Can you find area of the Semicircle? | (Triangles) |

Learn how to find the area of the Semicircle. Important Geometry and Algebra skills are also explained: Intersecting Chords theorem; area of the circle formula. Step-by-step tutorial by PreMath.com.
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Can you find area of the Semicircle? | (Triangles) | #math #maths | #geometry
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#FindSemiCircleArea #SemiCircle #IntersectingChordsTheorem #Chords #GeometryMath #PythagoreanTheorem #Euclid'sTheorem #EuclideanTheorem
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[@abeonthehill166]

Another clear and succinct demonstration Professor ………Thanks for sharing Man !

[@PreMath]

Glad you liked it! You are very welcome! Thanks for the feedback ❤️

[@laxmikantbondre338]

Steps 1. Assume CD = h 2. AD = 34/h, DB = 136/h ... Area of both ∆s is goven. 3. By Thales theorum ang ACB is 90 So CD is Altitude on the Hypotuneous of a Rt. traingle 4. So CD is Geometric Mean of AD and DB So h Sqr = 34/h * 168/h 5. So solving this CD = h = 2√17 Hence AD = 34/h = √17 and DB = 136/h = 4√17. So Diameter = AB = 5√17

[@PreMath]

Great! Thanks for sharing ❤️

[@allanflippin2453]

I took an approach more like the way Mr. PreMath usually does! That's similar triangles. The key is Thales theorem: angle ABC is a right angle. We can work out alpha and beta angles to show that yellow and green triangles are similar. Also, as pointed out in the video, AD and DB have the proportion 17:68. Adding them up, 2r = 5*AD or 5/4*DB. I'll call the height CD "h". The total of both triangles area is height * base / 2. Or h*2r/2 = (17+68). Simplified to h*r = 85. The ratio CD:AB equals DB:CD. Filling in the numbers, h * (2r/5) = (8r/5) / h. Cross multiplying, h^2 = 16 * r^2 / 25. Taking square root, h = 4r/5. Now consider h*r = 85, or r = 85/h. h = 4*85/h*5. or h = 68/h. Multiplying by h, h^2 = 68 or h = sqrt(68) or h = 2 * sqrt(17). From above, r = 5h/4. That means r = 10 * sqrt(17) / 4 or r = 5 * sqrt(17) / 2. Squaring, r^2 = 25 * 17 / 4. or r^2 = 425/4. The semicircle area is pi * r^2 / 2 or 425/8 pi.

[@PreMath]

Bravo! Thanks for sharing ❤️

[@anatoliy3323]

Simple but pleasant task. 👍💯Thank you so much, Professor

[@PreMath]

Glad to hear that! You are very welcome! Thanks for the feedback ❤️

[@PrithwirajSen-nj6qq]

(1/2*h*AD)/(1/2*h*BD) =17/68=1/4 >AD/BD=1/4 If AD =x BD =4x Diameter is 5x radius = 2.5 x h =√(x *4x ) geometric mean theorem h = 2x Take the 17 sq cm triangle 1/2*x*2x=17 > x =√17 Radius = 2.5*√17 Area of semicircle =1/2(π*17*25/4) sq cms

[@PreMath]

Excellent! Thanks for sharing ❤️

[@jamestalbott4499]

Thank you!

[@PreMath]

You are very welcome! Thanks for the feedback ❤️

[@murdock5537]

∆ ABC → AB = AD + BD = 2r/5 + 8r/5 = 2r; CD = h; sin⁡(ADC) = 1; hr = 85 h^2 = (2r/5)(8r/5) → h = 4r/5 → hr = (4r^2)/5 = 85 → r^2 = (25/4)17 → πr^2/2 = (25/8)17π

[@PreMath]

Excellent! Thanks for sharing ❤️

[@phungpham1725]

1/ Label CD= h, AD=a, BD= b and the diameter AB= d Consider the two right triangle ACD and BCD We have h. a=2x17 (1) and h.b=2x68=2x4x17(2) -> (1)x(2) sq h xaxb= sq 2xsq2xsq17 Because axb= sq h ( the right triangle altitude theorem) -> sqhxsqh= sq2xsq2sq17 h=2sqrt17 2/ hxd=2x85=170 -> d=170/(2sqrt17)=85/(sqrt17) Area= 1/2 pi . sq(85/sqrt17)/4 =1/8 pi . sq(85/sqrt17) = 166.9 sq cm

[@PreMath]

Excellent! Thanks for sharing ❤️

[@prossvay8744]

AD/BD=17/68=1/4 So AD=1a ; BD=4a Let CD=x So x°2=(1a)(4a) So x=2a Area of triangleABC=1/2(x)(5a)=85 1/2(2a)(5a)=85 So a=√17 Diameter of semisecle=1a+4a=5a=5√17 So Radius=5√17/2 Area of semicircle=1/2(π)(5√17/2)^2=425π/8cm^2=166.9cm^2.❤❤❤

[@PreMath]

Excellent! Thanks for sharing ❤️

[@hongningsuen1348]

Method using base side ratio = area ratio for equal height triangles and intersecting chords theorem: 1. Triangles ADC and BDC are equal height triangles. Hence AD:BD = 17:68 = 1:4 2. Let R be the radius. AB = AD + BD = 2R. Hence AD = (2/5)R and BD = (8/5)R 4. For complete circle, by intersecting chords theorem (CD)^2 = (AD)(BD) (CD)^2 = (2/5)(8/5)R^2 (equation 1) 5. In triangle ADC, area = 17 = (1/2)(AD)(CD) Hence CD = (17)(2)/[(2/5)R] = (5)(17)/R CD^2 = [(5)(17)/R]^2 (equation 2) 6. From equations (1) and (2) (2/5)(8/5)R^2 = [(5)(17)/R]^2 R^4 = (5^4)(17^2)/(2)(8) R^2 = (5^2)(17)/4 = 425/4 7. Area of semicircle = (1/2)(425/4)pi = (425/8)pi

[@PreMath]

Excellent! Thanks for sharing ❤️

[@joeschmo622]

I just saw them as similar triangles, ADC, CDB, and ACB. Yellow is a 1:2:sqrt5 triangle, green is 2:4:2sqrt5, and big is sqrt5:2sqrt5:5. Diameter is 5, radius is 5/2, area is 25pi/4. *Now* scale. So 1:2:sqrt5 has an area of 1unit, but is 17cm2 so that's the scaling factor. Circle area is (17)(25pi/4). Semicircle is half that, so (17)(25pi/8) or 425pi/8. ✨Magic!✨

[@PreMath]

Excellent! Thanks for sharing ❤️

[@juanalfaro7522]

Let CD=h and AD=x -> BD=2R-x. Since 17=x *h/2 and 68 = (2R-x) *h/2, then 2R-x = 4x -> 2R=5x -> R=5x.2. Chord Intersection Theorem: (2R-x) *x = h^2 = 4x*x = 4x^2 -> h=2x. Now [ADC] = x *h/2 = 2x^2/2 = x^2 = 17 -> x=sqrt (17) -> R=5*sqrt (17)/2 --> A(semicircle) = PI*R^2/2 = Pi/2 * [5*sqrt (17)/2] ^2 = Pi/2 *25*17/4 = Pi*425/8 cm^2

[@Waldlaeufer70]

17 * 4 = 68, this means, that the ratio of the legs is 1 : √4 = 1 : 2. x * 2x = 2 * 17 x² = 17 x = √17 d = AB = 1x + 4x = 5x = 5√17 r = 5/2 √17 A(semicircle) = 1/2 * r² * π = 1/2 * (5/2 √17)² π = 1/2 * 25/4 * 17 * π = (25 * 17) / 8 π = 53.125 π = 166.9 square units

[@cyruschang1904]

yellow hypotenuse = x gree hypotenuse = x✓(68/17) = 2x x(2x)/2 = 17 + 68 = 85 x = ✓85 circle diameter = ✓(x^2 + 4x^2) = x✓5 = 5✓17 semi circle area = π[(5✓17)/2]^2 ÷ 2 = 25(17)π/8 = 425π/8 (cm^2)

[@santiagoarosam430]

Razón entre áreas s²=68/17=4→ Razón de semejanza =s=√4=2→ Si AD=a→ CD=2a→DB=4a→ 2a*4a/2=4a²=68→ a=√17→ AB=AD+DB=2r =a+4a=5a=5√17 → r=5√17/2→ Área del semicírculo =πr²/2 =(π*25*17)/(4*2) =425π/8 =166,8971... Gracias y un saludo.

[@PreMath]

Excellent! You are very welcome! Thanks for sharing ❤️

[@LuisdeBritoCamacho]

STEP-BY-STEP RESOLUTION PROPOSAL USING THE LAW OF SIMILAR TRAINGLES : 01) Let AD = X cm 02) Let BD = Y cm 03) Let AB = (X + Y) cm 04) Let CD = h cm 05) X * h = 34 ; h = 34/X 06) Y * h = 136 ; h = 136/Y 07) 34/X = 136/Y ; Y/X = 4 ; Y = 4X 08) BD = 4X cm 09) AB = 5X cm 10) Triangle (ACD) is Similar to Triangle (BCD) ; So : 11) h/X = Y/h ; h^2 = (X * Y) ; As Y = 4X ; h^2 = 4X^2 ; sqrt(h^2) = sqrt(4X^2) ; h = 2X 12) 2X * X = 34 ; 2X^2 = 34 ; X^2 = 34/2 ; X^2 = 17 ; X = sqrt(17) ; Or : 13) 2X * 4X = 136 ; 8X^2 = 136 ; X^2 = 136/8 ; X^2 = 17 ; X = sqrt(17) 14) If : X = sqrt(17) ; Then 5*X = 5 * sqrt(17) 15) AB = 5 * sqrt(17) 16) Radius (R) = AB/2 cm ; R = (5 * sqrt(17)) / 2) ; R^2 = (25 * 17) / 4 ; R^2 = 425/4 17) Semi Circle Area (SCA) = Pi * R^2 / 2 18) SCA = ((425/2) * Pi) / 2 ; SCA = 425Pi / 8 ; SCA ~ 167 sq cm Thus, OUR PLANE AND SIMPLE ANSWER IS : Semicircle Area is approx. equal to 167 Square Centimeters. Greetings from Cordoba!!

[@PreMath]

Excellent!👍 Thanks for sharing ❤️

[@RondoCarletti]

for both triangles use A=1/2*g*h. h is equal to both triangles. So calculate 17 (and also 68) = 1/2*2*h (it works) or 1/2 *4*h (this is realistic) = Area 157,0796...

[@Ibrahimfamilyvlog2097l]

Very good sir❤❤

[@PreMath]

Excellent! Thanks for the feedback ❤️

[@quigonkenny]

Let ∠CAB = α and ∠ABC = β, where α and β are complementary angles that sum to 90°. As ∠ADC = 90°, ∠BCA = 90°-α = β and thus ∆BCA and ∆ADC are similar triangles. Similarly, as ∠BCD = 90°-β = α, ∆CDB is also aimilar to the above triangles. As ∆CDB and ∆ADC are similar triangles and the area of ∆CDB is 4 times the area of ∆ADC, then the lengths of the sides of ∆CDB are √4 = 2 times the length of the corresponding sides of ∆ADC. Thus DB = 2DC. And as the two triangles are similar, this means DC = 2AD. Let AD = x. AB = AD + DB AB = AD + 2DC AB = AD + 2(2AD) = 5AD = 5x Triangle ∆CDB: Aᴛ = bh/2 = DB(DC)/2 68 = 4x(2x)/2 = 4x² x² = 68/4 = 17 x = √17 AB = 5x 2r = 5√17 r = (5√17)/2 Semicircle O: Aₒ = πr²/2 = π((5√17/2)²/2 Aₒ = π(25)(17)/8 = 425π/8 ≈ 166.90 cm²

[@himo3485]

17 : 68 = 1: 4 = 1² : 2² AD=x CD=2x BD=4x (x+4x)*2x/2=17+68 5x²=85 x=√17 AO=BO=5x/2=5√17/2 Semicircle area = 5√17/2*5√17/2*π*1/2 = (425π/8)cm²

[@PreMath]

Excellent! Thanks for sharing ❤️

[@anthonycheng1765]

i use this method but just solve for x^2. need not for x, as area involves x^2.

[@rabotaakk-nw9nm]

2:40-5:00 Right triangle altitude theorem: CD²=AD•BD; h²=1k•4k => h=2k

[@PreMath]

Thanks for sharing ❤️

[@brettgbarnes]

△ABD ~ △ACD ~ △BCD (1/2)(2r)(CD) = [ABC] (1/2)(2r)(CD) = 17 + 68 CD = 85/r [BCD]/[ACD] = 68/17 = 4 BD = (CD)√([BCD]/[ACD]) BD = (85/r)√(4) BD = 170/r AD = (CD)√([ACD]/[BCD]) AD = (85/r)√(1/4) AD = 85/2r AB = AD + BD 2r = (85/2r) + (170/r) 2r = (85 + 340)/2r r² = 425/4 Area (semi-circle) = (πr²)/2 Area (semi-circle) = (π425/4)/2 Area (semi-circle) = 425π/8

[@MrPaulc222]

Although the numbers make it that DB is 4x the length of AD, the image makes it appear 3x the length. I will go with the numbers due to the usual warning that the image may not be accurate. Call CD, h. Split the diameter into (8/5)r and (2/5)r. Call CD, h. Intersecting chords: (16/25)r^2 = h^2, so r^2 = (25/16)h^2, r = (5/4)h, and h = (4/5)r. Green triangle: (4/5)r * (8/5)r = 136 (32/25)r^2 = 136 (16/25)r^2 = 68 (16/5)r^2 =340 (4/5)r^2 = 85 r^2 = 85*(5/4) r^2 = 425/4, so (425/4)pi for a full circle and (425/8)pi for a semicircle. 166.92 cm^2 I have now watched. Although we took different paths, there was plenty of overlap. I never bothered calculating r, because r^2 was more relevant. Thank you once again.

[@PreMath]

Excellent! You are very welcome! Thanks for the feedback ❤️

[@giuseppemalaguti435]

arctg h/(136/h)+arctg h/(34/h)=90..h^4=16*289..h=2√17...2r=34/2√17+136/2√17=5√17...r=5√17/2

[@PreMath]

Excellent! Thanks for sharing ❤️

[@wackojacko3962]

All Hail Euler and Set theory ...the hierarchy of the Right triangle within is the Isoceles and Scalene...the Oblique...the Acute and Obtuse...and Isoceles. Knowing the relationships of their angles and ratios of their sides allows one too readily calculate problems such as this one. How much fun can a person have in one day? 🙂

[@PreMath]

Excellent! Thanks for the feedback ❤️

[@misterenter-iz7rz]

Quite creative. but not too difficult 🎉. a,b=2a,2b=4a, then 34=2a^2, a^2=17, 2r=5a, 4r^2=25a^2, then the answer is 1/2 r^2 pi=1/2 × 25/4×17pi=425/8 pi.😅 ::: '

[@PreMath]

Glad to hear that! Thanks for sharing ❤️

[@unknownidentity2846]

Let's find the area: . .. ... .... ..... The areas of the two right triangles ACD and BCD can be calculated in the following way: A(ACD) = (1/2)*AD*CD A(BCD) = (1/2)*BD*CD ⇒ BD/AD = A(BCD)/A(ACD) = (68cm²/17cm²) = 4 ⇒ BD/(2*R) = BD/(AD + BD) = 4*AD/(AD + 4*AD) = 4*AD/(5*AD) = 4/5 ∧ AD/(2*R) = AD/(AD + BD) = 1*AD/(AD + 4*AD) = 1*AD/(5*AD) = 1/5 According to the theorem of Thales the triangle ABC is also a right triangle. Therefore we can apply the right triangle altitude theorem: CD² = AD*BD = AD*4*AD = 4*AD² ⇒ CD = 2*AD Now we are able to calculate the radius R and the area A of the semicircle: A(ACD) = (1/2)*AD*CD = (1/2)*AD*(2*AD) = AD² AD/(2*R) = 1/5 5*AD = 2*R ⇒ R = 5*AD/2 A = πR²/2 = π*(5*AD/2)²/2 = (25/8)π*AD² = (25/8)π*A(ACD) = (25/8)π*(17cm²) = (425π/8)cm² ≈ 166.90cm² Best regards from Germany

[@PreMath]

Excellent! Thanks for sharing ❤️

[@sorourhashemi3249]

122.656 the area. AD=a, so DB= 2r-a×CD/2=68, and a×CD/2=17 two equations and I got a=2.5 and CD=13.6, so 2r×13.6/2=68+17, r=6.25 and area of semicircle=122.656. Why didn't I got your answer?

[@verunes07]

Euclid teorem

[@PreMath]

Excellent! Thanks for the feedback ❤️

[@marcgriselhubert3915]

We can also use the Phytagorean theorem in triangle ODC to have CD.

[@PreMath]

Thanks for the feedback ❤️

[@burple_nida098]

Who is watching 1 day before exam😂😂