At 4:15, instead of computing the area of ΔAOB and deducting the area of ΔAPC, we can compute the area of trapezoid PCBO. To find the area of a trapezoid, find half the sum of the lengths of the two parallel sides ("bases"), then multiply that result by the distance between the parallel sides ("height"). The bases are PC = 2√2 and OB = 4√2, which sum to 6√2. The height is OP or 2√2. Area of PCBO = (1/2)(PC + OB)(OP) = (1/2)(6√2)(2√2) = 12. We deduct the quarter circle area = 2π, to find the yellow shaded area equals 12 - 2π, as PreMath also found.
You draw a horizontal line from the center of the semicircle to its circumference, it corresponds to the radius and the top of a trapezoid The yellow area = the trapezoid area - half of the semicircle area = (4/√2 + 8/√2)(4/√2)/2 - (π/4)(4/√2)^2 = 12 - 2π
Call AO and BO, R, and they are each 8/sqrt(2) which = 4*sqrt(2). The semicircle radius is half that, so r = 2*sqrt(2). POBC is a trapezoid whose area is (4*sqrt(2) + 2*sqrt(2))/2 * 2*sqrt(2). (6*sqrt(2))/2 = 3*sqrt(2) (3*sqrt(2))*(2*sqrt(2)) = 6*2 = 12. From this, subtract the area PCO. A full small circle would be 8pi, so PCO is 2pi. Yellow area = 12 - 2pi = 12 - (2(3.142)) = 12 - 6.284 = 5.716 un^2 (rounded). I've now looked. Yes, our methods were similar, except that I subtracted the small quadrant from the trapezoid. Thanks again. Great video.
I can only assume the purpose of the quarter circle AOB is to trick people into the wrong answer. Or force them into a long winded solution where they add back the area of the semicircle outside the triangle AOB.
We may use similarity of triangles theorem in case of triangles APC & AOB AP/AB=4/8=1/2 Hence PC = 1/2 OB =r/2 AP = 1/2 AO =r/2 Now In 🔺 APC 2*r^2/4 = 16 > r =4√2 units Now area AOB =1/2 *r^2 =16 sq units area of APC =1/2*(r/2)^2 = 4 sq units From triangle AOB or APC we notice those are isosceles rt triangles. Hence ang OAB or angle PAC is 45 degrees Hence central angle OPC is 45*2 =90 degrees Hence area of small quarter circle =[π*(r/2)^2]/4 =[22/7 *32/4]/4 =22*32/(7*4*4)=44/7 sq units Area of Yellow region = AOB - APC - small quarter circle =16 -4-44/7 =12-44/7 =5.7142 sq units (approx)
The radius R of the big circle is OA = 8/sqrt(2) = 4.sqrt(2) and the radius r of the small circle is OP = OA/2 = 2.sqrt(2). If we draw from A a parallel at (OB) and from C a parallel at (OA) we obtain a rectangle containing the small circle, tangent to the small circle in C and whose area is R.r = 16. The area inside this rectangle and outside the small circle is then 16 - (Pi.(r^2)/2) = 16 - 4.Pi. We separate the yellow shaded region into 2 regions: The first is the region left of the parallel at (OA) from C and the second is the right isosceles triangle on the right. The area of the first region is (16 - 4.Pi)/2 = 8 - 2.Pi and the area of the second one is (r^2)/2 = 4, so finally the area of the yellow shaded region is (8 - 2.Pi) + 4 = 12 - 2.Pi.
Another (simpler) method would be to consider the yellow shaded region as the trapezoïd OPCB less a quater of the small circle. (When R and r are known): Area of trapezoïd: ((PC + OB)/2).OP = ((r + R)/2).r = (((4.sqrt(2 + 2.sqrt(2))/2).(2.sqrt(2) = (3.sqrt(2)).(2.sqrt(2)) = 12 Area of quater of small circle: Pi.((r^2)/4) = Pi.(8/4) = 2.Pi. So our answer is 12 - 2.Pi.
In 🔺 AOB, mid points of AO and AB are P and C respectively. Hence PC ll OB (OB =r) ang OPC = 180 - ang AOB =90 degrees PC =1/2 *OB =r/2 Now from rt 🔺 AOB we may get r√2 =8 > r=4√2 units OB=4√2 units r/2=2√2 units PC =2√2 units Area of trapezium CPOB = 1/2(r/2 +r) sq units =1/2*2√2*(4√2+2√2)sq units =12 sq units Area of small quarter circle =π/4(2√2)^2=2π sq units Area of yellow region = area of trapezium - area of small quarter circle = (12-2π)sq units = 5.7142 sq units (approx) (putting π=22/7)
Triangle APC is similar to triangle AOB, the sides of APC are half as long as those of APC, so the area of APC is a quarter of the area of AOB. Do you never use the Ray Theorem?
Let's find the area: . .. ... .... ..... The triangle OAB is a right triangle, so we can apply the Pythagorean theorem. With R being the radius of the quarter circle and r being the radius of the inscribed semicircle we obtain: OA² + OB² = AB² R² + R² = 8² 2R² = 64 R² = 32 ⇒ R = √32 = 4√2 ⇒ r = PO = PA = OA/2 = R/2 = 4√2/2 = 2√2 Since AP/AO=r/R=(r/2)/r=1/2 and AC/AB=AC/(AC+BC)=AC/(2*AC)=1/2 and ∠PAC=∠OAB, we know that the triangles PAC and OAB are similar. Therefore PAC is also a right triangle. So we can conclude: PC/OB = AC/AB = 1/2 ⇒ PC = OB/2 = R/2 = r Since PAC is a right triangle (∠APC=90° ⇒ ∠OPC=90°), we know that OB and PC are parallel. Therefore we can conclude that OBCP is a trapezoid. Now we are able to obtain the area of the yellow region: A(yellow) = A(trapezoid OBCP) − A(quarter circle POC) = (1/2)*(OB + PC)*OP − πr²/4 = (1/2)*(R + r)*r − πr²/4 = (1/2)*(2r + r)*r − πr²/4 = 3r²/2 − πr²/4 = (6 − π)*r²/4 = (6 − π)*(2√2)²/4 = (6 − π)*8/4 = 12 − 2π ≈ 5.717 Best regards from Germany
@@PreMath Thank You Very Much. Can you, please, check my Resolution of the Last Problem? The one with Green Shaded Area. I am not certain about my Proposal. TYVM. Seasons Greetings From Portugal (Al - Andalus).
От площади ниже средней линии надо отнять четверть круга. Площадь этой прямоугольной трапеции составляет 3/8 от всего достроенного квадрата - 32, т. е. 3*32/8=3*4=12. Площадь круга πø²/4, делим ещё на 4, получаем πø²/16=32π/16=2π. Итого искомая площадь 12-2π.
Solution: Yellow Shaded Area = ∆ AOB Area - ∆ APC Area - Quarter Circle Area OPC ... ¹ OA = OB = r r = 8/√2 r = 8√2/2 r = 4√2 AP = PC = 4√2/2 = 2√2 Therefore ∆ AOB Area = ½ b h ∆ AOB Area = ½ 4√2 . 4√2 ∆ AOB Area = 16 ∆ APC Area = ½ b h ∆ APC Area = ½ 2√2 . 2√2 ∆ APC Area = 4 Quarter Circle Area OPC = ¼ π r² = ¼ π (2√2)² = 2π Replacing in Equation ¹ Yellow Shaded Area (YSA) = 16 - 4 - 2π YSA = 12 - 2π Square Units ✅ YSA ≈ 5,7168 Square Units ✅
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THANKU for Finding Spherical curved areas ! These r better than just finding straight lines areas bcuz NATURE uses Curvature ! Again Thanks as usual !
I'm glad you find it interesting! 😊
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At 4:15, instead of computing the area of ΔAOB and deducting the area of ΔAPC, we can compute the area of trapezoid PCBO. To find the area of a trapezoid, find half the sum of the lengths of the two parallel sides ("bases"), then multiply that result by the distance between the parallel sides ("height"). The bases are PC = 2√2 and OB = 4√2, which sum to 6√2. The height is OP or 2√2. Area of PCBO = (1/2)(PC + OB)(OP) = (1/2)(6√2)(2√2) = 12. We deduct the quarter circle area = 2π, to find the yellow shaded area equals 12 - 2π, as PreMath also found.
Very clever solution! Thank you so much😊😊😊
Excellent approach!😀
Thanks for sharing ❤️🙏
We can find area of trapezoid OPCB = (PC+OB)*PO/2 = 12and deduct quarter circle PCO area 2*pi.
Thanks for sharing your approach! ❤️
Solved it in my head!
Great video!
Excellent! Thanks for sharing!❤️
Amazing explanation, tank you teacher 👍.
It is honestly a pretty easy question. I solved it in my mind. Such questions should come in exams
D’Oh! I got it wrong, a bookkeeping mistake someplace and that with using pen and paper. 🥴 It looked easy to me going in.
@dantallman5345 every problem is difficult unless you try it and get a hold of it. So keep trying and never quit 🤞
You draw a horizontal line from the center of the semicircle to its circumference, it corresponds to the radius and the top of a trapezoid
The yellow area = the trapezoid area - half of the semicircle area = (4/√2 + 8/√2)(4/√2)/2 - (π/4)(4/√2)^2 = 12 - 2π
Excellent! Thanks for sharing ❤️
@ Thank YOU
Call AO and BO, R, and they are each 8/sqrt(2) which = 4*sqrt(2).
The semicircle radius is half that, so r = 2*sqrt(2).
POBC is a trapezoid whose area is (4*sqrt(2) + 2*sqrt(2))/2 * 2*sqrt(2).
(6*sqrt(2))/2 = 3*sqrt(2)
(3*sqrt(2))*(2*sqrt(2)) = 6*2 = 12.
From this, subtract the area PCO.
A full small circle would be 8pi, so PCO is 2pi.
Yellow area = 12 - 2pi = 12 - (2(3.142)) = 12 - 6.284 = 5.716 un^2 (rounded).
I've now looked. Yes, our methods were similar, except that I subtracted the small quadrant from the trapezoid.
Thanks again. Great video.
Excellent! Nice approach!
You are very welcome!
Thanks for the feedback ❤️🙏
Thank you!
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Thanks for the feedback ❤️
I can only assume the purpose of the quarter circle AOB is to trick people into the wrong answer. Or force them into a long winded solution where they add back the area of the semicircle outside the triangle AOB.
R= 8/√2=4√2cm ; r=½R=2√2 cm
A = A₁ - A₂ = ½bh - ½r²(α-sinα)
A = ½R½R - ½(2√2)²[π/2-sin(π/2)]
A = ¼(4√2)² - 4(π/2-1) = 8 - (2π-4)
A = 12-2π cm² = 5,717cm² ( Solved √ )
AO=OB=4√2 AP=PO=2√2
Yellow shaded area = 4√2*4√2*1/2 - 2√2*2√2*1/2 - 2√2*2√2*π*1/4 = 12 - 2π
Deduct area of quarter circle from area of trapezium PCBO.
That's a great approach! 👍
Thanks for the feedback ❤️
S=12-2π=2(6-π)≈5,71
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Thanks for sharing ❤️
We may use similarity of triangles theorem in case of triangles APC & AOB
AP/AB=4/8=1/2
Hence PC = 1/2 OB =r/2
AP = 1/2 AO =r/2
Now In 🔺 APC
2*r^2/4 = 16
> r =4√2 units
Now area AOB =1/2 *r^2
=16 sq units
area of APC =1/2*(r/2)^2
= 4 sq units
From triangle AOB or APC
we notice those are isosceles rt triangles.
Hence ang OAB or angle PAC is 45 degrees
Hence central angle OPC is 45*2 =90 degrees
Hence area of small quarter circle =[π*(r/2)^2]/4 =[22/7 *32/4]/4
=22*32/(7*4*4)=44/7 sq units
Area of Yellow region
= AOB - APC - small quarter circle
=16 -4-44/7
=12-44/7
=5.7142 sq units (approx)
Excellent!
Thanks for sharing ❤️
@@PreMath Thanks sir
The radius R of the big circle is OA = 8/sqrt(2) = 4.sqrt(2) and the radius r of the small circle is OP = OA/2 = 2.sqrt(2). If we draw from A a parallel at (OB) and from C a parallel at (OA) we obtain a rectangle containing the small circle, tangent to the small circle in C and whose area is R.r = 16. The area inside this rectangle and outside the small circle is then 16 - (Pi.(r^2)/2) = 16 - 4.Pi. We separate the yellow shaded region into 2 regions: The first is the region left of the parallel at (OA) from C and the second is the right isosceles triangle on the right. The area of the first region is (16 - 4.Pi)/2 = 8 - 2.Pi and the area of the second one is (r^2)/2 = 4, so finally the area of the yellow shaded region is (8 - 2.Pi) + 4 = 12 - 2.Pi.
Excellent!
Thanks for sharing ❤️
Another (simpler) method would be to consider the yellow shaded region as the trapezoïd OPCB less a quater of the small circle. (When R and r are known):
Area of trapezoïd: ((PC + OB)/2).OP = ((r + R)/2).r = (((4.sqrt(2 + 2.sqrt(2))/2).(2.sqrt(2) = (3.sqrt(2)).(2.sqrt(2)) = 12
Area of quater of small circle: Pi.((r^2)/4) = Pi.(8/4) = 2.Pi. So our answer is 12 - 2.Pi.
In 🔺 AOB, mid points of AO and AB are P and C respectively.
Hence PC ll OB (OB =r) ang OPC = 180 - ang AOB =90 degrees
PC =1/2 *OB =r/2
Now from rt 🔺 AOB we may get
r√2 =8 > r=4√2 units OB=4√2 units
r/2=2√2 units PC =2√2 units
Area of trapezium CPOB
= 1/2(r/2 +r) sq units
=1/2*2√2*(4√2+2√2)sq units
=12 sq units
Area of small quarter circle
=π/4(2√2)^2=2π sq units
Area of yellow region
= area of trapezium - area of small quarter circle
= (12-2π)sq units
= 5.7142 sq units (approx) (putting π=22/7)
Triangle APC is similar to triangle AOB, the sides of APC are half as long as those of APC, so the area of APC is a quarter of the area of AOB.
Do you never use the Ray Theorem?
R=4√2,r=2√2...Ay=(4√2)^2/2-(4π-(2π-4))=16-(2π+4)=12-2π
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Thanks for sharing ❤️
Nice! 2(6 - π) ≈ 40/7
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Thanks for sharing ❤️
Let's find the area:
.
..
...
....
.....
The triangle OAB is a right triangle, so we can apply the Pythagorean theorem. With R being the radius of the quarter circle and r being the radius of the inscribed semicircle we obtain:
OA² + OB² = AB²
R² + R² = 8²
2R² = 64
R² = 32
⇒ R = √32 = 4√2
⇒ r = PO = PA = OA/2 = R/2 = 4√2/2 = 2√2
Since AP/AO=r/R=(r/2)/r=1/2 and AC/AB=AC/(AC+BC)=AC/(2*AC)=1/2 and ∠PAC=∠OAB, we know that the triangles PAC and OAB are similar. Therefore PAC is also a right triangle. So we can conclude:
PC/OB = AC/AB = 1/2 ⇒ PC = OB/2 = R/2 = r
Since PAC is a right triangle (∠APC=90° ⇒ ∠OPC=90°), we know that OB and PC are parallel. Therefore we can conclude that OBCP is a trapezoid. Now we are able to obtain the area of the yellow region:
A(yellow)
= A(trapezoid OBCP) − A(quarter circle POC)
= (1/2)*(OB + PC)*OP − πr²/4
= (1/2)*(R + r)*r − πr²/4
= (1/2)*(2r + r)*r − πr²/4
= 3r²/2 − πr²/4
= (6 − π)*r²/4
= (6 − π)*(2√2)²/4
= (6 − π)*8/4
= 12 − 2π
≈ 5.717
Best regards from Germany
Excellent!
Thanks for sharing ❤️🙏
19.433
S= 12-2π= 5,716
Thanks sir!
You are very welcome!
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@ 8:33 you could have at least used 16 of the 105 trillion digits of pi we know. 😊
😀
Thanks for the feedback ❤️
I did it differently but I came up with the same answer in a few minutes.
STEP-BY-STEP RESOLUTION PROPOSAL :
01) AB = 8 ; AC = BC = 4
02) OA = OB = R (Radius)
03) 2R^2 = 64 ; R^2 = 32 ; R = sqrt(32) ; R = 4sqrt(2)
04) Triangle [AOB] Area = 16 sq un
05) Quarter of Circle Area (QCA) = 8Pi sq un
06) Semicircle Area (SA) = 4Pi sq un
07) Remainig Area = (QCA - SA) = (8 - 4)Pi = 4Pi sq un
08) Triangle [OBC] Area = 8 sq un
09) SCA - TA = (4Pi - 8) sq un
10) Yellow Shaded Area (YSA) = 8 - ((4Pi - 8) / 2)
11) YSA = 8 - (2Pi - 4) = 8 - 2Pi + 4 = (8 + 4) - 2Pi
12) YSA = (12 - 2Pi) sq un
13) YSA ~ 5,72 sq un
Therefore,
OUR BEST ANSWER IS :
Yellow Shaded Area is equal to (12 - 2Pi) Square Units or approx. equal to 5,72 Square Units.
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@@PreMath
Thank You Very Much. Can you, please, check my Resolution of the Last Problem? The one with Green Shaded Area. I am not certain about my Proposal.
TYVM. Seasons Greetings From Portugal (Al - Andalus).
@@LuisdeBritoCamacho
Thanks❤
[(1/2)(8²/2)(3/4)]-[(4/√2)²π/4] =12-2π u².
Gracias y saludos
От площади ниже средней линии надо отнять четверть круга. Площадь этой прямоугольной трапеции составляет 3/8 от всего достроенного квадрата - 32, т. е. 3*32/8=3*4=12. Площадь круга πø²/4, делим ещё на 4, получаем πø²/16=32π/16=2π. Итого искомая площадь 12-2π.
Excellent!
Thanks for sharing ❤️
@@PreMath u wlcm.👍
Solution:
Yellow Shaded Area = ∆ AOB Area - ∆ APC Area - Quarter Circle Area OPC ... ¹
OA = OB = r
r = 8/√2
r = 8√2/2
r = 4√2
AP = PC = 4√2/2 = 2√2
Therefore
∆ AOB Area = ½ b h
∆ AOB Area = ½ 4√2 . 4√2
∆ AOB Area = 16
∆ APC Area = ½ b h
∆ APC Area = ½ 2√2 . 2√2
∆ APC Area = 4
Quarter Circle Area OPC = ¼ π r² = ¼ π (2√2)² = 2π
Replacing in Equation ¹
Yellow Shaded Area (YSA) = 16 - 4 - 2π
YSA = 12 - 2π Square Units ✅
YSA ≈ 5,7168 Square Units ✅
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А этот маленький кусочек, выходит, 2π-32/8, итого 2π-4.
12-2pi
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4+1.7168=5.7168
8÷1.4142135624=5.656854×5.656854=32÷2=16+3.433=19.433