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Thank you professor for very nice solution.
This is a problem of Stewart's theorem in disguise. The mnemonic for Stewart's theorem is "DAD is the MAN who puts the BoMB into the s(C)iNk(C)." The equation is dad + man = bmb + cnc with the usual notations for the sides such that a = m + n, c is over m and b is over n in the triangle. The area of rectangle in this problem is just (an) in the equation. Hence transform the equation into an = (b^2 - d^2) + (c^2 - d^2)(n/m) In this problem a = m + n, b = 5, c = 3, d =3. Then an = (5^2 - 3^2) = 16. Note that n/m has no fixed value in this problem.
Nice method.
The disparity between the assertions made and the picture distracts from following the solution. An accurate figure would be so useful.
I think many of us will be able to get to the diagram at 2:35. When we then use the Pythagorean theorem to write equations, we have 2 equations, b² + h² = 3² and (a+b)² + h² = 5², but 3 unknowns, a, b, and h. We can't find a third equation. However, we note that we are solving for the area of the rectangle, a(a+2b), and do not need to solve for a, b and h separately. In fact, there may be many sets a. b and h which produce the same value for area = a(a + 2b). I like to try special cases. Let h = 3, collapsing ΔABM and ΔAFM into line segments of length 3, as alexbayan8302 did. Then, ΔACM is a right triangle with side 3 and hypotenuse 5, so the other side must be 4 (3 - 4- 5 Pythagorean triple). So, a = 4, b = 0, and BCDE is a square, which is a special case of a rectangle, and its area is 4² = 16. If this were a multiple choice test, we would assume that the area of the rectangle does not change as h varies over a valid range, and mark 16 as our answer. To support our suspicion of other solutions for a, b and h, let's let h = 2. Then, b = √5 and (a + b) = √(21), so a = √(21) - √5 and (a + 2b) = √(21) - √5 + 2√5 = √(21) + √5. Area of square = a(a + 2b) = (√(21) - √5)(√(21 + √5) = (√21)² - (√5)² = 21 - 5 = 16. We get the same result, 16, supporting our suspicions. So, we see if our set of 2 equations will produce an integer value for a(a + 2b):. b² + h² = 3² = 9, so h² = 9 - b². (a + b)² + h² = 5² expands to a² +2ab +b² + h² = 25, h² = 25 - (a² +2ab +b²) and h² = 25 - a² -2ab -b². Equating h², 9 - b² = 25 - a² -2ab -b² and a² + 2ab =16. We factor a² + 2ab into a(a + 2b) and find that the area of the rectangle is 16 for all valid (a, b) pairs. (Pairs are valid if a and b are both positive and produce a positive h which satisfies the 2 equations provided by the Pythagorean theorem.)
Let FC=x ,BF=y Apply Stewart’s theorem in triangle ABC: AB²⋅FC+AC²⋅BF=BC⋅(AF²+BF⋅FC) 9x+25y=(x+y)(9+xy) 9x+25y=9x+x²y+9y+xy² => x² y+xy²=16y => x²+xy=16 (y>0) x(x+y)=16 => area (BCDE)=16 square units
This is a problem of Stewart's theorem in disguise. The mnemonic for Stewart's theorem is "DAD is the MAN who puts the BoMB into the s(C)iNk(C)." The equation is dad + man = bmb + cnc with the usual notations for the sides such that a = m + n, c is over m and b is over n in the triangle. The area of rectangle in this problem is just (an) in the equation. Hence transform the equation into an = (b^2 - d^2) + (c^2 - d^2)(n/m) In this problem a = m + n, b = 5, c = 3, d =3. Then an = (5^2 - 3^2) = 16. Note that n/m has no fixed value in this problem.
3rd method: Calculate areas of triangles AFE, BCF and DEC, then substract them from the full square area to get S[CEF]=9x²-3x²/2-3x²/2-9x²/4=15x²/4. By Pythagoras EC=√10x and FC=3√5x/2. Draw a height h from F to base EC. then h(EC)/2=15x²/4, h√10x/2=15x²/4 then h=3√10x/4. By Pythagoras calculate segment m at EC from C to the height, to get m=3√10x/4. Then m=h and obviously θ=45º.
If this is just a short answer question, collapse AB and AF then AFC is a right triangle with sides 3-4-5 and CD=FC=BC=4 so 4*4=16. To prove it for the general case is simple too if you put angle ABF=AFB=theta;angle ACF=phi. BC = 3*cos theta + 5+ cos phi ; FC = 5*cos phi - 3* cos theta Also 3 * sin theta = 5 * sin phi; square both sides and 16=25 - 9=25 * (cos phi)^2 - 9 * (cos theta)^2 = BC * FC = BC * CD This is essentially the same as your solution.
Col teorema dei seni e coseno risulta FC=√10,BF=6/√10,CD=√10...Arett=(√10+6/√10)√10=16...ma non sono sicuro
Let DC = CF = x. Drop a perpendicular from A to BF at G. Let AG = h. As AB = FA, ∆FAB is an isosceles triangle, so AG bisects ∆FAB and creates two congruent right triangles, ∆BGA and ∆AGF. Therefore, BG = GF = y. Triangle ∆BGA: BG² + AG² = AB² y² + h² = 3² h² = 9 - y² Triangle ∆AGC: AG² + GC² = CA² h² + (x+y)² = 5² h² = 25 - x² - 2xy - y² 9 - y² = 25 - x² - 2xy - y² x² + 2xy = 16 x(x+2y) = 16 Note that the length of ED, which is the width of rectangle BEDC, is equal to x+2y. Also, the length of DC, which is the height of rectangle BEDC, is equal to x. Therefore, the area of rectangle BEDC is equal to x(x+2y), which is equal to 16.
Muito boa, essa!
Math Booster, I subscribed because your videos are super cool!
sin ( 2 z) = cos (3 z) 2 sin ( z) cos ( z) = 4 cos^3 ( z) - 3 cos ( z) 2 sin ( z) = 4 ( 1 - sin^2 ( z)) - 3 4 sin^2 ( z) + 2 sin ( z) = 1 2 sin ( z) =√( 1 + 1/4 ). - 1/2 sin ( z) =( √5 - 1) /4
This is probably the first time I have intuitively understood how HL congruency can be used to solve the area of a rectangle. All that is required is to draw the midpoint in the first triangle so that HL is justified and then the h is indented. After doing some algebra and Pythagoras Theorem is it simply substituting in what b is.
I actually meant substitute in h squared after constructing the triangle where the side adjacent the right angle is a + b. I forgot point that out.
S=16
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*I want to prove that θ=45.* *It is enough to prove that θ is an acute angle of an isosceles right triangle* !!!! EK⊥FC construction Area (EFC)=(ABCD)-(AEF)-(BFC)-(EDC) = 9x²-(3x²)/2-(9x²)/4-(3x²)/2 (EFC)=(15x²)/4 (1) In orthogonal triangle apply Pythagoras theorem : FC=(3√5 x)/2 Now area (EFC)=(EK⋅FC)/2 => (EFC)=EK (3√5 x)/4 (2) (1),(2)=> EK (3√5 x)/4 = (15x²)/4 => *EK= x√5* (3) Apply Pythagoras theorem in ΔEKC => KC²=EC²-EK² =10x²-5x² *KC=x√5* (4) (3),(4) => EK=KC => right triangle EKC is isosceles = > θ=45° finish
2:56 tan(alpha) ....You are using trigonometry, the video title said without trigonometry !
Ιf I was a civil engineer 😊 Apply cosines formula in triangle EFC: EF²=EC²+FC²-2EC⋅FC cosϑ => 2⋅EC⋅FC⋅cosϑ = EC²+FC²- EF² (1) EC²=10x² => EC=x√10 FC²=9x²+9x²/4=45x²/4 => FC=(3x√5)/2 EF²=4x²+9x²/4= 25x²/4 (1)=> 2⋅ x√10⋅(3x√5)/2 ⋅cosϑ= 10x²+45x²/4 - 25x²/4 => 3x²⋅5√2 ⋅cosϑ=15x² => cosθ= 1/√2 => cosθ=√2/2 => θ=π/4
Wtf only now you are saying ABCD is a square. Why did you not state it originally asso
Should the statement of the problem not specify that circle is tangent to 3 sides of rectangle, and that point P is a point of tangency? Otherwise we are making all sorts of assumptions to solve.
Area of ABC = sqrt 3 a^2/2 Area of ABD= sqrt 3*ax/4 Area of ADC = sqrt 3*ax/2 sqrt 3 a^2/2 = sqrt 3*ax/4 + sqrt 3*ax/2 a^2/2=ax/4+ax/2 2a^2=ax+2ax 2a^2=3ax 2a=3x a=3x/2 law of cosines in ABD 4=a^2+x^2-2ax cos 60º 4= (3x/2)^2+x^2-2(3x/2)x*1/2 (4= 9x^2/4 +x^2-3x^2/2)×4 16=9x^2+4x^2-6x^2 16=7x^2 16/7=x^2 x=4/sqrt 7
Area of ABC = sqrt 3 a^2/2 Area of ABD= sqrt 3*ax/4 Area of ADC = sqrt 3*ax/2 sqrt 3 a^2/2 = sqrt 3*ax/4 + sqrt 3*ax/2 a^2/2=ax/4+ax/2 2a^2=ax+2ax 2a^2=3ax 2a=3x a=3x/2 law of cosines in ABD 4=a^2+x^2-2ax cos 60º 4= (3x/2)^2+x^2-2(3x/2)x*1/2 (4= 9x^2/4 +x^2-3x^2/2)×4 16=9x^2+4x^2-6x^2 16=7x^2 16/7=x^2 x=4/sqrt 7 x= 4*sqrt 7/7
ABD is similar to ABC since both have theta and 90º angles so AB/BC=BD/AB AB^2=3x*x AB=x*sqrt 3 tan @= opp/adj = x*sqrt 3/3*x= sqrt 3/3 arctan (sqrt 3/3)= arctan (tan @) since 0<@<90 30º=@
Outstanding problem with 2 elegant solutions. Success wjith problems of this caliber require Dedication/Practice, starting with easier ones, gradually increasing the difficulty. Inherent problem solving ability and being able to process at higher levels surely help, but they don't guarantee anything.
φ = 30°; ∆ ABC → AB = x + y; AC = y; ABC = θ = ?BC = BD + CD = z + x; AD = a; ADC = 8φ/3 DCA = 4φ/3 → CAD = 2φ; CT = AC + TA = y + a → ∆ ATD → TA = DA → DAT = 6φ - 2φ = 4φ → ATD = TDA = φ AB = AM + BM → sin(DMB) = 1 → MDC = φ + 8φ/3 = 11φ/3 → BDM = (φ/3)(18 - 11) = 7φ/3 → θ = 3φ - 7φ/3 = (φ/3)(9 - 7) = 2φ/3 → TBA = ABD = θ or: φ = 30°; ∆ ADC → AD = a; CD = x; AC = y = AS + CS → sin(DSA) = 1; ADC = 8φ/3 DCA = DCA = 4φ/3 → CAD = 2φ → ADS = φ → AS = a/2 → DS = a√3/2 → CS = y - a/2 ∆ CSD → DCS = 4φ/3; CSD = 3φ → SDC = 5φ/3 CD = x → sin(4φ/3)/(a√3/2) = 2sin(4φ/3)/a√3 = 2√3sin(4φ/3)/3a = sin(5φ/3)/(y - a/2) = 2cos(4φ/3)/(2y - a) = 1/x → x = a√3/2sin(4φ/3) (2y - a)/2cos(4φ/3) = a√3/2sin(4φ/3) → y = (a/2sin(4φ/3))(cos(4φ/3)√3 + sin(4φ/3)) → x + y = (a/2sin(4φ/3))(sin(4φ/3) + √3cos(4φ/3) + 1) ∆ ABC → BC = BN + CN → sin(BNA) = 1 → AN = h → sin(4φ/3) = h/y → h = ysin(4φ/3) sin(θ) = h/(x + y) → h = (x + y)sin(θ) = ysin(4φ/3) → sin(θ) = ysin(4φ/3)/(x + y)↔ m ∶= sin(4φ/3) + cos(4φ/3)√3 ≈ 1,9696↔sin(θ) = msin(4φ/3)/(m + √3) ≈ 0,342 → θ = 2φ/3 🙂
Rotate a copy of the Square by 90° Clockwise about C. EF = FE', and ∠E'CF=90-θ Since ∆ECF is Congruent to ∆E'CF by SSS, 90-θ = θ Hence, θ = 45°
Nice solution, but you have to show that EF = FE', which is not too complicated: EF is hypotenuse of △AEF, where AF = 3X/2, AE = 2X = 4X/2.This is a 3-4-5 triangle with EF = 5X/2 Since D' = B, then FBE' is a straight line, and FE' = FB + BE' = FB + D'E' = FB + DE = 3X/2 + X = 5X/2 And of course CF = CF and CE = CE' (since we are rotating about C, distances to C remain unchanged)
Join E toF. Find the areas of the 3triangles afe,fbc and edc and add.That gives (21/4)x^2. The area of ABCD=(9) x^2. Therefore the area of triangle efc equals (15/4)x^2. |ec|=root 10 times x and |fc|= (root 45 times x)/2. Area of triangle efc= ( root10/2)x mult. By (root 45/2)x mult by sine theta. Equate the two values for area of triangle efc and you get (root 450/4)by (x^2) by sine theta= (15/4)by(x^2). So sine theta = 1/(root2) giving theta = 45 degrees.
Rataa marke aaye ho kya bhai... bas direct amswer de rhe ho
Rotate triangle CDE 90 degrees around point C. Point E will move to position E1. Obviously, ∡ECE1 = 90°. FE1 = x/2 + x/3 = 5x/6. By the Pythagorean theorem: EF = sqrt(x^2/4 + 4x^2/9) = 5x/6. It follows that EFE1C is a deltoid. The diagonal CF is the bisector of the angle EСE1, so Θ = 90°/2 = 45°.
(10√5/7)^2
Как видно из решений, второй метод дополнительных графических построений, даёт быстрый 100% результат решения, так как он наглядный и более понятный для восприятия. сложнее найти точные значения прилежащих углов a и в, так как тригонометрический калькулятор на даст точные значения этих углов по тангенсам, синусам, и прочей дребедени. Если при расчётах космических полётов, ошибиться на тысячную долю угла запуска, относительно плоскости эклиптики, то без коррекции орбиты полёта. можно улететь не туда!
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At 11:33, QM has been determined to be equal to X. Since PQ = 3X, that leaves PM = 2X. Construct ME. ΔPME and ΔQCM are congruent by side-angle-side (PE = QM = X, <EPM = <MQC = 90°, PM = QC = 2X). By congruency and corresponding sides, ME = CM. Therefore ΔEMC is isosceles. Let <PEM = α. Then, by corresponding angles of congruent triangles, <QMC = <PEM = α. Let <PME = ß. Then, by corresponding angles of congruent triangles, <QCM = <PME = ß. The sum of the acute angles in a right triangle is 90°, so α + ß = 90°. Note that <QMC + <EMC + <PME is a straight angle or 180°, so α + <EMC + ß = 180°, substitute 90° for α + ß and 90° + <EMC = 180°, therefore <EMC = 90°. ΔEMC is already found to be isosceles and is now found to be a right triangle, so it must be an isosceles right triangle, which has the properties that both acute angles are 45°. Therefore, Θ = 45°.
Without loss of generality put AF=FB=3; AE=4; ED=2 Area of triangle EFC = 6*6 - 1/2 * (3*6+3*4+2*6)= 15 Let M be the perpendicular foot from E to FC then 15 =0.5 * FC * EM; Since FC= sqrt(9+36), EM= 30/sqrt45 = 10/sqrt 5 = 2*sqrt 5 ; EC = sqrt(4+36)= 2 * sqrt 10 = EM * sqrt 2 So Theta is 45 degree(since CM^2=EC^2-EM^2 => CM=EM if you absolutely want to avoid trig). I was really not going to comment here but it is a simple problem that can be deduced almost by inspection.
Also note that in the second method Triangles PME is congruent to Triangle CQM so angle PME + angle CMQ=90; angle EMC=90 and EM=MC; Theta is 90/2= 45 degree. This involves very little calculation whatsoever and not even Pythagorean.
well, i have 6 squares, 3 x 2; and the lines to be extended to the corners, then, at a glance
I think that I can do that WITHOUT trigonometry. And I think that this looks easier than expected. All that you have to do is to make a chord that intersects with on of the other chords, subtend right angles, then use the angle bisector extension of the circle theorem to use AA similarity of two right angled triangles, then make second and chord that forms two CONGRUENT triangles that are congruent simply by the definition of what a supplementary angle is. The congruency is justified from applying the Pythagoras theorem twice AFTER you determine the sides of the two similar triangles. I think that that is the best summary that can think of.
Also the first chord has to have the same sides in order for right angles to be subtended. And you have to use the definition of a supplementary angle once you establish equal sides and angles. And this is done in order to check the triangle that subtends theta is equilateral equiangular or a right scalene triangle. And because the right scalene pair has congruent angles, you need the definition of a supplementary angle, correct?
asnwer= 45
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From Pythagoras and Thales theorems : BC^2 = AB^2 + AC^2 = 9^2 + 18^2 ➔ BC = 9sqrt(5) AM/AB = AN/AC ➔ AN / AM = AC/AB = 18/9 = 2 Now use of trig, express the tangent of corresponding angles tan(B) = MP/BP = a/ BP and tan(B) = AC/ AB = 18/9 = 2 ➔ BP = a/2 tan(C) = OC / ON = OC/a and tan(C) = AB / AC = 9 / 18 = 1/2 ➔ OC = 2a BC = 9sqrt(5) = BP + PO + OP = a/2 + a + 2a ➔ a = 18sqrt(5) / 7 ➔ a^2 = 1620 /49
φ = 30°; ∆ ABC → AC = 3; BC = CD + BD = x + 5; sin(ACB) = 1; AB = ? DAC = α; BAD = 3φ/2; AD = y = DE → AB = AE + BE = y√2 + BE sin(ADE) = 1; CBA = θ → DF = 3 → BF = 2 → sin(θ) = √5/5 = 3/AB → AB = 3√5 or: tan(θ) = 3/(x + 5) = x/2 → x = 1 → y = √10 → AB = 2√5 + √5 or: BAC = 3φ/2 + α → CBA = 3φ/2 - α → tan(3φ/2) = tan(α + θ) → tan(α) = 1/3 → tan(θ) = 1/2
Дополнительное построение показывает, что AD, является средней линией треугольника ВСЕ, СЕ=5, АD=5/2. По теореме Пифагора, находим Х. Х^2=5^2+(5/2)^2=125/4. Х=5\/5/2!
It can be done using sine law x/sin(60) = y/sin(80) x/y = sin(60)/sin(80) (x+y)/sin(40) = y/sin(theta) (x+y)/y = sin(40)/sin(theta) x/y + 1 = sin(40)/sin(theta) sin(60)/sin(80) + 1 = sin(40)/sin(theta) (sin(60) + sin(80))/sin(80) = sin(40)/sin(theta) sin(theta)/sin(40) = sin(80)/(sin(60) + sin(80)) sin(theta) = sin(80)sin(40)/(sin(60) + sin(80)) sin(theta) = sin(80)sin(40)/(2sin(70)cos(10)) sin(theta) = sin(80)sin(40)/(2sin(70)sin(80)) sin(theta) = sin(40)/(2sin(70)) sin(theta) = sin(40)/(2cos(20)) sin(theta) = 2sin(20)cos(20)/(2cos(20)) sin(theta) = sin(20) , but we know that theta must be acute angle so the only one solution is theta = 20 degrees
∆ ABC → AB = 9; AC = 18; sin(CAB) = 1 → BC = 9√5 ∎MNOQ → MN = NO = OQ = MQ = a; AB = BM + AM = 9; AC = AN + CN = 18; BC = BD + a + CO QMB = BCA = δ → tan(δ) = 1/2 → CO = 2a → BD = a/2 → BC = 7a/2 = 9√5 → a = 18√5/7 → a^2 = 5(18/7)^2
Same method using minimal symbols while presentation remains complete and clear (OK as exam answer). 1. Let side of square be a, segment of BC to left of square be x and segment of BC to right of square be y. Hence BC = x + a + y. In triangle ABC, BC = 9√5 (by Pythagoras theorem) Hence x + a + y = 9√5 2. Angles B and C are complementary angles of the right-angled triangle ABC. 3. Lower triangles on left and right side of square are right-angled triangles as sides of square give them right angles. 4. These 2 triangles have the same complementary angles B and C by rule of angle sum of triangles equal 180. Hence they are similar to triangle ABC (AAA). 5. Ratio (side opposite angle B : side opposite angle C) is equal in triangle ABC, left triangle, right triangle as follows: 18/9 = a/x = y/a Hence x = a/2 and y = 2a 6. Substitute x = a/2 and y = 2a into x + a + y = 9√5. 2a + a + a/2 = 9√5 (7/2) a = 9/√5 Hence area of square a^2 = [(9/√5)(2/7)]^2 = 1620/49
Sorry, typing mistake in step 6, there should be no / in 9/√5.
Let CE =b, CB = a. The equation of the line EA (С(0, 0)): aу - (a - b)x - ab = 0. Coordinates of the center of the circle: O(r, r) The distance from the point O to the line EA is equal to the radius: r = [-a∙r + (a - b)∙r + a∙b]/√[a^2 + (a - b)^2]. Hence we get a quadratic equation whose solution is : r = b∙[√[a^2 + (a - b)^2] - b]/[2∙(a - b)]. Substituting a =4, b = 1 gives : r = 1∙[√[4^2 + (4 - 1)^2] - 1]/[2∙(4 -1)] = 2/3 .
Or, side * (18/9 + 9/18 + 1) = hypotenuse which is: 20.12 Thus, Side = 5.74 Area: Side^2 = 33.06 Formula's a real time saver.
From the Pythagorean theorem: BC=9*sqrt(5). Note that all the triangles in the graph are similar. For this BP=1/2 MP or BP=a/2 , analogously OC=2 ON or OC=2a. So, BP+PO+OC=BC or a/2 + a + 2a = 9*sqrt(5) so, a= 18*sqrt(5)/7 and Area = a * a = 1620/49 .
If |AM|= a then |AN| = 2a.Then |BM|= 9-a and |NC|= 18-2a. Also |BC|=9 root 5. The 3 small triangles are similar with the triangle ABC. So using triangles ABC and MBP you get 9 root5\ (9-a)=18/x where x = length of side of square. Taking the triangles ABC and NOC we get (9root 5)/x=9/a. Then working that out you get [18-(root5)x]\2= x/(root 5). So x = (18times root 5)/7. X^2= 1620/49.