1×(sqrt(3)-1))=2(pi/6)-sqrt(3)+1=0.315147 approximately 😊

@PreMath 10 месяцев назад

Thanks ❤️

@AdamjacobiGIA 8 месяцев назад

I was so stumped when I saw the thumbnail - as soon as you drew the lines to the radius it all clicked! What a clever puzzle to solve, thank you for sharing it!

@DB-lg5sq 10 месяцев назад

شكرا يمكن استعمال حساب التكامل......

@PreMath 10 месяцев назад

Thanks ❤️

@mohabatkhanmalak1161 10 месяцев назад

A very interesting problem. Enjoyed watching for the solution, I didn't get but I will now try it out myself as an exercise. Thank you teacher!☀

@PreMath 10 месяцев назад

Thanks my dear friend ❤️🌹

@santiagoarosam430 10 месяцев назад

Si llamamos T a la proyección ortogonal de Q sobre BG→ QT=1 y QB=2→ El ángulo QBG=30º→ Por simetría, el ángulo en B está trisecado y PBQ=30º→ BT=√3→ CQ=√3 -1 → Área verde = (⊓2²/12)-2(1*CQ/2) =(⊓/3)-(√3 -1)=(⊓/3)+1-√3 =0.3151 Gracias y un saludo cordial.

@PreMath 10 месяцев назад

Thanks❤️🌹

@nassernasser879 10 месяцев назад

Excellent and awesome! Nice question and great explanation. Thanks a million!👍👏

@PreMath 10 месяцев назад

Glad you liked it! ❤️🌹

@MrPaulc222 10 месяцев назад

Before viewing: a quick very rough estimate to ensure my final answer is in the correct ballpark. It looks about 1/3 sq un or about (1/10)pi, sq un, depending on the final formatting. Quadrant radius is 2, so quadrant area is (4 pi)/4 = pi sq un. BC = sqrt(2) and C to the quadrant’s closest circumference is 2 - sqrt(2). Take C as a new circle centre with radius of 2 - sqrt(2). Green area is ((2 - sqrt(2))^2 * pi ((2 - sqrt(2))^2 = 4 - 4*sqrt(2) + 2. Simplify to 6 - 4*sqrt(2), then multiply by pi. I'm getting 1.078... which is clearly wrong. Aha. It's a smaller quadrant. I forgot to divide by 4. Oops! It's 0.2695... A bit less than my visual, very rough estimate, but close enough for me to go with it.

@PreMath 10 месяцев назад

Thanks ❤️

@MrPaulc222 10 месяцев назад

I went wrong with this one but, more importantly, I now see where and why. It might have been a good idea if I'd realised sooner that C cannot be the circle centre of the smaller quadrant. Ho-hum. I move on :)

@comdo777 9 месяцев назад

asnwer=1 cm isit 1

@normanc918 10 месяцев назад

After we find that the sector PBQ is 30 degree, then we can conclude that its area is one third of the sector ABQ which area is pi. The total areas of PBC and QBC = sq root 3 -1, therefore the green area is pi - (sq root 3 -1) = pi - 3 + 1.

@NaceerAlassady-x2l 10 месяцев назад

Good afternoon....how can drive the relation which make me find the third line of non-right angle triangle.I have the value of the other tow lines and the angle between there. Please sir....not the low..... I want driving the low. Thank you very much

@ybodoN 10 месяцев назад

Since DB = 1 and the hypotenuse BP = 2, △DBP is a special 30° - 60° - 90° triangle ⇒ PD = √3. By symmetry, △BCP ≅ △BCQ with a base of √3 − 1 and a height of 1 for a total area of √3 − 1......➀ Since the right angle ABG is trisected, the area of the circular sector BPQ is ¹⁄₁₂ π 2² or ⅓ π .........➁ The green shaded region is ➁ − ➀, that is ⅓ π − (√3 − 1) or 1 − √3 + ⅓ π ≈ 0.315 square units.

@Okkk517 10 месяцев назад

After establishing the sector PQ subtends angle 30. We calculate PQ from which we infer that PC=CQ=sqrt(3)-1. Then, the result follows as ( sector PBQ - triangle BPQ + triangle PQC) area=pi/3-sqrt(3)+1.

@pk2712 10 месяцев назад

Your geometry skills are superb , and so are your detailed solutions .

@PreMath 10 месяцев назад

Glad you like them! Thanks a lot ❤️🌹

@dhsanj8877 10 месяцев назад

Basis triangle BEP the side BE is sqrt3 Now sides AB and GH are parallel with right angles and sides equal making it square Thus, at the centre of square .the intersection will be at right angles meaning green part is quarter circle Since mid pointscare connecting Further side CP can now be deduced as sqrt3 -1 Therefore area of green section is pi * (sqrt3 -1)^2 /4