Can you find area of the Yellow shaded semicircle? | (Triangle) |
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- Опубликовано: 30 сен 2024
- Learn how to find the area of the Yellow shaded semicircle. Important Geometry skills are also explained: circle theorem; Heron's formula; triangle area formula; circle area formula. Step-by-step tutorial by PreMath.com
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Can you find area of the Yellow shaded semicircle? | (Triangle) | #math #maths | #geometry
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Thanks Sir
Very nice and enjoyable
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With my respects.
Many many thanks, dear 🌹
12√5=1/2(7)(r)+1/2(8)(r)
So r=8√5/5
Yellow area=1/2(π}(8√5/5)^2=32π/5=20.1 square units.❤
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Beginning @ 9:32 It's funny how we are always making irrational numbers rational again by judgment of rounding off numbers. ...Just sayin. 🙂
😀
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Nice good sar❤❤❤
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STEP-BY-STEP RESOLUTION PROPOSAL :
01) Triangle Area = 12*sqrt(5) ; Using Heron's Formula
02) R = Radius of Semicircle
03) (7R + 8R) / 2 = 12*sqrt(5)
04) 15R = 24*sqrt(5)
05) R = 24*sqrt(5) / 15
06) R = 8*sqrt(5) / 5
07) R ~ 3,6
08) Yellow Area = Pi * R^2 / 2
09) YA = (Pi * (64 * 5) / 25) / 2
10) YA = (320 * Pi) / 50
11) YA = (32 * Pi) / 5
12) YA ~ 100,531 / 5
13) YA ~ 20,106
ANSWER : The Yellow Area equal to 20,106 Square Units.
Best Regards from The Islamic International Institute of Universal Knowledge
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I have a better solution to find the coordinates of point O.
(AO) is the bissector of angleBAC, so BO/BA = CO/CA, so BO/7 =CO/8 = (BO + CO)/15 = 9/15,
So, BO/7 = 9/15 and BO = 21/5, and O(21/5; 0)
Now: R = distance from O to (AB) to finish.
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Проведём перпендикуляры из центра полуокружности к сторонам 7 и 8, а из вершины треугольника проведём биссектрису в центр полуокружности, получим два треугольника, площади которых равны 7R/2 и 8R/2, в сумме, они дают площадь треугольника АВС,=15R/2, которая, в свою очередь равна Корню квадратному из произведения полупериметра на разность полупериметра с каждой из сторон треугольника. Площадь равна 12\/5, которую мы приравняем к площади, выраженной через радиус полуокружности, 12\/5=15R/2, откуда R=8\/5/5. Площадь полуокружности равна 3,1416*R^2/2=20,1.
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R/sinABC+R/sinACB=9...dalle formule di..Briggs cosABC/2=√(16/21)..cosACB/2=√(5/6)... svolgo i calcoli risulta R=8√5/5
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Draw radii DO & EO, they are tangent to sides AB & AC and therefore perpendicular to the sides by the Circle Theorem.
Find the semi-perimeter of △ABC and use Heron's Formula.
s = (a + b + c)/2
= (9 + 8 + 7)/2
= 24/2
= 12
A = √[s(s - a)(s - b)(s - c)]
= √[12(12 - 9)(12 - 8)(12 - 7)]
= √(12 * 3 * 4 * 5)
= √720
= (√144)(√5)
= 12√5
Draw segment AO. It forms two triangles, △AOB & △AOC. These triangles combine to form △ABC.
Find their areas.
A = (bh)/2
= (7r)/2
A = (8r)/2
= 4r
△ABC Area = △AOB Area + △AOC Area
12√5 = (7r)/2 + 4r
(15r)/2 = 12√5
15r = 24√5
r = (24√5)/15
= (8√5)/5
Now find the area of the semicircle.
A = (πr²)/2
= 1/2 * π * [(8√5)/5]²
= 1/2 * π * 320/25
= (160π)/25
= (32π)/5
So, the area of the yellow semicircle is (32π)/5 square units (exact), or about 20.11 square units (approximation).
Complete the circle. Thereafter draw tangents from points B and C.
Then a tangential quadrilateral will be formed with sides 8, 7,7,8 units consecutively.
Area of quadrilateral = 24√5
Inradius =24√5/(7+8)=8√5/5=8/√5
Area of semicircle =1/2*64π/5=32π/5
Comment please
(7+8+9)/2=12 △ABC=√[12･(12-7)(12-8)(12-9)]=√[12･5･4･3]=√720=12√5
7r/2 + 8r/2 = 12√5 15r = 24√5 r=8√5/5
Yellow Area = 8√5/5 * 8√5/5 * π * 1/2 = 32π/5
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Before looking at the video, I'm wondering if the triangle area could be 3.5r for ABO and 4r for ACO, making 7.5r. Then calculate the area in numbers via Heron's Formula. This will give the value for r.
Heron's Formula:
7+8+9=24, so semiperimeter is 12
sqrt(12(12-9)(12-8)(12-7))
sqrt(12*3*4*5)
sqrt(720)
sqrt(4)*sqrt(180)
sqrt(4)*sqrt(4)*sqrt9)*sqrt(5)
4*3*sqrt(5), so 12*sqrt(5)
7.5r = 12*sqrt(5)
15r = 24*sqrt(5)
r = (24/15)*sqrt(5)
r = (8/5)*sqrt(5)
r^2 = (64/25)*5 = 320/25 = 64/5, so the area of the full circle would be (64/5)pi
As a semicircle it is (32/5)pi un^2
32*3.142 = 100.55 (rounded)
100.55/5 = 20.11 un^2 (rounded)
Just watched the video and we went the same route, albeit with some variation on how we calculated. I haven't used Heron's much, but I can see it's very useful. Thanks again.
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Another approach: Let radius of semi-circle be R and let EB = X, then AE = 7 - X. Let BO = Y, then OC = 9 - Y. Also DC + AD = 8 and AD = AE = 7 - X so DC = 8 - AD = 1+ X. Next we apply Pythagoras theorem to the right angled triangle EBO giving
Y^2 = X^2 + R^2 --------(1)
and also to right angled triangle OCD giving
(9 - y)^2 = (1+ X)^2 +R^2 ---------(2).
Subtracting equation (1) from equation (2) gives
Y = (40 - X)/9 -----(3)
Now we can apply the Cosine rule to triangle ABC to get the Cosine of angle ABC as
(7^2 + 9^2 - 8^2)/2*7*9 = 0.52381.
Now Cosine of angle ABC = X/Y = 0.52381--------(4)
We solve equations (3) and (4) to get X= 2.2 and Y = 4.2 and then use these values in equation (1) to get R = 3.57777 and area of semi=circle = 1/2*pi*R^2 = 20.1061 squared units
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Something different: (I don't copy details)
Let's use an orthonormal center B, first axis (BA)
B(0;0) c(9;0)
Equation of the circle center B, radius7: x^2 + y^2 = 49
Circle center C, radius 8: (x -9)^2 + y^2 = 64
Intersection (with positive ordinate):
Point A(11/3; (8/3).sqrt(5))
Equation of (BA): 8.sqrt(5).x -11.y = 0
Distance from M(x; y) to (BA):
(Abs(8.sqrt(5).x -11.y))/21
Equation of (CA): sqrt(5).x +2.y -9.sqrt(5) = 0
Distance M to (CA): (Abs(sqrt(5).x +2.y -9.sqrt(5))/21
(AO) is the bissector of angleBAC)
Its equation is obtained when writing that distance from M to (BA) is equal to distance from M to (CA)
We obtain two equations of straight lines and choose the one: 15.sqrt(5).x +3.y -63.sqrt(5) =0
Then point O is the intersection with the first axis.
Then point O(21/5; 0)
The radius of the yellow circle is the distance from O to (BA) (or to CA): (8/5).sqrt(5)
Then the area is evident.
I know this is long and complicate, but I wanted to find "something else".
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It seems to be quite difficult puzzle, semicircle inscribed in an irregular triangle for too many parameters. However making use area formulas, parameters are cut down.😮😮😮😮
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Let's find the area:
.
..
...
....
.....
First of all we calculate the area of the triangle ABC according to the formula of Heron:
a = BC = 9
b = AC = 8
c = AB = 7
s = (a + b + c)/2 = (9 + 8 + 7)/2 = 24/2 = 12
A(ABC)
= √[s * (s − a) * (s − b) * (s − c)]
= √[12 * (12 − 9) * (12 − 8) * (12 − 7)]
= √(12 * 3 * 4 * 5)
= 12√5
AB and AC are tangents to the yellow semicircle. Therefore we known that ∠AEO=∠BEO=∠ADO=∠CDO=90°. As a consequence there exists another way to calculate the area of the triangle ABC, that enables us to obtain the radius R of the yellow semicircle:
A(ABC)
= A(ABO) + A(ACO)
= (1/2)*AB*h(AB) + (1/2)*AC*h(AC)
= (1/2)*AB*OE + (1/2)*AC*OD
= (1/2)*AB*R + (1/2)*AC*R
= (1/2)*R*(AB + AC)
⇒ R = 2*A(ABC)/(AB + AC) = 2*12√5/(7 + 8) = 24√5/15 = 8√5/5
Now we are able to calculate the area of the yellow semicircle:
A(yellow) = πR²/2 = π*(8√5/5)²/2 = π*(64*5/25)/2 = (32/5)*π ≈ 20.11
Best regards from Germany
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Very good question! 👌👍
Glad you think so!
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Tem como saber as distâncias BO e CO?
OB=4.2