At 6:00, the algebra is more straightforward if we observe that ΔABC and ΔCDO are similar. The equation OE/AE = CD/OD gets replaced by BC/AB = CD/OD. We substitute BC = 3, AD = 4, CD = 3 - r and OD = r to get 3/4 = (3 - r)/r and cross multiply: 3r = 4(3 - r), 3r = 12 - 4r, 7r = 12 and r = 12/7. Area of semicircle = (1/2)πr² = (1/2)π(12/7)² = 72π/49, as PreMath also found.
Once we've found that AB = 4 (at 2:18), we can consider triangles AOB and BOC OE is perpendicular to AB, therefore △AOB has base AB = 4 and height OE = r OD is perpendicular to BC, therefore △BOC has base BC = 3 and height OD = r Area(△AOB) + Area(△BOC) = Area(△ABC) (1/2 * 4 * r) + (1/2 * 3 * r) = 6 7r/2 = 6 r = 12/7 Area(semi-circle) = 1/2 πr² = 1/2 π(144/49) = *72π/49*
Triangle ∆ABC: A = bh/2 = AB(BC)/2 6 = AB(3)/2 3AB = 2(6) = 12 AB = 12/3 = 4 Draw radii OD and OE. As AB and BC are tangent to semicircle O at E and D respectively, then ∠OEB = ∠BDO = 90°. Since ∠EBD = 90°, then ∠DOE must equal 90° as well, and OEBD is a swuare with side length OD = OE = r. As ∠ODC = ∠ABC = 90° and ∠C is common, ∆ODC and ∆ABC are similar triangles. As ∠AEO = ∠ABC = 90° and ∠A is common, ∆AEO is also similar to ∆ABC, and thus also similar to ∆ODC. OE/AE = DC/OD = BC/AB r/AE = DC/r = 3/4 3AE = 4r => AE = 4r/3 3r = 4DC => DC = 3r/4 AE + EB = AB 4r/3 + r = 4 7r/3 = 4 r = 4(3/7) = 12/7 Semicircle O: A = πr²/2 = π(12/7)²/2 = (144π/49)/2 [ A = 72π/49 ≈ 4.616 sq units ]
Something else: Let's use an orthonormal center B and first axis (BA). We have B(0; 0) A(4; 0) C(0; 3), as AB = (2.area of ABC)/BC = 12/3 = 4. VectorAC(-4; 3), then A parametric equation of (AC) is x = 4 -4.k; y = 3.k. So O(4 -4.k; 3.k) for a certain real k. Distance from O to (BC) = OD = abscissa of O = 4 -4.k; and distance from O to (AB) = OE = ordinate of O = 3.k. These two distances are equal (to the radius R of the circle), so 4 -4.k = 3.k and so k = 4/7, and the radius of the circle is R = 4 -4.(4/7) = 3.(4/7) = 12/7. Finally, the green area is then Pi. (R^2).(1/2) = (72/49).Pi
AB = 4, making AC = 5 due to 3,4,5. ODBE is a square with sides r. CDO is similar to ABC. Rough estimate for r is 2 or a little under. (3-r)/r = 3/4 12-4r = 3r 12-7r=0. r=12/7 Full circle area is (144/49)pi, so semicircle is (72/49)pi 4.62
,👍 (1/2)(3 AB) = 6 AB = 4 we take reflection of triangle in AC , we obtain a kite with area 12 and semi perimeter 7 r = area/semiperimter =12/7 semi circle area = (π/2)(144/49) = 72 π/49
triangle area = 6 height = 3 6=1/2 * B * 3 B=4 let line be y=3/4 * x + b y intercept = (0,3) x intercept = (4,0) sub (0,3) -> y=3/4x + b 3=3/4 * (0) + b b=3 y=(3/4)x+3 let equation of circle = (x+h)^2+(y-k)^2=(r)^2 because the circle is located in the 2nd quadrant h=k=r because the horizontal and vertical radii are equal (x+r)^2+(y-r)^2=(r)^2 for y=(3/4)x+3, let -x=y; this is the point found in the line which is the center of the circle -x=(3/4)x+3=-12/7 take the absolute = 12/7 this is the radius of the circle r= 12/7 (x+12/7)^2+(y-12/7)^2=(12/7)^2 Area of the circle = 144/49 * pi Area of the semicircle = 72/49 * pi
Solution: A ∆ABC = ½ b h 6 = ½ b 3 3b = 12 b = 4 Proportion ∆ ABC and ∆ ODC (the triangles are similar) 4/3 = r/3-r 12 - 4r = 3r 12 = 7r r = 12/7 Green Semicircle Area (GSA) = ½ π r² GSA = ½ π (12/7)² GSA = ½ π 144/49 GSA = 72π/49 Square Units ✅ GSA ≈ 4.6162 Square Units ✅
1/ AB= 4 2/ Connect OB Consider the two triangle COB and AOB: 1/2 (3r+4r) =6 --> 7r/2= 6 -> r=12/7 Area of the semi green circle= pi/2 x sq(12/7) = 72pi/49=4.62 sq units😅😅😅
I came back. A = (bh)/2 6 = (3b)/2 3b = 12 b = 4 So, AB = 4. Draw radii DO & EO. Label DO = EO = r. By the Tangent Line to Circle Theorem, ∠BDO & ∠BEO are right angles. By the Two-Tangent Theorem, BD = BE. So, BDOE is a square. BD = BE = r. So, CD = BC - BD = 3 - r & AE = AB - BE = 4 - r. Label α & β as complementary angles. Let m∠A = α. Then, m∠AOE = m∠C = β. Therefore, m∠COD = α. So, △ABC ~ △AEO ~ △CDO by AA. By definition of similarity, (4 - r)/r = r/(3 - r). (4 - r) = r²/(3 - r) (4 - r)(3 - r) = r² r² = (4 - r)(3 - r) r² = 12 - 3r - 4r + r² r² = r² - 7r + 12 0 = -7r + 12 7r = 12 r = 12/7 Find the area of the semicircle. A = (πr²)/2 = [π(12/7)²]/2 = [π(144/49)]/2 = (144π)/98 = (72π)/49 So, the area of the green shaded semicircle is (72π)/49 square units (exact), or about 4.62 square units (approximation).
Let's find the area: . .. ... .... ..... From the given area of the right triangle ABC and the given side length BC we can conclude: A(ABC) = (1/2)*AB*BC ⇒ AB = 2*A(ABC)/BC = 2*6/3 = 4 The right triangles AEO and CDO are similar (∠AEO=∠CDO=90° and ∠EAO=∠COD). With r being the radius of the semicircle we obtain: EO/AE = CD/DO EO/(AB − BE) = (BC − BD)/DO EO/(AB − DO) = (BC − EO)/DO r/(4 − r) = (3 − r)/r r² = (3 − r)(4 − r) r² = 12 − 7r + r² 0 = 12 − 7r ⇒ r = 12/7 Now we are able to calculate the area of the green semicircle: A(semicircle) = πr²/2 = π*(12/7)²/2 = π*(144/49)/2 = (72/49)π ≈ 4.616 Best regards from Germany
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At 6:00, the algebra is more straightforward if we observe that ΔABC and ΔCDO are similar. The equation OE/AE = CD/OD gets replaced by BC/AB = CD/OD. We substitute BC = 3, AD = 4, CD = 3 - r and OD = r to get 3/4 = (3 - r)/r and cross multiply: 3r = 4(3 - r), 3r = 12 - 4r, 7r = 12 and r = 12/7. Area of semicircle = (1/2)πr² = (1/2)π(12/7)² = 72π/49, as PreMath also found.
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Once we've found that AB = 4 (at 2:18), we can consider triangles AOB and BOC
OE is perpendicular to AB, therefore △AOB has base AB = 4 and height OE = r
OD is perpendicular to BC, therefore △BOC has base BC = 3 and height OD = r
Area(△AOB) + Area(△BOC) = Area(△ABC)
(1/2 * 4 * r) + (1/2 * 3 * r) = 6
7r/2 = 6
r = 12/7
Area(semi-circle) = 1/2 πr² = 1/2 π(144/49) = *72π/49*
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A = 6 cm² = ½bh (given data)
b = 2A/h = 2A/3 = 4 cm
c² = b² + h² --> c = 5 cm
(4-r)/cosα + (3-r)/sinα = 5
5/4 (4-r)+ 5/3 (3-r) = 5
5 - 5/4 r + 5 - 5/3 r = 5
5/4 r + 5/3 r = 5
¼ r + ⅓ r = 1
7r = 12 --> r = 12/7 cm
A = ½πr² = 4,616 cm² ( Solved √ )
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3*AB/2=6 AB=4 OE=EB=r AE=4-r
r/(4-r)=3/4 4r=3(4-r) 7r=12 r=12/7
Green Semicircle area = 12/7*12/7*π*1/2=72π/49
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Triangle ∆ABC:
A = bh/2 = AB(BC)/2
6 = AB(3)/2
3AB = 2(6) = 12
AB = 12/3 = 4
Draw radii OD and OE. As AB and BC are tangent to semicircle O at E and D respectively, then ∠OEB = ∠BDO = 90°. Since ∠EBD = 90°, then ∠DOE must equal 90° as well, and OEBD is a swuare with side length OD = OE = r.
As ∠ODC = ∠ABC = 90° and ∠C is common, ∆ODC and ∆ABC are similar triangles. As ∠AEO = ∠ABC = 90° and ∠A is common, ∆AEO is also similar to ∆ABC, and thus also similar to ∆ODC.
OE/AE = DC/OD = BC/AB
r/AE = DC/r = 3/4
3AE = 4r => AE = 4r/3
3r = 4DC => DC = 3r/4
AE + EB = AB
4r/3 + r = 4
7r/3 = 4
r = 4(3/7) = 12/7
Semicircle O:
A = πr²/2 = π(12/7)²/2 = (144π/49)/2
[ A = 72π/49 ≈ 4.616 sq units ]
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Something else:
Let's use an orthonormal center B and first axis (BA). We have B(0; 0) A(4; 0) C(0; 3), as AB = (2.area of ABC)/BC = 12/3 = 4.
VectorAC(-4; 3), then A parametric equation of (AC) is x = 4 -4.k; y = 3.k. So O(4 -4.k; 3.k) for a certain real k.
Distance from O to (BC) = OD = abscissa of O = 4 -4.k; and distance from O to (AB) = OE = ordinate of O = 3.k.
These two distances are equal (to the radius R of the circle), so 4 -4.k = 3.k and so k = 4/7, and the radius of the circle is R = 4 -4.(4/7) = 3.(4/7) = 12/7.
Finally, the green area is then Pi. (R^2).(1/2) = (72/49).Pi
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Треугольник египетский
(3-r)/3=r/4; r= 12/7
S= πr^2/2=π(12/7)^2/2= 4,62
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AB = 4, making AC = 5 due to 3,4,5.
ODBE is a square with sides r.
CDO is similar to ABC.
Rough estimate for r is 2 or a little under.
(3-r)/r = 3/4
12-4r = 3r
12-7r=0.
r=12/7
Full circle area is (144/49)pi, so semicircle is (72/49)pi
4.62
Green semicircle area=1/2(12/7)^2(π)=72π/49=4.62 square units.❤❤❤
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,👍
(1/2)(3 AB) = 6
AB = 4
we take reflection of triangle in AC , we obtain a kite with area 12 and semi perimeter 7
r = area/semiperimter
=12/7
semi circle area = (π/2)(144/49) = 72 π/49
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triangle area = 6
height = 3
6=1/2 * B * 3
B=4
let line be y=3/4 * x + b
y intercept = (0,3)
x intercept = (4,0)
sub (0,3) -> y=3/4x + b
3=3/4 * (0) + b
b=3
y=(3/4)x+3
let equation of circle = (x+h)^2+(y-k)^2=(r)^2 because the circle is located in the 2nd quadrant
h=k=r because the horizontal and vertical radii are equal
(x+r)^2+(y-r)^2=(r)^2
for y=(3/4)x+3, let -x=y; this is the point found in the line which is the center of the circle
-x=(3/4)x+3=-12/7
take the absolute = 12/7
this is the radius of the circle
r= 12/7
(x+12/7)^2+(y-12/7)^2=(12/7)^2
Area of the circle = 144/49 * pi
Area of the semicircle = 72/49 * pi
The triangle area = 6 tells us that it's a 3 - 4 - 5 triangle
3 = 3r/4 + r = 7r/4
r = 12/7
The area of the half circle = (12/7)^2 x π ÷ 2 = 72π/49
Solution:
A ∆ABC = ½ b h
6 = ½ b 3
3b = 12
b = 4
Proportion ∆ ABC and ∆ ODC (the triangles are similar)
4/3 = r/3-r
12 - 4r = 3r
12 = 7r
r = 12/7
Green Semicircle Area (GSA) = ½ π r²
GSA = ½ π (12/7)²
GSA = ½ π 144/49
GSA = 72π/49 Square Units ✅
GSA ≈ 4.6162 Square Units ✅
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1/ AB= 4
2/ Connect OB
Consider the two triangle COB and AOB:
1/2 (3r+4r) =6
--> 7r/2= 6
-> r=12/7
Area of the semi green circle= pi/2 x sq(12/7) = 72pi/49=4.62 sq units😅😅😅
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I came back.
A = (bh)/2
6 = (3b)/2
3b = 12
b = 4
So, AB = 4.
Draw radii DO & EO. Label DO = EO = r.
By the Tangent Line to Circle Theorem, ∠BDO & ∠BEO are right angles.
By the Two-Tangent Theorem, BD = BE.
So, BDOE is a square. BD = BE = r.
So, CD = BC - BD = 3 - r & AE = AB - BE = 4 - r.
Label α & β as complementary angles.
Let m∠A = α. Then, m∠AOE = m∠C = β.
Therefore, m∠COD = α.
So, △ABC ~ △AEO ~ △CDO by AA.
By definition of similarity, (4 - r)/r = r/(3 - r).
(4 - r) = r²/(3 - r)
(4 - r)(3 - r) = r²
r² = (4 - r)(3 - r)
r² = 12 - 3r - 4r + r²
r² = r² - 7r + 12
0 = -7r + 12
7r = 12
r = 12/7
Find the area of the semicircle.
A = (πr²)/2
= [π(12/7)²]/2
= [π(144/49)]/2
= (144π)/98
= (72π)/49
So, the area of the green shaded semicircle is (72π)/49 square units (exact), or about 4.62 square units (approximation).
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My way of solution ▶
A(ΔABC)= [AB]*[BC]/2
A(ΔABC)= 6 square units
[BC]= 3
⇒
6= 3*[AB]/2
⇒
[AB]= 4 length units
b) By considering the green semicircle we can write:
[OD]= [OE]= r
∠COD= ∠CAB
∠ODC= ∠ABC
∠DCO= ∠BCA
⇒
ΔODC ~ ΔABC
⇒
[OD]/[AB] = [DC]/[BC] = [CO]/[CA]
[DB]=[OE]= r
⇒
[DC]= 3-r
⇒
r/4= (3-r)/3
3r= 12-4r
7r= 12
r= 12/7
Agreen= πr²/2
Agreen= π*(12/7)²*(1/2)
Agreen= (72/49)π
Agreen ≈ 4,62 square units ✅
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S=72π/49≈4,62
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STEP-BY-STEP RESOLUTION PROPOSAL :
01) Triangle Area (TA) = 12 / 2 ; TA = (B * h) / 2 ; 6 = (B * 3) / 2 ; B * 3 = 12 ; B = 12 / 3 ; B = 4
02) Triangle [ABC] with Sides (3 ; 4 ; 5)
03) OE = OD = EB = BD = R
04) [BDOE] is a Square with Area = R^2
05) AE = (4 - R)
06) CD = (3 - R)
07) CD / R = R / AE
08) (3 - R) / R = R / (4 - R)
09) (3 - R) * (4 - R) = R^2
10) 12 - 3R - 4R + R^2 = R^2
11) 12 - 7R = 0
12) 7R = 12
13) R = 12 / 7
14) Green Semicircle Area (GSA) = (Pi * R^2) / 2
15) GSA = (Pi * (12/7)^2) / 2
16) GSA = 144Pi / 98
17) GSA = 72Pi/49
18) GSA ~ 4,62
Thus,
OUR BEST ANSWER :
Green Semicircle Area equal to 72Pi/49 Square Units.
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Is there another way if dont use similar trianngle
Yes, use trigonometry.
Tan(alpha) = r/(4-r) =3/4
That gives r = 12/7
*SIMPLE Solution:*
AB×BC/2 = [ABC]→AB×3/2 =6 → AB=4.
[OBA] + [BOC] = [ABC]
r×AB/2 + r×BC/2 = 6
4r + 3r = 12 → *r=12/7*
Therefore, the area S of the semicircle is given by:
S = πr²/2 → *S = 72π/49 square units*
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Let's find the area:
.
..
...
....
.....
From the given area of the right triangle ABC and the given side length BC we can conclude:
A(ABC) = (1/2)*AB*BC ⇒ AB = 2*A(ABC)/BC = 2*6/3 = 4
The right triangles AEO and CDO are similar (∠AEO=∠CDO=90° and ∠EAO=∠COD). With r being the radius of the semicircle we obtain:
EO/AE = CD/DO
EO/(AB − BE) = (BC − BD)/DO
EO/(AB − DO) = (BC − EO)/DO
r/(4 − r) = (3 − r)/r
r² = (3 − r)(4 − r)
r² = 12 − 7r + r²
0 = 12 − 7r
⇒ r = 12/7
Now we are able to calculate the area of the green semicircle:
A(semicircle) = πr²/2 = π*(12/7)²/2 = π*(144/49)/2 = (72/49)π ≈ 4.616
Best regards from Germany
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Are∆ = (1/2)•4•r + (1/2)•3•r = 6
7•r = 12 ➡️ r = 12/7
No Need Square Equation. 😉
Complicated.
Semicircle = (1/2)•π•r² = (0.5)•(3.1416)•(12/7)² = 4.616