This Question Stumps 90% of Students - Russian Test Problem

[@gikasmith5511]

A very good exercise, but I think I have a simpler solution. I used the property of the bisector: BD/DC = sqrt(2329)/27. The same applies to areas of ABD and ADC, since the triangles have the common height. Then A(ABD)/A(ABC) = sqrt(2329)/(sqrt(2329)+27), and since the area of ABC is 540, then comes the solution: A(ABD) = 540*sqrt(2329)/(sqrt(2329)+27)

[@michaeldoerr5810]

This problem seems a lot similar to previus problems that had tangent 2theta. And not only that but I shall practice and time myself. And given that this problem seems like a refresher on how to use the tangent double angle identity and on how to combine with quadratic formula, I hope that this makes me part of the 10%.

[@devondevon4366]

Answer 346. 269 or 346.27 Pythagorean Theorem, Law of Sines (three times) then Heron's Formula. First, find the hypotenuse, AB = 48.26 Pythagorean Theorem (AB = sqrt (40^2 + 27^2) Second, Find angle A (2 thetas) using the Law of Sines. 48.26/sine 90 = 40/ Sine A Sine A = 40/48.26 * sine 90 Sine A = 0.88828 * 1 A= 55.980147 degrees Hence, 2 thetas = 55.980147 degrees Hence, angle B = 34.019852 degrees ( 180- (90 + 55.980147) Hence, theta = 55.980147/2 = 27.99007 degrees Hence, for triangle ACD, angle D = 62.01 degrees (180-90 + 27.99007) Let's use the Law of Sines to find the length AD: 27/ Sine 62.01 degrees = AD/sine 90 27* sine 90/sine 62.01 = AD 30.57 = AD For triangle ABD, let's find the length BD, again using the Law of Sines. BD/Sine 27.99007 = 30.57/ 34.019852 degrees BD = 30.577 * Sine 27.99007/sine 34.019852 BD = 25.64944 So the dimensions of triangle ABD are 25.64944, 30.577, and 48.26 Hence, the area using Heron's Formula is 346.269 Answer

[@spacer999]

Let a=BD, b=AB=√(40^2+27^2)=√2329. a/b=(40-a)/27 --> a=40b/(27+b) Area=27a/2=540b/(27+b)~346.27

[@quigonkenny]

Triangle ∆ACB: AC² + CB² = BA² 27² + 40² = BA² BA² = 729 + 1600 = 2329 BA = √2329 As AD is an angle bisector of ∠CAB, then BD/DC = AB/CA. Let BD = x. BD/DC = AB/CA x/(40-x) = √2329/27 √2329(40-x) = 27x 40√2329 - √2329x = 27x 27x + √2329x = 40√2329 x(27+√2329) = 40√2329 x = 40√2329/(27+√2329) x = 40√2329(27-√2329)/(27+√2329)(27-√2329) x = (1080√2329-93160)/(729-2329) x = (93160-1080√2329)/1600 x = (2329-27√2329)/40 Aᴛ = ((2329-27√2329)/40)(27)/2 Aᴛ = (27(2329)-729√2329)/80 Aᴛ = (62883-729√2329)/80 Aᴛ ≈ 346.27 sq units

[@prossvay8744]

Yellow area=1/2(27)(40)-1/2(27)(27√2329-729)/40=346.27 square units.

[@Irtsak]

Let AB=x Apply Pythagoras theorem in right triangle ABC and you will estimate x. x= √2329. Now AD is bisector so BD=(BC⋅AB)/(AB+AC)=(40⋅x)/(x+27) (known formula) Area of triangle (ABD) = (BD·AC)/2 = 20x/(x+27) Now young men Substitute x with √2329 and you have finished ! I am too old to do such things 😊

[@andrec.2935]

Permita-me uma sugestão: a bissetriz interna divide o lado oposto em segmentos proporcionais aos lados adjacentes ao ângulo. É uma alternativa a excelente solução apresentada!

[@alexkirchoff5286]

The puzzle in itself is almost trivial; it's that √2329 which makes the whole thing sooo annoying to deal with. What makes a lot of difference is if you have at your disposal a calculator or not. Anyway,even a non-scientific calculator should have a basic sqrt function.

[@quigonkenny]

You always have the option of leaving the solution in the exact format, so it can certainly be done without a calculator. But for me the hardest part was keeping track of the various multiples of 2329 required for calculation and not mistyping anything or leaving off a trailing 0 in a product (which I did once but caught).

[@user-sw9lb2zs6e]

27/x equals 48,26/(40-x) hence x=14,35

[@danmimis4576]

You draw a perpendicular from D on AB: problem solved.

[@murdock5537]

φ = 30° → sin⁡(3φ) = 1; ∆ ABC → AB = b; AC = 27 BC = BD + CD = (40 - a) + a; AD = k; sin⁡(BCA) = 1; CAD = DAB = θ → CAB = 2θ → cos⁡(2θ) = 27/b → b = √(40^2 + 27^2) → cos⁡(θ) = 27/k = √((1/2)(1 + cos⁡(2θ))) → k = 27/cos⁡(θ) → sin⁡(θ) = √((1/2)(1 - cos⁡(2θ))) → area ∆ ABD = (1/2)sin⁡(θ)bk ≈ 346,3

[@devondevon4366]

346.269

[@giuseppemalaguti435]

40/27=tg2θ=2tgθ/(1-(tgθ)^2)...tgθ=x/27...20/27=(x/27)/(1-x^2/729)...20(1-x^2/729)=x...20(729-x^2)=729x...20x^2+729x-14580=0...x=-729+√(729^2+1600*729)/40=(-729+27√2329/40..b=40-x=(2329-27√2329)/40..Ay=27b/2...mah,ho fatto i calcoli a mente...