Can you find area of the Semicircle? | Trapezoid | (Trapezium) |#math #maths | #geometry

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Can you find area of the Semicircle? | Trapezoid | (Trapezium) |#math #maths | #geometry

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[aliturkseven]

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

[MegaSuperEnrique]

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

[genagg]

triangle DAC leads to the radius of the semi-circle? Afraid that's where you lost me

[nandisaand]

That solution is WAY too complicated. Here's a simpler method:
After determining h=4, I made 2 right triangles:
<center of AB>, O, B
And
<center of DC>, 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

[phungpham]

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😅

[rms]

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?

[quigonkenny]

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 <-- [1] = [2]
8x = 36 - 25 = 11
x = 11/8

(11/8)² + 36 = r² <-- [1]
r² = 121/64 + 36 = (121+2304)/64
r² = 2425/64

Semicircle O:
Aᴏ = πr²/2 = π(2425/64)/2
Aᴏ = 2425π/128 cm² ≈ 59.52 cm²\

[benlap]

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

[santiagoarosam]

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.

[sergioaiex]

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

[murdock]

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)

[aljawad]

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.

[prossvay]

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 thanks sir.

[unknownidentity]

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

[rgcriu]

teorema de cuerdas 3*9=4*X => X=27/4 => teorema de faure 4R²=a²+b²+c²+d²

[RachelClark-en]

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.

[marcgriselhubert]

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

[bakrantz]

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.

[LuisdeBritoCamacho]

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.

[Snazamanaz]

I don't understand why b/sin beta = 2r. If it's not too much trouble, please explain.