Can you solve this 10th grade tricky geometry problem

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[@soli9mana-soli4953]

Great!! Very nice explanation! An alternative solution is applying II Euclid's theorem on ADB and ACB right triangles considering their heights DH and CK. Triangle DEH is an 30°,60°,90° right triangle so setting EH = a, follows that DE = 2a and DH = a√ 3, then (a√ 3)² = (4 - a)*(12 + a) (Euclid's th. on ADB) a = √ 13 - 1 => 2a = DE = 2√ 13 - 2 Triangle CEK also is an 30°,60°,90° right triangle so setting EK = b, follows that CE = 2b and CK = b√ 3, then (b√ 3)² = (12 - b)*(4 + b) (Euclid's th. on ABC) b = √ 13 + 1 => 2b = CE = 2√ 13 + 2 Now applying cosine law on DEC with angle DEC = 60° then cos 60° = 1/2 we can write: X² = (2√ 13 + 2)² + ( 2√ 13 - 2)² - 2*(2√ 13 + 2)*(2√ 13 - 2)*1/2 X² = 64 => X = 8

[@giuseppemalaguti435]

Il raggio R=8,EC=6 col teorema dei seni..l'equazione risolutiva è quella degli angoli in D e C come termini del raggio...essi sono ovviamente uguali .arcsin 3√3/x-arcsin√3/4=120-arcsin3√3/x+arcsin√3/4….risulta facilmente x=8

[@d.m.7096]

Can be solved using EPT (Extension of Pythagoras Theorem) Let M be the center. Hence, MD = MC = 8. In ∆DEM, DM = 8 = side opposite to 60°. By EPT, DM^2 = EM^2 + ED^2 - EM*ED Hence, 64 = 16 + DE^2 - 4*DE Solving this quadratic we get, DE = √52 + 2 Similarly, in ∆CEM, CM is side opposite vto 120°. Hence, by EPT, CM^2 = EC^2 + EM2 + EC*EM Hence, 64 = EC^2 + 16 + 4*CE Solving this quadratic gives CE = √52 - 2 In ∆CDE, CD = x is opposite to 60°. Hence by EPT, x^2 = CE^2 + DE^2 - CE*DE Substituting above values of CE and DE, We get x = 8

[@comdo777]

asnwer= 9.5 cm 120

[@АрнольдЕльбиздыкоевич]

8

[@khaitomretro]

This is one of the semi-circle theorums I was taught many years ago. Providing the two angles either side of the angle DEC are the same (i.e. the bisector of DEC is perpendicular to AB) then X remains the same lenght regardless of where E is along the line AB, so it equals the length you'd get if E was at the circle's origin. In this instance DEC is 60° so X equals the radius. If the outer angles are 45° and DEC 90° then X would be radius * ✓2. Simple.

[@antonioruzzarin4550]

Qqqq4454555qw

[@kextrz]

WTF? Diameter is 4 + 12 = 16, radius is 8, so EO is 8 - 4 = 4. The problem is that EO is also supposed to equal 8 / tan (60°), or 8 × tan (30°), which is 4,6188. Unless I'm missing something, this is the worst blunder I've ever seen in a math video.

[@quigonkenny]

Assuming O is the center, EO is going to be 4. The angles at any other points are immaterial. Now, if you want to drop a perpendicular from C to AB and measure from E to that point, then 4.6188 sounds about right. But that's not going to be O. OC isn't perpendicular to AB, even though it kind of looks that way on the diagram. Thankfully it doesn't need to be for CD to be a 60° arc and thus for x to equal the radius.

[@kextrz]

@@quigonkenny You are correct. The graph is visually misleading. The right side of the equilateral triangle is leaning right, 4,34° past vertical. According to my calculations, segment AC is 11,734; BC is 10,877; CE is 9,211. The angle opposite segment CE is 85,86°.

[@holyshit922]

I used cosine law 7 times (two times it was the special case of cosine law - Pythagorean identity) ACB right triangle and Pythagorean identity in this triangle Cosine law in triangle AEC and in triangle CEB with supplementary angles (120 and 60) Add up results from cosine law and i have quadratic equation ADB right triangle and Pythagorean identity in this triangle Cosine law in triangle AED and in triangle DEB with supplementary angles (60 and 120) Add up results from cosine law and i have quadratic equation Cosine rule in triangle EDC

[@AntoninaShushkova]

When we find inscribed angleDFC=30, we can use 2R= x/Sin 30. R=8, so x=8

[@sergeyvinns931]

DC=2\/15!

[@dickroadnight]

I think this is a simple coordinate geometry problem.