Can you find area of the Semicircle? | Trapezoid | (Trapezium) |

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Learn how to find the area of the Semicircle. Important Geometry and Algebra skills are also explained: Trapezoid area formula; area of the circle formula; Pythagorean theorem; Law of Sines. Step-by-step tutorial by PreMath.com.
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Комментарии : 59

@yalchingedikgedik8007 Месяц назад

Thanks Sir That’s very nice ❤❤❤❤❤

@PreMath Месяц назад

You are very welcome!😀 Thanks for the feedback ❤️

@aliturkseven Месяц назад

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

@PreMath Месяц назад

Excellent! Thanks for sharing ❤️

@tedn6855 Месяц назад

I did same way. Much better

@MegaSuperEnrique Месяц назад

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

@jarikosonen4079 Месяц назад

a=43/8, r=5*sqrt(97)/8, r^2=18+121/128 ???

@MegaSuperEnrique Месяц назад

@@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.

@PreMath Месяц назад

Thanks for sharing ❤️

@genagg5248 Месяц назад

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

@PreMath Месяц назад

I'll prove the extended sine rule pretty soon. Thanks for asking❤️

@raytawa Месяц назад

I could not work that one out either. All good up until then.

@nandisaand5287 Месяц назад

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

@phungpham1725 Месяц назад

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😅

@PreMath Месяц назад

Excellent! Thanks for sharing ❤️

@benlap1977 Месяц назад

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

@quigonkenny Месяц назад

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

@bupera Месяц назад

¿Porqué a/sen x=2r ? No lo entiendo.

@tatertot4810 4 дня назад

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

@aljawad Месяц назад

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.

@PreMath Месяц назад

Excellent! Thanks for sharing ❤️

@rms3 Месяц назад

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?

@antonydalmeida1169 15 дней назад

😂,. Strange method 😂😂

@santiagoarosam430 Месяц назад

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.

@PreMath Месяц назад

Excellent! Thanks for sharing ❤️

@prossvay8744 Месяц назад

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.

@PreMath Месяц назад

Excellent! You are very welcome! Thanks for sharing ❤️

@RachelClark-e6n 29 дней назад

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.

@murdock5537 Месяц назад

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)

@PreMath Месяц назад

Excellent! Thanks for sharing ❤️

@marcgriselhubert3915 Месяц назад

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

@PreMath Месяц назад

Excellent! Thanks for sharing ❤️

@giuseppemalaguti435 Месяц назад

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

@PreMath Месяц назад

Excellent! Thanks for sharing ❤️

@rgcriu2530 Месяц назад

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

@PreMath Месяц назад

Thanks for the feedback ❤️

@bakrantz Месяц назад

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.

@PreMath Месяц назад

Excellent! Thanks for sharing ❤️

@sergioaiex3966 Месяц назад

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

@johnbrennan3372 Месяц назад

Very nice solution.

@unknownidentity2846 Месяц назад

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

@PreMath Месяц назад

Excellent! Thanks for sharing ❤️

@sgarstin Месяц назад

Where does alpha and 2r come from ?

@jamestalbott4499 Месяц назад

Thank you!

@Snazamanaz Месяц назад

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

@himo3485 Месяц назад

(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)

@PreMath Месяц назад

Excellent! Thanks for sharing ❤️

@michaeldoerr5810 Месяц назад

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.

@PreMath Месяц назад

Thanks for the feedback ❤️

@AndreyDanilkin Месяц назад

AE=4, DE=3, EC=9 -> 4*x=3*9 -> x=27/4 -> diametr^2=(x+AE)^2 + 6^2 .... r=5V97/8

@RAG981 Месяц назад

Clever method.

@PreMath Месяц назад

Thanks for sharing ❤️

@LuisdeBritoCamacho Месяц назад

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.

@misterenter-iz7rz Месяц назад

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😅

@PreMath Месяц назад

Thanks for the feedback ❤️

@dom8316 Месяц назад

It's wrong, prof.! What is alpha? You have not specified! Correct, please.

@malcolmdale9607 29 дней назад

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.