Hi professor that was very nice solution . but i suggest other way Draw radius of circle is perpendicular to chord DC In point H and it will be perpendicular to AB in point K as well then HD=HC=DC/2=12/2=6 and KA=KB=AB/2 = 6/2 =3 Let assume OH=x then Ok=4+x with using Pythagoras theorem in Triangle OHC : x*x+36=R*R using Pythagoras theorem in Triangle OKB :(x+4)(x+4)+9=R*R x^2+36= x^2+8*x +16+9 => 8*x=11 then x=11/8 x^2+36=R^2 and x=11/8 => R*R =(2425/64) semi circle area = Pi*R*R/2= (2425*Pi)/128
@@jarikosonen4079 ??? is correct. You don't need to solve for r at all, just r². And your a was the height from origin to the top of the rhombus, mine was to the base.
That solution is WAY too complicated. Here's a simpler method: After determining h=4, I made 2 right triangles: , O, B And , O, B Enter Mr Pythagoras: (3)²+(4+X)²=R² (6)²+X²=R² (Where X=distance from O to center of DC) Solve for X: X=9/8 Plug this into second Pythagorean formula: (6)²+(9/8)²=R² ... R=6.1
Completely baffled by introducing "2r" on the right hand side of the sine law equation at 6:46. Where does this come from? I do not see the basis or relationship? Can someone please explain?
I went through a completely different route using the formula for chord length using perpendicular distance from the center and the radius: Chord Length = 2 × √(r2 − d2) where "d" is the perpendicular distance and "r" the radius. We know both chords. We don't know d, but we know from the height of the trapezium that one is 4 cm longer than the other. Therefore, we can do simultaneous equations. - For chord CD we have 12 = 2√(r²-d²) - For chord AB we have: 6 = 2√(r²-(d+4)²) Carefully solve it to find d (perpendicular distance of chord CD from center) is 11/8 cm. Then I used pythagoras to find r
1/ h=4 cm 2/ From the center O, draw OH perpendicular to chord CD intersecting AB at point K. Because AB //CD-> OH is perpendicular to both AB and CD at the midpoint of the two chords. Label OH= x and R= radius Focus on the triangles OKB and OHC SqR=sq3+sq(4+x) =sqx+8x+25 (1) And sqR=sqx+36 (2) (1)-(2) -> x=11/8 -> sqR=sq(11/8) + 36=2425/64 Area of the semicircle=1/2 .pi.2425/64=2425/128 x pi😅
Trapezoid ABCD: Aᴛ = h(a+b)/2 36 = h(6+12)/2 = 9h h = 36/9 = 4 Draw OM, where M is the midpoint of AB. As AB is a chord, OM and AB are perpendicular. As AB and CD are parallel, OM also bisects CD perpendicularly. Let N be the intersection point of CD and OM. Let ON = x. Triangle ∆CNO: CN² + ON² = OC² 6² + x² = r² x² + 36 = r² --- [1] Triangle ∆BMO: BM² + OM² = OB² 3² + (x+4)² = r² 9 + x² + 8x + 16 = r² x² + 8x + 25 = r² --- [2] x² + 8x + 25 = x² + 36
You don’t need law of sines Draw triangle DOA. Angle DOA is DOUBLE angle DCA (inscribed angle vs central angle) Use law of cosines with angle DOA whose sides are OD,OA, and DA OD=OA= radius BOOM
Trapezoid area=1/2(6+12)(h)=36 So h=4cm Connect O to E and F that E middle CD and F middle AB Let OE=x In ∆ AOF AF^2+OF^2==R^2 AF=x+h=x+4 3^2+(x+4)^2=R^2 9+16+8x+x^2=R^2 x^2+8x+25=R^2 (1) connect O to D in ∆ OED OE^2+DE^2=OD^2 x^2+6^2=R^2 R^2-x^2=36 (2) (1) R^2-x^2=8x+25 (1)&(2) 8x+25=36 8x=36-25=11 So x=11/8cm (2) R^2-(11/8)^2=36 R^2=(36+121/64) R^2=2425/64 cm So semicircle area=1/2(π)(2425/64=2425π/128cm^2=59.52cm^2.❤❤❤ thanks sir.
Solution: Area of Trapezoid (AT) = ½ h (a + b) 36 = ½ h (12 + 6) 36 = 9 h h = 4 By The Chords Theorem, we have 4 . x = 3 . 9 x = 27/4 4 + 27/4 = 43/4 So half of 43/4 is 43/8 3² + (43/8)² = r² 9 + 1849/64 = r² r² = 2425/64 A = πr²/2 A = 2425π/128 cm² Or A ~= 59,518 cm²
The mistake comes with using a/Sina = 2r. This would only work if 2r (ie the diameter) was one of the sides of the triangle. Then you would have a triangle inscribed in a semicircle with a right angle at the circumference giving sin 90 =1. In this problem this is not the case, so cannot be used. I agreed with Mega’s solution.
Nice! I solved it differently after obtaining the value of the height, by assuming a displacement (d) from DC to the base of the semicircle, and then applied the Pythagorean theorem to points B and C, where the hypotenuse in each case is the radius of the circle. Equating the two expressions yield the displacement (d), and from any of the expressions the radius can be found, hence the area of the semicircle.
We use an orthonormal center P (middle of [D,C]) and first axis (PC). The height of the trapezoïd is h and ((12 +6)/2).h = 36, so h = 4, and we have C(6;0) D(-6;0) B(3;4) and A(-3;4) The equation of the circle is x^2 + y^2 + a.x + b.y + c = 0 C is on the circle, so 36 + 6.a + c = 0 D is on the circle, so 36 - 6.a + c = 0 These equations give that a = 0 and c = -36 The equation of the circle is now w^2 + y^2 + b.y - 36 = 0 B is on the circle, so 9 + 16 +4.b - 36 = 0, giving b = 11/4 The equation of the circle is x^2 + y^2 +(11/4).y - 36 = 0 or x^2 + (y +(11/8))^2 = 36 + (11/8)^2 = 2425/64 If R is the radius of the circle, then R^2 = 2425/64 Finally the area of the semi circle is (Pi/2).(R^2) = (2425/128).Pi
Distancia vertical entre las dos cuerdas: [(12+6)/2]h=9h=36→ h=4 → Llamamos "P" a la proyección ortogonal de A sobre DC, y "Q" la proyección sobre el diámetro horizontal → DC=DP+PC=3+9→ Potencia de Prespecto a la circunferencia =3*9=27=4(4+2PQ)→ PQ=11/8→ Si M es el punto medio de AB→ MO=4+(11/8)=43/8→ Potencia de M respecto a la circunferencia =3²=[r-(43/8)][r+(43/8)]→ r=5√97/8→ Área semicírculo =πr²/2=2425π/128 =59,51845... ud². Gracias y saludos.
Draw perpendicular bisector through CD and AB. Set up two Pythagorean Theorem relationships using the radius to the points B and C where an unknown value of 'a' is ascribed to the perpendicular distance between O and the 12 cm segment. On Pythagorean relationship, 3^2 + (4 + a)^2 = r^2 which simplifies to 25 + 8a + a^2 = r^2. The other Pythagorean is 6^2 + a^2 = r^2 which simplifies to 36 + a^2 = r^2. Make these two simplified equations equal given they are both in terms of r^2 to solve for a. 25 + 8a + a^2 = 36 + a^2 where a^2 cancels and a = 11/8. Plug a into the second Pythagorean relationship, 36 + (11/8)^2 = r^2. Solving for r^2 = 2425/64. Area of semicircle is (pi * r^2)/2. Plugging in r^2 gives 2425pi/128.
The semicircle area is 2425pi. Also at the 2:50 mark I have noticed that focus on delta AEC is 16 plus 81. Is that because we halved 6??? I just want to make sure.
STEP-BY-STEP RESOLUTION PROPOSAL : 01) Let the Vertical Distance between Point O and Line DC equal "X" cm 02) OC^2 = X^2 + 6^2 ; R^2 = (X^2 + 36) 03) The Vertical Distance between Point O and Line AB = (X + 4) cm 04) (X + 4)^2 + 3^2 = R^2 05) (X + 4)^2 + 3^2 = X^2 + 36 06) X^2 + 8X + 16 + 9 = X^2 + 36 07) 8X + 25 = 36 08) 8X = 36 - 25 09) 8X = 11 10) X = 11/8 cm 11) X = 1,375 cm 12) R^2 = X^2 + 36 13) R^2 = (121 / 64) + 36 ; R^2 = (121 + 2.304) / 64 ; R^2 = 2.425 / 64 14) Area = 2.425Pi / 128 sq cm 15) Area ~ 59,52 sq cm ANSWER : Semicircle Area equal to (2.425Pi/128) Square Cm ( ~ 59,52 Square Cm). P.S. - I jumped the height of the Trapezoid 'cause it's useless calculation and very simple, h = 4.
Let's find the area: . .. ... .... ..... First of all we calculate the height h of the trapezoid: A(ABCD) = (1/2)*(AB + CD)*h ⇒ h = 2*A(ABCD)/(AB + CD) = 2*(36cm²)/(12cm + 6cm) = 2*(36cm²)/(18cm) = 4cm Now let M and N be the midpoints of AB and CD, respectively. Then the triangles OAM and ODN are right triangles and we can apply the Pythagorean theorem. With r being the radius of the semicircle we obtain: OA² = OM² + AM² OD² = ON² + DM² r² = (ON + h)² + (AB/2)² r² = ON² + (CD/2)² (ON + h)² + (AB/2)² = ON² + (CD/2)² [ON + (4cm)]² + (6cm/2)² = ON² + (12cm/2)² [ON + (4cm)]² + (3cm)² = ON² + (6cm)² ON² + (8cm)*ON + 16cm² + 9cm² = ON² + 36cm² (8cm)*ON = 11cm² ⇒ ON = (11/8)cm r² = (ON + h)² + (AB/2)² = [(11/8)cm + 4cm]² + (6cm/2)² = [(11/8)cm + (32/8)cm]² + (3cm)² = [(43/8)cm]² + [(24/8)cm]² = (1849/64)cm² + (576/64)cm² = (2425/64)cm² r² = [(11/8)cm]² + (12cm/2)² = [(11/8)cm]² + [(48/8)cm]² = (121/64)cm² + (2304/64)cm² = (2425/64)cm² ✓ Now we are able to calculate the area of the semicircle: A = πr²/2 = (2425π/128)cm² ≈ 59.52cm² Best regards from Germany
Not only can I not find the area but I don't want to. Absolutely useless. I will never ever want to calculate the area of a semi-circle as long as I live.
Thanks Sir
That’s very nice
❤❤❤❤❤
You are very welcome!😀
Thanks for the feedback ❤️
Hi professor that was very nice solution . but i suggest other way
Draw radius of circle is perpendicular to chord DC In point H and it will be perpendicular to AB in point K as well
then HD=HC=DC/2=12/2=6 and KA=KB=AB/2 = 6/2 =3
Let assume OH=x then Ok=4+x
with using Pythagoras theorem in Triangle OHC : x*x+36=R*R
using Pythagoras theorem in Triangle OKB :(x+4)(x+4)+9=R*R
x^2+36= x^2+8*x +16+9 => 8*x=11 then x=11/8
x^2+36=R^2 and x=11/8 => R*R =(2425/64)
semi circle area = Pi*R*R/2= (2425*Pi)/128
Excellent!
Thanks for sharing ❤️
I did same way. Much better
I used chords to get 6*6=r²-a² and 3*3=r²-(a+4)². Solve to get a=11/8 and r²=36+121/64, then solve for area from r². Edit: Corrected equations sorry
a=43/8, r=5*sqrt(97)/8, r^2=18+121/128 ???
@@jarikosonen4079 ??? is correct. You don't need to solve for r at all, just r². And your a was the height from origin to the top of the rhombus, mine was to the base.
Thanks for sharing ❤️
triangle DAC leads to the radius of the semi-circle? Afraid that's where you lost me
I'll prove the extended sine rule pretty soon.
Thanks for asking❤️
I could not work that one out either. All good up until then.
That solution is WAY too complicated. Here's a simpler method:
After determining h=4, I made 2 right triangles:
, O, B
And
, O, B
Enter Mr Pythagoras:
(3)²+(4+X)²=R²
(6)²+X²=R²
(Where X=distance from O to center of DC)
Solve for X:
X=9/8
Plug this into second Pythagorean formula:
(6)²+(9/8)²=R²
...
R=6.1
Completely baffled by introducing "2r" on the right hand side of the sine law equation at 6:46. Where does this come from? I do not see the basis or relationship? Can someone please explain?
😂,. Strange method 😂😂
I went through a completely different route using the formula for chord length using perpendicular distance from the center and the radius: Chord Length = 2 × √(r2 − d2) where "d" is the perpendicular distance and "r" the radius.
We know both chords. We don't know d, but we know from the height of the trapezium that one is 4 cm longer than the other. Therefore, we can do simultaneous equations.
- For chord CD we have 12 = 2√(r²-d²)
- For chord AB we have: 6 = 2√(r²-(d+4)²)
Carefully solve it to find d (perpendicular distance of chord CD from center) is 11/8 cm.
Then I used pythagoras to find r
1/ h=4 cm
2/ From the center O, draw OH perpendicular to chord CD intersecting AB at point K.
Because AB //CD-> OH is perpendicular to both AB and CD at the midpoint of the two chords.
Label OH= x and R= radius
Focus on the triangles OKB and OHC
SqR=sq3+sq(4+x) =sqx+8x+25 (1)
And sqR=sqx+36 (2)
(1)-(2) -> x=11/8
-> sqR=sq(11/8) + 36=2425/64
Area of the semicircle=1/2 .pi.2425/64=2425/128 x pi😅
Excellent!
Thanks for sharing ❤️
Trapezoid ABCD:
Aᴛ = h(a+b)/2
36 = h(6+12)/2 = 9h
h = 36/9 = 4
Draw OM, where M is the midpoint of AB. As AB is a chord, OM and AB are perpendicular. As AB and CD are parallel, OM also bisects CD perpendicularly. Let N be the intersection point of CD and OM. Let ON = x.
Triangle ∆CNO:
CN² + ON² = OC²
6² + x² = r²
x² + 36 = r² --- [1]
Triangle ∆BMO:
BM² + OM² = OB²
3² + (x+4)² = r²
9 + x² + 8x + 16 = r²
x² + 8x + 25 = r² --- [2]
x² + 8x + 25 = x² + 36
You don’t need law of sines
Draw triangle DOA.
Angle DOA is DOUBLE angle DCA (inscribed angle vs central angle)
Use law of cosines with angle DOA whose sides are OD,OA, and DA
OD=OA= radius
BOOM
Where does alpha and 2r come from ?
Thank you!
Trapezoid area=1/2(6+12)(h)=36
So h=4cm
Connect O to E and F that E middle CD and F middle AB
Let OE=x
In ∆ AOF
AF^2+OF^2==R^2
AF=x+h=x+4
3^2+(x+4)^2=R^2
9+16+8x+x^2=R^2
x^2+8x+25=R^2 (1)
connect O to D
in ∆ OED
OE^2+DE^2=OD^2
x^2+6^2=R^2
R^2-x^2=36 (2)
(1) R^2-x^2=8x+25
(1)&(2)
8x+25=36
8x=36-25=11
So x=11/8cm
(2) R^2-(11/8)^2=36
R^2=(36+121/64)
R^2=2425/64 cm
So semicircle area=1/2(π)(2425/64=2425π/128cm^2=59.52cm^2.❤❤❤ thanks sir.
Excellent!
You are very welcome!
Thanks for sharing ❤️
Nice, many thanks, Sir! ∆ ADC → AD = 5; CD = 12; AC = √97; DCA = δ →
25 = 144 + 97 - 2(12)(√97)cos(δ) → cos(δ) = 9√97/97 → sin(δ) = 4√97/97 →
cos(2δ) = cos^2(δ) - sin^2(δ) = 77/97 → ∆ ADO → AO = DO = r; DOA = 2δ
AD = 5 → 25 = 2r^2(1 - cos(2δ)) → r^2 = 25(97)/64 → semicircle area = 25π97/128
btw: This „another version of law of sines“ is a simple rearrangement of
law of cosines using the additional theorem:
ADC = α → AOC = 2α; AO = CO = r; AC = √97 = a →
a^2 = 2r^2(1 - cos(2α)) = 2r^2(1 - (cos^2(α) - sin^2(α))) = 4r^2sin^2(α) → 2r = a/sin(α)
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Solution:
Area of Trapezoid (AT) = ½ h (a + b)
36 = ½ h (12 + 6)
36 = 9 h
h = 4
By The Chords Theorem, we have
4 . x = 3 . 9
x = 27/4
4 + 27/4 = 43/4
So half of 43/4 is 43/8
3² + (43/8)² = r²
9 + 1849/64 = r²
r² = 2425/64
A = πr²/2
A = 2425π/128 cm²
Or
A ~= 59,518 cm²
Very nice solution.
The mistake comes with using a/Sina = 2r. This would only work if 2r (ie the diameter) was one of the sides of the triangle. Then you would have a triangle inscribed in a semicircle with a right angle at the circumference giving sin 90 =1. In this problem this is not the case, so cannot be used. I agreed with Mega’s solution.
¿Porqué a/sen x=2r ?
No lo entiendo.
Nice! I solved it differently after obtaining the value of the height, by assuming a displacement (d) from DC to the base of the semicircle, and then applied the Pythagorean theorem to points B and C, where the hypotenuse in each case is the radius of the circle. Equating the two expressions yield the displacement (d), and from any of the expressions the radius can be found, hence the area of the semicircle.
Excellent!
Thanks for sharing ❤️
We use an orthonormal center P (middle of [D,C]) and first axis (PC).
The height of the trapezoïd is h and ((12 +6)/2).h = 36,
so h = 4, and we have C(6;0) D(-6;0) B(3;4) and A(-3;4)
The equation of the circle is x^2 + y^2 + a.x + b.y + c = 0
C is on the circle, so 36 + 6.a + c = 0
D is on the circle, so 36 - 6.a + c = 0
These equations give that a = 0 and c = -36
The equation of the circle is now w^2 + y^2 + b.y - 36 = 0
B is on the circle, so 9 + 16 +4.b - 36 = 0, giving b = 11/4
The equation of the circle is x^2 + y^2 +(11/4).y - 36 = 0
or x^2 + (y +(11/8))^2 = 36 + (11/8)^2 = 2425/64
If R is the radius of the circle, then R^2 = 2425/64
Finally the area of the semi circle is (Pi/2).(R^2) = (2425/128).Pi
Excellent!
Thanks for sharing ❤️
teorema de cuerdas 3*9=4*X => X=27/4 => teorema de faure 4R²=a²+b²+c²+d²
Thanks for the feedback ❤️
h=2*36/(12+6)=4...r^2=6^2+x^2=3^2+(4+x)^2...11=8x..x=11/8..r^2=36+121/64=(2304+121)/64=2425/64
Excellent!
Thanks for sharing ❤️
Distancia vertical entre las dos cuerdas: [(12+6)/2]h=9h=36→ h=4 → Llamamos "P" a la proyección ortogonal de A sobre DC, y "Q" la proyección sobre el diámetro horizontal → DC=DP+PC=3+9→ Potencia de Prespecto a la circunferencia =3*9=27=4(4+2PQ)→ PQ=11/8→ Si M es el punto medio de AB→ MO=4+(11/8)=43/8→ Potencia de M respecto a la circunferencia =3²=[r-(43/8)][r+(43/8)]→ r=5√97/8→ Área semicírculo =πr²/2=2425π/128 =59,51845... ud².
Gracias y saludos.
Excellent!
Thanks for sharing ❤️
Draw perpendicular bisector through CD and AB. Set up two Pythagorean Theorem relationships using the radius to the points B and C where an unknown value of 'a' is ascribed to the perpendicular distance between O and the 12 cm segment. On Pythagorean relationship, 3^2 + (4 + a)^2 = r^2 which simplifies to 25 + 8a + a^2 = r^2. The other Pythagorean is 6^2 + a^2 = r^2 which simplifies to 36 + a^2 = r^2. Make these two simplified equations equal given they are both in terms of r^2 to solve for a. 25 + 8a + a^2 = 36 + a^2 where a^2 cancels and a = 11/8. Plug a into the second Pythagorean relationship, 36 + (11/8)^2 = r^2. Solving for r^2 = 2425/64. Area of semicircle is (pi * r^2)/2. Plugging in r^2 gives 2425pi/128.
Excellent!
Thanks for sharing ❤️
I don't understand why b/sin beta = 2r. If it's not too much trouble, please explain.
(6+12)*h/2=36 18h=72 h=4
(12-6)/2=3 AD=√[3^2+4^2]=√25=5
3*9 = 4x x = 27/4
(4+27/4)/2=43/8
r^2=(43/8)^2+3^2=1849/64+576/64
=2425/64
Semicircle area = 2425/64*π*1/2=2425π/128(cm^2)
Excellent!
Thanks for sharing ❤️
If there is a difficult way, why do we choose the easy way?
The semicircle area is 2425pi. Also at the 2:50 mark I have noticed that focus on delta AEC is 16 plus 81. Is that because we halved 6??? I just want to make sure.
Thanks for the feedback ❤️
AE=4, DE=3, EC=9 -> 4*x=3*9 -> x=27/4 -> diametr^2=(x+AE)^2 + 6^2 .... r=5V97/8
Clever method.
Thanks for sharing ❤️
STEP-BY-STEP RESOLUTION PROPOSAL :
01) Let the Vertical Distance between Point O and Line DC equal "X" cm
02) OC^2 = X^2 + 6^2 ; R^2 = (X^2 + 36)
03) The Vertical Distance between Point O and Line AB = (X + 4) cm
04) (X + 4)^2 + 3^2 = R^2
05) (X + 4)^2 + 3^2 = X^2 + 36
06) X^2 + 8X + 16 + 9 = X^2 + 36
07) 8X + 25 = 36
08) 8X = 36 - 25
09) 8X = 11
10) X = 11/8 cm
11) X = 1,375 cm
12) R^2 = X^2 + 36
13) R^2 = (121 / 64) + 36 ; R^2 = (121 + 2.304) / 64 ; R^2 = 2.425 / 64
14) Area = 2.425Pi / 128 sq cm
15) Area ~ 59,52 sq cm
ANSWER : Semicircle Area equal to (2.425Pi/128) Square Cm ( ~ 59,52 Square Cm).
P.S. - I jumped the height of the Trapezoid 'cause it's useless calculation and very simple, h = 4.
Let's find the area:
.
..
...
....
.....
First of all we calculate the height h of the trapezoid:
A(ABCD) = (1/2)*(AB + CD)*h
⇒ h = 2*A(ABCD)/(AB + CD) = 2*(36cm²)/(12cm + 6cm) = 2*(36cm²)/(18cm) = 4cm
Now let M and N be the midpoints of AB and CD, respectively. Then the triangles OAM and ODN are right triangles and we can apply the Pythagorean theorem. With r being the radius of the semicircle we obtain:
OA² = OM² + AM²
OD² = ON² + DM²
r² = (ON + h)² + (AB/2)²
r² = ON² + (CD/2)²
(ON + h)² + (AB/2)² = ON² + (CD/2)²
[ON + (4cm)]² + (6cm/2)² = ON² + (12cm/2)²
[ON + (4cm)]² + (3cm)² = ON² + (6cm)²
ON² + (8cm)*ON + 16cm² + 9cm² = ON² + 36cm²
(8cm)*ON = 11cm²
⇒ ON = (11/8)cm
r² = (ON + h)² + (AB/2)² = [(11/8)cm + 4cm]² + (6cm/2)² = [(11/8)cm + (32/8)cm]² + (3cm)² = [(43/8)cm]² + [(24/8)cm]² = (1849/64)cm² + (576/64)cm² = (2425/64)cm²
r² = [(11/8)cm]² + (12cm/2)² = [(11/8)cm]² + [(48/8)cm]² = (121/64)cm² + (2304/64)cm² = (2425/64)cm² ✓
Now we are able to calculate the area of the semicircle:
A = πr²/2 = (2425π/128)cm² ≈ 59.52cm²
Best regards from Germany
Excellent!
Thanks for sharing ❤️
Plenty of computation equired? 72=(6+12)h=18h, h=4, sqrt(r^2-9)=4+sqrt(r^2-36), r^2-9=16+r^2-36+8sqrt(r^2-36), 121=64(r^2-36), r^2=121/64+36, therefore the answer is 1/2×r^2 pi😅
Thanks for the feedback ❤️
It's wrong, prof.! What is alpha? You have not specified!
Correct, please.
Not only can I not find the area but I don't want to. Absolutely useless. I will never ever want to calculate the area of a semi-circle as long as I live.
Area=(2425/128)pi or 18.95pi or 59.503cm^2
Excellent!
Thanks for sharing ❤️