Can you the Perimeter a+b+c? | (Semicircle) |

[@jamestalbott4499]

Thank you!

[@PreMath]

You are very welcome! 😊🙏

[@imetroangola17]

*_Nota de Aula:_* c² = a²+ b² = (a+b)² - 2ab (a+b)² = c² + 2ab. Ora, c²=175 e ab= 2×[ABC] = 2×35=70. Daí, (a+b)² = 175 + 140 = 315 = 9×35. Assim, a + b = 3√35. Portanto, *a + b + c = 3√35 + 5√7*

[@himo3485]

7 : 28 = 1 : 4 = 1² : 2² AD=k CD=2k DB=4k AB=AD+DB=5k 5k*2k*1/2=7+28 5k²=35 k2=7 k=√7 a²=(√7)²+(2√7)²=35 a=√35 b²=(2√7)²+(4√7)²=140 b=2√35 c=AB=5√7 a + b + c = 3√35 + 5√7 (cm)

[@cyruschang1904]

The two small triangles are similar, their linear ratio = √7 : √28 = 1 : 2 This tells us also the two small triangles and the large triangle are all 1 - 2 - √5 triangles So we can write a = x, b = 2x, c = (√5)x ab = 2x^2 = 2(7 + 28) cm^2 => x = √35 cm a + b + c = (1 + 2 + √5)(√35 cm) = (3√35 + 5√7) cm

[@alexniklas8777]

I solved this problem with your method. Thanks sir!

[@Abdelfattah-hr8tt]

Im hear following you dear

[@alexundre8745]

Bom dia Mestre Acertei Graças ao Sr estou aprendendo Geometria

[@marcgriselhubert3915]

Without thinking outside the box: Be h = CD. We have AD.h = 2.7 = 14 and BD.h = 2.28 = 56, so BD = 4.AD, and AD as AD +BD = c, we get that AD= (1/5).c and BD = (4/5).c In triangle ADC: h^2 = a^2 -(1/25).(c^2), and in triangle BDC : h^2 = b^2 - (16/25).(c^2), so b^2 - a^2 = (15/25).(c^2) = (3/5).(c^2) As a^2 + b^2 = c^2 in triangle ABC, we get then that b^2 = (4/5).(c^2) and a^2 = (1/5).(c^2), giving b = (2.sqrt(5))/5).c and a = (sqrt(5)/5).c We had h^2 = a^2 - (1/25).(c^2), so h^2 = (1/5).(c^2) - (1/25).(c^2) = (4/25).(c^2), giving that h = (2/5).c We had AD.h = 14, so ((1/5).c).((2/5).c) = 14, and so c^2) = (14.25)/2 = 7.25, giving that c = 5.sqrt(7) Now we get a = (sqrt(5)/5).c = sqrt(35) and b = ((2.sqrt(5))/5).c = 2.sqrt(35), and finally a + b + c = sqrt(35) + 2.sqrt(35) + 5.sqrt(7) = (sqrt(7)).(5 + 3.sqrt(5)).

[@marcelowanderleycorreia8876]

A questão é mais difícil do que parece. Muito inteligente!! Parabéns mestre!! 👍

[@quigonkenny]

As ∆ADC = 7cm² and ∆CDB = 28cm², which is a ratio of 1:4, and the two triangles share the same height (CD), then their bases (AD and DB) must share the same ratio. Thus AD = c/5 and DB = 4c/5. As ∠BCA = 90° by Thales' Theorem, then as ∠ADC = ∠CDB = 90° and ∠CAB = ∠CAD = ∠BCD = 90°-∠B, then ∆BCA, ∆ADC, and ∆CDB are similar triangles. AD/CD = CD/DB (c/5)/CD = CD/(4c/5) CD² = (c/5)(4c/5) = 4c²/25 CD = √(4c²/25) = 2c/5 Triangle ∆CDB: A = bh/2 = DB(CD)/2 28 = (4c/5)(2c/5)/2 28 = (8c²/25)/2 = 4c²/25 c² = 28(25/4) = 7(25) = 175 c = √175 = 5√7 CA/DC = BC/DB a/2√7 = b/4√7 2√7b = 4√7a b = 4√7a/2√7 = 2a a² + b² = c² a² + (2a)² = (5√7)² a² + 4a² = 175 5a² = 175 a² = 175/5 = 35 a = √35 b = 2a = 2√35 a + b + c = √35 + 2√35 + 5√7 [ a + b + c = 3√35 + 5√7 ≈ 30.977cm ]

[@davidseed2939]

my method since the left and right triangles have an area ratio of 4 and because those triangles are similar. (a property of rt angled triangles divided by the perpendicular) then corresponding lengths in the triangles will have a ratio of √4 =2 hence b=2a. now the area of the whole triangle is 35 and this is calculated as ab/2 35=a(2a)/2 a^2=35, b^2=140 and by Pythagoras c^2 =a²+b²=175 it follows that a=√35, b=2√35 c=5√7 so a+b+c=3√35+5√7 Incidentally, this diagram can be used to prove pythagoras. First using angle sums, the two constituent triangles and the whole triangle are all similar. For any similar triangles, the area of each triangle is in proportion to the square of corresponding lengths in the triangle. The chosen corresponding lengths are the hypotenuses of each triangle (a,b,c) ie area of ΔACD =ka² ΔDCB=kb² ΔABC=kc² now the two component triangles have a total area equal to the area of the large triangle. ie. ka²+kb²=kc² hence a² +b ² =c² QED 😊

[@PrithwirajSen-nj6qq]

ABC & ACD are similar triangles. Hence a^2/c^2 =7/35=1/5 > a/c =1/√5--(1) ABC & BCD are similar triangles. Hence b^/c^2=28/35=4/5 > b/c =2/√5 --(2) From (1) & (2) we get a ঃ b ঃ c = 1ঃ 2ঃ√5=kঃ2k ঃ √5 k The area of the 🔺 ABC = 1/2*ab =35 1/2*k*2k =35 >k =√35 Hence a =√35 b =2√35 c =5√7 Hence Perimeter =(√35+2√35+5√7) units =(3√35 +5√7) units

[@wasimahmad-t6c]

A=5.353 b=13.076696 c=14

[@AnonimityAssured]

The title needs correcting.

[@sergioaiex3966]

Solution: Triangles with the same height will have areas whose ratios is the same as the ratio of their bases In other words: ∆ ADC Area/∆ BDC Area 7/28 = 1/4 Therefore If AD = k, BD = 4k Applying Chords Theorem h² = k . 4k h² = 4k² h = 2k ∆ ADC A = ½ base × heigth 7 = ½ k × 2k 14 = 2k² k = √7 ===== Applying Pythagorean Theorem in ∆ ADC (k)² + (2k)² = a² k² + 4k² = a² a² = 5k² a = k √5 a = √7 . √5 a = √35 ====== c = k + 4k c = 5k c = 5√7 ====== Applying Thales's Theorem in ∆ ABC and, at the same time, Pythagorean Theorem, to calculate "b" a² + b² = c² (√35)² + b² = (5√7)² 35 + b² = 175 b² = 140 b = 2√35 ======= Final Step a + b + c = √35 + 2√35 + 5√7 a + b + c = 3√35 + 5√7 cm ✅ a + b + c = 30.9769 cm ✅

[@PrithwirajSen-nj6qq]

BDC /ADC = (1/2*BD*CD)/(1/2*AD*CD)= 28/7 >BD /AD = 4/1=4x/x In respect of the semicircle CD is the geometric mean of AD and BD Then CD ^2=AD*BD=x*4x > CD =2x rt 🔺 ADC =1/2 *AD *CD =1/2*x*2x=x^2 Now x ^2 =7 x =√7 AD =√7 BD =4x =4√7 c =AD +BD =5√7--(1) In rt 🔺 ADC a =√(AD^2 + CD ^2) = √(7+4*7)=√35 --(2) In rt 🔺 BDC b =√(BD^2+CD^2) =√(16*7+4*7) =2√35 ---(3) From (1),(2) & (3) a +b+c =√35 +2√35+ 5√7 =(3√35 +5√8) units

[@unknownidentity2846]

Let's face this Christmas challenge: . .. ... .... ..... The triangle ABC can be divided into the right triangles ACD and BCD. So we can conclude: A(ABC) = A(ACD) + A(BCD) = (1/2)*AD*CD + (1/2)*BD*CD = (1/2)*(AD + BD)*CD A(ACD) = (1/2)*AD*CD ∧ A(BCD) = (1/2)*BD*CD ⇒ BD/AD = A(BCD)/A(ACD) = (28cm²/7cm²) = 4 ⇒ BD = 4*AD According to the theorem of Thales the triangle ABC is a right triangle. So we can apply the right triangle altitude theorem: CD² = AD*BD CD² = AD*(4*AD) CD² = 4*AD² ⇒ CD = 2*AD A(ACD) + A(BCD) = (1/2)*(AD + BD)*CD 7cm² + 28cm² = (1/2)*(AD + 4*AD)*(2*AD) 35cm² = 5*AD² 7cm² = AD² ⇒ AD = √(7cm²) = (√7)cm CD = 2*AD = (2√7)cm BD = 4*AD = (4√7)cm Now we apply the Pythagorean theorem to the right triangles ACD and BCD: AC² = AD² + CD² = (√7)²cm² + (2√7)²cm² = 7cm² + 28cm² = 35cm² ⇒ AC = √( 35cm)² = (√35)cm BC² = BD² + CD² = (4√7)²cm² + (2√7)²cm² = 112cm² + 28cm² = 140cm² ⇒ BC = √(140cm)² = (2√35)cm Now we are able to calculate the perimeter of the triangle ABC: P(ABC) = AB + AC + BC = (AD + BD) + AC + BC = (√7)cm + (4√7)cm + (√35)cm + (2√35)cm = (5√7)cm + (3√35)cm ≈ 30.98cm And now (as a Christmas gift) let's check the side lengths by applying Heron's formula: s = (AB + AC + BC)/2 = [(5√7 + 3√35)/2]cm s − AB = [(5√7 + 3√35)/2 − 5√7]cm = [(3√35 − 5√7)/2]cm s − AC = [(5√7 + 3√35)/2 − √35]cm = [(5√7 + √35)/2]cm s − BC = [(5√7 + 3√35)/2 − 2√35]cm = [(5√7 − √35)/2]cm s * (s − AB) * (s − AC) * (s − BC) = [(5√7 + 3√35)/2]*[(3√35 − 5√7)/2]*[(5√7 + √35)/2]*[(5√7 − √35)/2]cm⁴ = [(9*35 − 25*7)*(25*7 − 35)/16]cm⁴ = [(315 − 175)*(175 − 35)/16]cm⁴ = [140*140/16]cm⁴ = 35²cm⁴ ⇒ A(ABC) = √[s * (s − AB) * (s − AC) * (s − BC)] = 35cm² ✅ Best regards from Germany

[@HERRANDIAZ]

jajajajajaj x2

[@Thampuran-o9o]

👍👍👍