Find Area of Triangle ABC | Maths Olympiad | Important Geometry Skills Explained | 2 Methods

Find Area of Triangle ABC | Maths Olympiad | Important Geometry Skills Explained | 2 Methods
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[@ghhdcdvv5069]

تمرين جيد جميل. رسم واضح مرتب. شرح واصح مرتب. شكرا جزيلا لكم والله يحفظكم ويحميكم ويرعاكم جميعا. تحياتنا لكم من غزة فلسطين

[@alinayfeh4961]

Area of Traingle ABC equal 231/16√(15)

[@chaosredefined3834]

Let BC = s The area is equal to 1/2 * 8 * s * sin(2x), and also equal to 1/2 * s * 14 * sin(x). Equating these, we get 4 s sin(2x) = 7 s sin(x) Divide both sides by s 4 sin(2x) = 7 sin(x) Substitute the identity sin(2x) = 2 sin(x) cos(x) 4 * 2 sin(x) cos(x) = 7 sin(x) Note that x =/= 0 or 180, so sin(x) =/= 0. Divide by sin(x) 8 cos(x) = 7 cos(x) = 7/8 Square both sides, and apply that cos^2(x) = 1 - sin^2(x) 1 - sin^2 (x) = 49/64 sin^2(x) = 15/64 sin(x) = sqrt(15) / 8 Another way of expressing the area is 1/2 * 8 * 14 * sin(180 - 3x) = 56 * sin(180 - 3x). Note that sin(180 - 3x) = sin(3x), so the area = 56 * sin(3x) sin(3x) = 3 cos^2(x) sin(x) - sin^3(x) sin(3x) = 3 (49/64) (sqrt(15)/8) - 15 sqrt(15)/ 512 sin(3x) = 147 sqrt(15) / 512 - 15 sqrt(15) / 512 sin(3x) = 132 sqrt(15) / 512 sin(3x) = 33 sqrt(15) / 128 Area = 56 * sin(3x) = 56 * 33 * sqrt(15) / 128 Area = 231 * sqrt(15) / 16 So, the area is 231/16 * sqrt(15), or approximately 55.92

[@andreasandre4756]

I have never seen so long calculation. 8x14=112/2=56 Sabc=56,. Are you in your mind?

[@chaosredefined3834]

@@andreasandre4756 Except it's slightly less than 56. So... your method is flawed.

[@Alessandro-1977]

Genius solution, I admit I couldn' t solve it without trigonometry, i.e. h = 8*sin(2*x) = 14*sin(x), etc...till to solution area = 231/16*√15 with a few passages

[@sharonmarshall3671]

I think having determined the value of, it is easier to substitute for x in 180-3x to apply the formula Area = 1/2 bcSinA where A = 180-3x

[@rajanpanikkare5917]

Wewww2ww2😅i

[@timeonly1401]

@8:08 Once we get a=17/4, the third side of the big triangle is 2a+8 = 2(17/4)+8 = 33/2 = 16.5. To get the area of the big triangle, we can use Heron's formula, with a=8, b=14, c=16.5: The semi-perimeter s = (a+b+c)/2 = (8+14+16.5)/2 = 38.5/2 = 19.25, so that the area is: A = √[s(s-a)(s-b)(s-c)] = √[(19.25)(11.25)(5.25)(2.75)] = √[(77/4)(45/4)(21/4)(11/4)] = (1/16)√[(3^3)(5)(7^2)(11^2)] = (231/16)√15 Done!! 😁

[@dr.essamel-dinabdel-ghafou3383]

⁶😊

[@wilfredolunaholguin1975]

Usando la fórmula de Heron tienes que usar una calculadora para realizar las oportunidades.En un examen de ingreso o una competencia no se permite el uso de calculadora,por eso no es conveniente usar la fórmula de Heron

[@timeonly1401]

@@wilfredolunaholguin1975 As you can see from my work... I did all of this with each step clearly doable by hand and without a calculator. My answer is an EXACT one (same answer as the video presenter's), NOT a decimal approximate as one would expect if one had used a calculator (The square root of 15 is irrational, since 15 is not a perfect square; so, calculating it would've resulted in a non-terminating, non-repeating decimal tail....)

[@elmer6123]

Your boss says to you, "I need the area of this triangle right now in my office." Have you memorized Heron's formula? I sure as hell haven't. But I do know the sine law and double angle formula. Which means that I can use my pocket calculator to give her the answer right there in her office in a couple minutes.

[@RodrigoBaltuilhedosSantos]

You can get the third size using (14²-8²)/8=16.5. This formula is valid because of α and 2α angles

[@santiagoarosam430]

Trazamos AD simétrico de AB respecto a la altura “a” por A del ∆ABC → BC=(BD)+(DC)=(b+b)+(8) → ∆ADC es isósceles y tiene lados 8/8/14 → d= Altura de ∆ADC por D → d²=8²-(14/2)²→ d=√15 → Área ∆ADC =(14√15)/2 = 8a/2 → a=(7√15)/4 → b²=8²-a² → b=17/4 → BC=2b+8=33/2 → Área ∆ABC =(BC)a/2 =(231√15)/16 =55.9162 Gracias y saludos cordiales.

[@prime423]

Simply use a right triangle whose Cosine is known. The Sine is easily found. The rest is trivial.

[@drnandkishorbagul3087]

We can also solve by Application of Pythagoras theorem for obtuse angle theorem.

[@LogintoMaths]

I solved it by sin rule And I got the correct answer

[@quigonkenny]

Let P be the point on BC where ∠BPA = 2X. As ∠ABP = ∠BPA, ∆PAB is isoscemeles, and PA = BA = 8. As ∠BPA is an external angle to ∆APC at P, its value equals the sum of the two opposite angles, ∠PCA and ∠CAP. As ∠PCA = X, ∠CAP = 2X-X = X as well. This means ∆APC is isosceles, and PC = AP = 8. Draw PQ, where Q is the midpoint of CA. As ∆APC is isosceles, PQ is perpendicular to CA, and this creates two new congruent right triangles, ∆CQP and ∆PQA. As Q is the midpoint, CQ = QA = 7. Triangle ∆CQP: QP² + CQ² = PC² QP² + 7² = 8² QP² = 64 - 49 = 15 QP = √15 Draw AT, where T is the midpoint of BP. As ∆PAB is isosceles, AT is perpendicular to BP, and this creates two new congruent right triangles, ∆BTA and ∆ATP. As T is the midpoint, BT = TP = Y. As ∠ATC = ∠CQP = 90° and ∠TCA and ∠PCQ are the same angle, ∆ATC and ∆CQP are similar, along with ∆PQA. Triangle ∆ATC: TC/CA = CQ/PC (8+Y)/14 = 7/8 8(8+Y) = 14(7) 64 + 8Y = 98 8Y = 98 - 64 Y = 34/8 = 17/4 AT/CA = QP/PC AT/14 = √15/8 AT = 14√15/8 = 7√15/4 Triangle ∆ABC: [ABC] = bh/2 = (8+2(17/4))(7√15/4)/2 [ABC] = (4+(17/4))(7√15/4) [ABC] = (33/4)(7√15/4) [ABC] = (231√15)/16 ≈ 55.916

[@hongningsuen1348]

Your 1st method can be simplified by another construction: 1. Construct isosceles triangle ADC by adding new side AD of 14 (= AC) to the left of AB. Hence angle ADB = x. 2. Triangle ADB is also an isosceles triangle and angle BAD = x (by ext. angle of triangle). Hence DB = 8 (equal sides of isosceles triangle). 3. Area of triangle ABD by Heron's formula = 7 Sqrt(15) with sides 14, 8, 8. 4. Triangles ADC and BAD are similar triangles (AAA). Hence DC:AC = 14:8. Given AC = 14. Hence DC = 14x14/8 = 196/8 = 49/2 5. BC = DC - DB = (49/2) - 8 = 33/2 6. Area of triangle ABC:Area of triangle ABD = BC:BD = (33/2):8 (equal height triangles) Hence area of triangle ABC = [(33/2) /8] x 7 Sqrt(15) = (231/16) x Sqrt(15). Your 2nd method can be simplified by getting angle x from Cos(x) then Sin(3x) to avoid using complicated trigonometric identities other than law of sines. This is certainly the method of choice when solving MCQ in exam.

[@BLACKICE-m3d]

1/2×8×14×sin(180-3x)

[@User-jr7vf]

I did by the second method (without looking at the solution)

[@avanishpandey4680]

It is easy question

[@lucas0_03]

​@@avanishpandey4680 u right bro. You just need tô know practically arithmetic

[@user-hiraaidi]

In a triangle ABC,if a=50m, b=30m and C=30°, then area of the triangle is...... Kasari ho sir

[@libardouribe7617]

👍

[@harrymatabal8448]

What if I say the area is 1/2×8×14sin3x==56sin 3x

[@harrymatabal8448]

I like to hear what you have to say Thanks

[@adgf1x]

Area=231(15^0.5) /16

[@mohamadtaufik5770]

55,916

[@محمدالنجار-و2ف]

I have a simple 6:05 solution

[@fujinly9558]

I can not see

[@giuseppetornabene989]

50 passaggi quando ne servono solo 3

[@integer9655]

in many places you just wanted to make a lengthy video , which was a poor idea. Don't do that. 😟

[@elmer6123]

I am 80 years old and was afraid I'd die before getting to the end. Without watching the video I just used the law of sines, double angle formula, and my pocket calculator to get the answer in a couple minutes.

[@BLACKICE-m3d]

1/2×8×9×sin(180-3x)