Can you find area of the largest semicircle? | (Semicircles) | #math #maths | #geometry

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Can you find area of the largest semicircle? | (Semicircles) | #math #maths | #geometry

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[allanflippin]

Around 7:20 where you find the length of FB = 4+8+6 = 18, I found a simpler way to calculate R. Consider right triangle OFE. OF = 18-R, EF = 6 and EO = R. R can be found using Pythagorean theorem. (18-R)^2 + 6^2 = R^2. Multiplying out, 324 - 36R + R^2 + 36 = R^2. Eliminating R^2, and combining numbers we get 36R = 360 or R = 10.

[danmike]

You're definitely a knowledgeable person. Thanks 😊

[KenW-kbuk]

Could also extend big semi-circle to full circle and use intersecting chords to deduce AF length: 6 x 6 = (6+8+4) x AF => AF = 2

[prossvay]

Green semicircle area=18cm^2
1/2(π)r1^2=18π
r1=6cm
Yellow semicircle area=8cm^2
1/2(π)r2^2=8
So r2=4cm
Connect P to Q and C to Q
In CPQ
CQ^2+CP^2=PQ^2
6^2+CP^2=10^2
So CP=8cm
CB=CP+BP=8+4=12cm
CF=OE=6cm
FB=12+6=18cm
AF=2R-18
Connect O to F and E to F
AF=2R-18 OF=R-(2R-18)=18-R
In∆ OEF
EF^2+OF^2=OE^2
6^2+(18-R)^2=R^2
So R=10cm
BIg sermicle area=1/2(π)(10^2)=50π cm^2=157.08 cm^2.❤❤❤

[himadrikhanra]

R1= 6, R2 =4
Length of transverse common tangent is radius = R ( out)
Area = pi R ^2 /2

[misterenter-izrz]

(4+8-r+6)^2+36=r^2, (18-r)^2+36=r^2, 18^2-36r+36=0, r=(18^2+36)/36=10, thus the answer is 50pi.😊

[unknownidentity]

Let's find the area:
.
..
...
....


Let's use the labels R, R(g) and R(y) for the radii of the big semicircle, the green semicircle and the yellow semicircle, respectively. From the known areas of the green and the yellow semicircle we obtain:

A(g) = πR²(g)/2
(18π)cm² = πR²(g)/2
36cm² = R²(g)
⇒ R(g) = √(36cm²) = 6cm

A(y) = πR²(y)/2
(8π)cm² = πR²(y)/2
16cm² = R²(y)
⇒ R(y) = √(16cm²) = 4cm

The yellow and the green semicircle have exactly one point of intersection. Therefore the distance PQ of their centers is equal to the sum of their radii:

PQ = PD + QD = R(y) + R(g) = 4cm + 6cm = 10cm

Since AB is a tangent to the green semicircle, the triangle PQC is a right triangle and we can apply the Pythagorean theorem:

PQ² = QC² + PC²
PQ² − QC² = PC²
PQ² − R²(g) = PC²
(10cm)² − (6cm)² = PC²
100cm² − 36cm² = PC²
64cm² = PC²
⇒ PC = √(64cm²) = 8cm

Now let's add point F on AB such that CFEQ is a square. Then we obtain:

BF = PB + PC + CF = R(y) + PC + R(g) = 4cm + 8cm + 6cm = 18cm

According to the theorem of Thales the triangle ABE is a right triangle, so we can apply the right triangle altitude theorem:

AF*BF = EF²
(AB − BF)*BF = EF²
(2*R − BF)*BF = R²(g)
(2*R − 18cm)*(18cm) = (6cm)²
(36cm)*R − 324cm² = 36cm²
(36cm)*R = 360cm²
⇒ R = (360cm²)/(36cm) = 10cm

Now we are able to calculate the area of the big semicircle:

A = πR²/2 = π*(10cm)²/2 = (50π)cm²

[marcgriselhubert]

The radius of the green semi circle is 6 and the radius of the yellow semi circle is 4.
We use an orthonormal center P and first axis (PB). We have B(4; 0) and E(-14; 6) on the big semi circle whose equation is (x -a)^2 + y^2 = R^2
or x^2 +y^2 - 2.a.x + a^2 - R^2 = 0 (with a the abscissa of its center O and R its radius).
B is on it, so: 16 - 8.a + a^2 - R^2 = 0, E is on it, so: 196 + 36 + 28.a +a^2 - R^2 = 0. We substract and obtain: 36.a + 216 = 0, and so a = -6.
We can now have a^2 - R^2 and then R, but simpler: abs(a) = 6 = OP, and then R = 6 + 4 = 10.
Finally the area of the big semi circle is then (1/2).Pi.(10^2) = 50.Pi.

[sergioaiex]

Solution:

A = π a²
18π = π a²/2
π a² = 36π
a² = 36
a = 6

A = π b²
8π = π b²/2
π b² = 16π
b² = 16
b = 4

Triangle CPQ
CQ = 6
PQ = 4 + 6 = 10
CP = 8 (Pythagorean Triple)

Then, CB = 12
CF = 6
EF = h = 6
AF = m = ?
FB = n = 6 + 12 = 18

h² = m . n
6² = m . 18
36 = m . 18
m = 2

Diameter = 2 + 18
Diameter = 20
Radius = 10

A = π r²
A = π 10²/2
A = 50π Square Units

[phungpham]

My first attempt is almost the same as yours😅
So it is more fun to suggest another solution.
1/ Extend EQ to build the diameter EE’ and BP to build the diameter BB’.
We have: EE’=12 and BB’=8

2/ Focus on the tangent point D:
We have: the three points P, D, andQ are collinear( tangent theorem)
EE’//BB’—> angle E’QD=angle B’PD
Because the two angles mentioned above are the exterior angles of the two isiosceles triangles EQD and BPD
So, angle QDE=angle BDP
—> E, D, and B are collinear
—-> similarly E’D and B’ are collinear
—> E’B’//=AE ( both are perpendicular to EB)
—> AB’=12 and AB=20
Radius of the big semicircle=10
Area=1/2 pix100= 50 pi sq units😊😅😅

[LucasBritoBJJ]

I solved almost the same way…. Just got h²=mn

h=6
m=18
n=x

And it is easy to find x = 2

[philipkudrna]

I managed to get to the point where the last piece was missing (by only calculating in my head), but got stuck then, as I failed to see that you can solve it by proportion via constructing a „Thales triangle“. Once you realize that, you can also calculate the pretty solution in your head (18/6=6/2!). The radius of the large semi circle is 10 and thus the area is 50pi.

[rey-dqnx]

QC=6
QP=10
so CP=8
Let CO=x
Draw a tangent line from O to D
then DO=CO=x
From triangle ODP
x²+4² =(8-x)²
x²+16=64-16x+x²
16x=48
x=3


OP=CP-CO
OP=8-3=5
Therefore
OB=OP+BP
OB=5+4=9
OB=9=radius of big semi circle
where did I go wrong, professor or anyone ?

[s]

R = 5+4 = 9
4= r of the yellow, 5 = po

[santiagoarosam]

Los radios verde y amarillo son 6 y 4---> CP²=(6+4)²-6²=64---> CP=8. --->. Potencia de F =6²=(6+8+4)*AF---> AF=2---> AO=(2+6+8+4)/2=10---> Área semicírculo grande =10²π/2=50π.
Gracias y saludos.

[LuisdeBritoCamacho]

The Answer is that Big Semicircle Area is equal to 50Pi Square Cm.

01) Big Diameter Diameter (D) : D = 4 + 4 + 6 + 6 = 8 + 12 ; D = 20

02) Moving the Green Circle to left until Line AE is orthogonal to AB reduces 2 cm, that is the projected intersection of Green and Yellow Circle, in Line AB.

03) CP = 8. CP^2 = 10^2 - 6^2 ; CP^2 = 100 - 36 ; CP^2 = 64 ; CP = 8.

04) Moving Point C, 2 cm to the Left there will be no Diagonal Tangent. Now the two Circles are Tangent in a Vertical Line.

05) So : The Big Circle Diameter must be equal to the Sum of the Green Diameter and the Yellow Diameter. D = 12 + 8 = 20 cm

06) So : The Big Circle Radius = 10 cm

07) Semicircle Area = 100*Pi / 2 = 50Pi sq cm

ANSWER:

Big Semicircle area equal 50Pi Square Centimeters.

[Birol]

Second way of solution ▶
radius of the largest semicircle: R
radius of the small semi circle: r₁
radius of the middle semicircle: r₂

By making an accurate drawing, I found out that the line [PQ] is parallel to [EO] (also equal: = R) ❗

[PQ] // [EO]

A₁= π*r₁²/2
A₁= 8π cm²
⇒
8π= πr₁²/2
16= r₁²
r₁= 4 cm

A₂= π*r₂²/2
A₂= 18π cm²
⇒
18π= πr₂²/2
36= r₂²
r₂= 6 cm

Let's consider the right triangle ΔCPQ
[QC]= r₂= 6 cm
[PQ]= r₁ + r₂
[PQ]= 4+6
[PQ]= 10 cm

By applying the Pythagorean theorem:
[QC]²+[CP]²= [PQ]²
6²+[CP]²= 10²
⇒
[CP]= 8 cm

If we move the [QP] to the left side of r₂ cm, we get the line [EO]
⇒
[PO] would be equall to [QE]
[PO]= [QE]
[PO]= 6 cm❗
⇒
[CO]= 8-6
[CO]= 2 cm

Let's consider the right triangle ΔOSE
S is a point on the radius r₂ so that [ES] ⊥ [SO]
By applying the Pythagorean theorem we get:
[OS]²+[SE]²= [EO]²
[OS]= 6 cm
[SE]= [EQ]+[QS]
[QS]= [CO]
[QS]= 2 cm
[SE]= 6+2
[SE]= 8 cm
⇒
6²+8²= R²
R= 10 cm

Alarge-semicircle= π*10²/2
Alarge-semicircle= 50π
Alarge-semicircle≈ 157, 08 cm²

[quigonkenny]

Let Rɢ be the radius of green semicircle Q, Rʏ the radius of yellow semicircle P, and R the radius of big semicircle O.

Green semicircle Q:
Aɢ = πRɢ²/2
18π = πRɢ²/2
Rɢ² = 18(2) = 36
Rɢ = √36 = 6

Yellow semicircle P:
Aʏ = πRʏ²/2
8π = πRʏ²/2
Rʏ² = 8(2) = 16
Rʏ = √16 = 4

As the centers of two circles are collinear with their point of tangency, D is on PQ. As PD = Rʏ = 4 and QD = Rɢ = 6, PQ = 6+4 = 10.

Triangle ∆QCP:
QC² + PC² = PQ²
6² + PC² = 10²
PC² = 100 - 36 = 64
PC = √64 = 8

Drop a perpendicular from E to AB at M. As EQ is parallel to AB and QC is perpendicular to AB as C is the tangent point between AB and semicircle Q, EM = QC = 6 and CM = QE = 6.

MB = CM + PC + PB = 6 + 8 + 4 = 18

Triangle ∆EMO:
EM² + OM² = OE²
6² + (18-R)² = R²
36 + 324 - 36R + R² = R²
36R = 360
R = 10

Big semicircle O:
Aꜱ = πR²/2 = π10²/2 = 100π/2
Aꜱ = 50π sq units ≈ 157.08 sq units

[Birol]

My way of solution ▶
1) first way:
radius of the largest semicircle: R
radius of the small semi circle: r₁
radius of the middle semicircle: r₂

A₁= π*r₁²/2
A₁= 8π cm²
⇒
8π= πr₁²/2
16= r₁²
r₁= 4 cm

A₂= π*r₂²/2
A₂= 18π cm²
⇒
18π= πr₂²/2
36= r₂²
r₂= 6 cm

Let's consider the right triangle ΔCPQ
[QC]= r₂= 6 cm
[PQ]= r₁ + r₂
[PQ]= 4+6
[PQ]= 10 cm

By applying the Pythagorean theorem:
[QC]²+[CP]²= [PQ]²
6²+[CP]²= 10²
⇒
[CP]= 8 cm

Let's say:
[CO]= x
[OP]= 8-x
⇒
[OP]+[PB]= R
8-x+r₁ = R
8-x+4= R
12-x=

Let's consider the right triangle ΔOSE
S is a point on the radius r₂ so that [ES] ⊥ [SO]
By applying the Pythagorean theorem we get:
[OS]²+[SE]²= [EO]²
[OS]= r₂ = 6 cm
[SE]= x+r₂ = 6+x
[EO]= R
⇒
R²= 6²+(6+x)²
R²= 36+36+12x+x²
R²= 72+12x+x²

R=
⇒
(12-x)²= 72+12x+x²
144-24x+x²= 72+12x+x²
72= 36x
x= 2 cm
⇒
[CO]= 2 cm

R= 12-x
R= 10 cm

Alarge-semicircle= πR²/2
R= 10 cm
⇒
Alarge-semicircle= π*10²/2
Alarge-semicircle= 50π
Alarge-semicircle≈ 157, 08 cm²

[michaeldoerr]

The radius of the larger circle is 10 units. Also at the 9:25, I think that I have noticed a fact regarding AA similarity and right angles: apart from aides different sides being similar, there also has to be a comparison between different triangles that are are formed. The first triangle is formed by beta-alpha and the second triangle is formed by alpha beta. I am also wondering the way that I actually did that is just good as yours. I set the original sides as AF/EF=FE/FB. I am hoping that my construction is too far off.