An alternate simpler approach will be: after getting equation 1, consider triangle CFB, and get the equation 2 as, alpha + 2x = 90 deg.. then solve these two simultaneous equations.🙏🏼

Thank you!

Put point P on line AB so that CP bisects the angle C. Then there triangles ADP DCP CBP are congruent. Therefore angle C is 6x and x is 24

This isn't math it's meth

Thank you very much sir 😊

The aux lines were tricky but the algebra part became easy

Ótima solução! 🎉🎉🎉

Per il teorema dei seni risulta √(2a^2-2a^2cos4x)/sin2x=a/sin(5x-90)...a lato uguale che si semplifica,rimane l'equazione in incognita x..risulta cos5x=(-1/2).x=24

Let's find x: . .. ... .... ..... First of all we add point E on AB such that ADE is an isosceles triangle (AD=DE). Then we can conclude: ∠AED = ∠DAE = 3x ⇒ ∠ADE = 180° − ∠DAE − ∠AED = 180° − 3x − 3x = 180° − 6x ⇒ ∠CDE = ∠ADC − ∠ADE = 4x − (180° − 6x) = 4x − 180° + 6x = 10x − 180° Since AD=DE and AD=CD, the triangle CDE is also an isosceles triangle (CD=DE), so we obtain: ∠DCE = ∠CED = (180° − ∠CDE)/2 = [180° − (10x − 180°)]/2 = (180° − 10x + 180°)/2 = (360° − 10x)/2 = 180° − 5x ⇒ ∠BEC = 180° − ∠AED − ∠CED = 180° − 3x − (180° − 5x) = 180° − 3x − 180° + 5x = 2x = ∠CBE The triangle BCE is an isosceles triangle as well and we finally obtain: CE = BC = CD = AD = DE ⇒ CD = CE = DE Therefore the triangle CED is not only an isosceles triangle, it is an equilateral triangle and we can conclude: ∠DCE = ∠CED = 180° − 5x = 60° ⇒ 120° = 5x ⇒ x = 24° ∠CDE = 10x − 180° = 60° ⇒ 240° = 10x ⇒ x = 24° ✓ Best regards from Germany

Very good.

If you draw point E on segment AB such that ∠CEB＝9X and connect it to CE, ⊿CEB becomes an isosceles triangle. ∴CB＝CE. In this case, ∠BCE＝180-18X, so ∠DCE＝30X-180°. Also, since CD＝CE, triangle CDE becomes an isosceles triangle, and ∠CDE＝∠CED＝180-15X. Also, ∠DEA＝180-(∠CEB+∠CED)＝180-(9X+180-15X)＝6X, ∠DEA＝∠DAE＝6X, so DA=DE. Since DA=DC, CD＝CE＝DE and ⊿CED becomes an equilateral triangle. ∴∠CED＝180-15X＝60° ∴15X＝120°　X=8°.

@ 2:35 , droppin that perpendicular auxiliary is genius! ...imoh 🙂.

AC = 25 due to Pythagorean triple. Point E is the far right of the diameter (Thales) and CE is also 25. 24^2 = (25+7)^2 = 576 + 1024 = 1600. sqrt(1600) = 40 for d, so 20 for r. 25^2 - 20^2 = (CB)^2. (CB)^2 = 225 CB = 15 Checked video. Yep.

I just looked at it and came up with the answer. I thought that I made some logic flaw when I saw it was 10 minutes. I jumped to the end and saw my answer.

Wouldn’t triangle COA be a similar triangle to triangle CBA?

y+56 or y+96?

What math sofware do you use to draw the figer?

(8)^2(8)^2={64+64}={128 81}=209 {90°A90°B+90°C+90°D}=360°ABCD/209=1.159 3^17 3^8^9 3^8^3^2 1^2^3^3^1 1^1^2^3^1 2^3 (ABCD ➖ 3ABCD+2).

It's simplest to use the sine theorem.

π×20/sin45=π20✓2

And I Solved it withn a min in my mind 😂

Me watching it 30 min before exam 😂

sigma boy

Hmmm... Haven't we done this one before? As OC and OD are both radii of quarter circle O, OD = OC = 10. As A is the midpoint of OD, OA = AD = 10/2 = 5. Triangle ∆COA: OC² + OA² = AC² 10² + 5² = AC² AC² = 100 + 25 = 125 AC = √125 = 5√5 Mirror quarter circle O about OD, such that OE is the radius opposite OC. As C and E are opposite ends of a diameter and ∠B is a 90° angle on the circumference, then if BA is extended, it will pass through E, by Thales' Theorem. As OC = OE, ∠COA = ∠AOE = 90°, and OA is common, then ∆COA and ∆AOE are congruent triangles by SAS. As ∠ACO = ∠OEB and ∠COA = ∠EBC = 90°, then ∆EBC is similar to both triangles. BC/CE = OA/AC BC/20 = 5/(5√5) = 1/√5 BC = 20(1/√5) = 4√5 EB/BC = OC/OA EB/4√5 = 10/5 = 2 EB = (4√5)2 = 8√5 EA + AB = 8√5 5√5 + AB = 8√5 AB = 8√5 - 5√5 = 3√5 Triangle ∆ABC: A = bh/2 = BC(AB)/2 A = 4√5(3√5)/2 = 12(5)/2 = 30 sq units

Since size of circle is not specified, we can calculate this quickly if there is no need to show work. For circle with radius = 0, point O = A, so triangle is half the area of square: Area(△OBD) = 1/2 * (d^2/2) = 1/4 * 8^2 = 16 -------- Since I haven't seen this used here, I'll use coordinate geometry to find solution for all cases. Let x = side length of square, r = radius of circle. x^2 + x^2 = 8^2 → 2x^2 = 64 → x^2 = 32 → *x = 4√2* Using coordinate geometry with A at origin, we get: A = (0, 0), B = (4√2, 0), C = (4√2, 4√2), D = (0, 4√2), O = (−r, r) Line BD: x + y = 4√2 Triangle OBD has base *b* = BD *= 8* and height h = Perpendicular distance from O (−r, r) to line BD (x + y − 4√2 = 0) *h =* |−r + r − 4√2| / √(1^2 + 1^2) *= 4* *Area(△OBD) = 1/2 * b * h = 1/2 * 8 * 4 = 16*

I thought out of the box as Premath did. Instead of going the Pythagorean route, I used the proportions method. 5/5*sqrt5 = a/20. a = 4*sqrt5. Then you can see you have 2/3 of the sides needed for a Pythagorean triplet 3/4/5. Therefore the last remaining segment is 3*sqrt5. At this point just calculate the area same as Premath did.

My guy saving me from failing, THANK YOU SO MUCH

شكرا لكم على المجهودات يمكن استعلمال OBK قائم الزاوية في K من(AD) وOBH قائم الزاوية في Hمن(OC) منتصف [AC] هيI وILB قائم الزاوية فيL من( BH) OB^2=OH^2 + HB^2 IB^2 = IL^2 + BL^2 .......... OH=8 ; HB=6 AB^2=9×5 BC^2 =16×5 S=1/2 AB BC =30

Angle CAB is double angle OCA, so we can calculate AB and CB with trig identities. OCA = atan(5/10) b = 10/cos(OCA) a = b.sin(2OCA) c = b.cos(2OCA) Area = c.a/2‎ = 30

Draw radii AO & BO. We get a central angle ∠AOB, which subtends the same arc AB with ∠C, an inscribed angle. By the Measure of an Inscribed Angle Theorem, m∠AOB = 90°, so ∠AOB is a right angle. This makes △AOB an isosceles right triangle. Solve for the radius of ⊙O. c = a√2 20 = r√2 r = 20/(√2) = (20√2)/2 = 10√2 Find the circumference of ⊙O. C = 2πr = 2 * π * (10√2) = 20π√2 So, the circumference of the circle is 20π√2 units (exact), or about 88.86 units (approximation).

OC=OD=OF=10 => CF=2•10=20; OA=½•10=5 ΔAOF=Δ(1/2/vʼ5)(5/1) => AF=vʼ5(5/1)=5vʼ5 ΔCBF~ΔAOF(AA) => => ΔCBF=Δ(1/2/vʼ5)(20/vʼ5) => => BC=1(20/vʼ5)=4vʼ5; BF=2(20/vʼ5)=8vʼ5 AB=BF-AF=8vʼ5-5vʼ5=3vʼ5 [ABC]=½AB•BC=½3vʼ5•4vʼ5=30 sq.un. 😁

OCA has sides 5, 10, 5*sqrt(5). ABC's hypotenuse is 5*sqrt(5). <OAC = 63.435(rounded) <CAD = 118.565 BA extended to the end of a new quadrant will have length 5*sqrt(5) + AB (call this x), so x + 5*sqrt(5). Call CB, y. (x+5*sqrt(5))^2 + y^2 = 400 and x^2 + y^2 = 125 x^2 + 10x*sqrt(5) + 125 + y^2 = 400 (equation 1) x^2 + y^2 = 125 (equation 2) Re-jig (1): x^2 + 10x*sqrt(5) - 275 = -y^2 Re-jig (2): x^2 - 125 = -y^2 Therefore, x^2 + 10x*sqrt(5) - 275 = x^2 - 125 Then: 10x*sqrt(5) - 275 = -125 Then: 10x*sqrt(5) - 150 = 0 Then: 10x*sqrt(5) = 150 Then: x*sqrt(5) = 15 Then: x = 15/sqrt(5) Rationalise to x = (15*sqrt(5))/5 = 3*sqrt(5) ABC: Hypotenuse = 5*sqrt(5). Short side = 3*sqrt(5). Using 3,4,5, y is 4*sqrt(5) (3*sqrt(5))*(4*sqrt(5)) = 12*5=60 Green area is half that so 30un^2 I bet you did it much more simply LOL. Yes, I did complicate it a bit, but I went simpler than you by noting that the green area was a 3,4,5. My calculating angles at the beginning was wasteful of my limited brainpower.

Once known one of the 2 cathetii you can see that sides of ABC are a pythagorean triplet with factor √ 5

F in AB?

Sir We just complete a semicircle with a diameter CF and join BF. Now 🔺s FCB & FOA are similar. [ argument -- I) ang CBF =ang AOF (both r 90 degs II) ang CFB = ang OFA (same angle) and ang FCB =ang OAF (the third angle) ] Hence CF /AF =BC /OA=BF/OF Now take the second and the third ratios we get BC/5 =BF/10 >BC /BF =5/10=1/2=x/2x Now rt 🔺 FCB BC ^2 +BF ^2 = 20^2 > 5x^2=20^2 > x ^2=400/5=80 Area of 🔺 FCB =1/2*x*2x =x^2=80 -- (1) Area of 🔺 AFC =1/2*20*5 =50 -(2) Area of green triangle = (1)-(2)=30 sq units.

Thank you!

I started the same way by extending the quarter circle and because angle B is 90 degrees, extending BA to the other side of the circle (point E) must form a 180 degree angle on COE by the Interior Angle Theorem. Then I used coordinate geometry to show that B was at position (8,6) relative to O. The line EA had slope 2 and intersection at E (0, -10) to give y = 2x - 10, and it intersects with the circle x^2 + y^2 = 10^2. Now I have two new triangles: ECB and ECA, the difference of which is the triangle ABC. ABC = ECB - ECA ABC = 1/2 b(EC) h(B) - 1/2 b(EC) h(A) ABC = 1/2 * 20 * 8 - 1/2 * 20 * 5 ABC = 80 - 50 ABC = 30 square units

30

32.151699

❤ excellent job sir AB = c , BC = a a^2+c^2 = 125 OABC is cyclic quadrilateral hence 5a+10c = (5√5)(10) a = 10√5 - 2c c^2 - 8√5 c +75 = 0 c = 3√5 , 5√5 c = 3√5 because hypotenuse is 5√5 then a = 4√5 Area = 30

CA=√(10²+5²)=5√5→ Si CBAº=90º→ Recta BA pasa por P, extremo inferior del diámetro vertical y por simetría AC=AP → Potencia de A respecto a la circunferencia =5*(5+10)=AB(5√5)→ AB=3√5 y AC=5√5→ CB=4√5→ Área ABC =AB*BC/2=3*4*5/2=30 u². Gracias y saludos.

Otima solução professor! 🎉🎉🎉 Uma outra forma, possível: Sen F=AO/OF = 5/5√5 = 1/√5. Daí, Sen F = BC/CF 1/√5= BC/20 → *BC = 20/√5.* se cos² F + sen² F = 1 Cos ² F = 1 - 1/5 = 4/5 Cos F = 2/√5. Cos F = BF/FC = BF/20= 2/√5 BF= 8√5 → AB= BF - AF AB = 8√5 - 5√5 → *AB = 3√5* A[ABC] = BC× AB/2 A[ABC] = 20/√5 × 3√5/2 *A[ABC] = 30 unidades quadradas*

Solution: In Triangle OAC we have a right triangle, therefore we will applying Pythagorean Theorem: 5² + 10² = AC² 125 = AC² AC = 5√5 Drawing some auxiliary lines, we will going to use Chords Theorem, after Thales's Theorem 15 × 5 = 5√5 × AB AB = 3√5 In Triangle ABC, once again, we will applying Pythagorean Theorem (3√5)² + BC² = (5√5)² 45 + BC² = 125 BC² = 125 - 45 BC = √80 BC = 4√5 Green Triangle Area = ½ (3√5) × (4√5) Green Triangle Area = ½ 60 Green Triangle Area = 30 Square Units ✅

tan α = 1/2 --> α = 26,565° β = 90° - α = 63,435° Segment CA: a = 10/cosα = 10*√5/2= 5√5cm Segment CB: b = a cos (β-α) = 4/5 a = 4√5cm Shaded area: A = ½ a.b.sin(β-α) A = ½.(5√5).(4√5).(3/5) A = 30 cm² ( Solved √ )

We use an orthonormal center O and first axis (OD). We have A(5: 0) C(0; 10) B(10.cos(t); 10.sin(t)) with t = angleDOB (0°<t<90°). VectorAB(10.cos(t) -5; 10.sin(t)); VectorCB(10.cos(t); 10.sin(t) -10). They are orthogonal, so (10.cos(t) -5).(10.cos(t)) + (10.sin(t)).(10.sin(t) -10) = 0 We develop and simplify by 100 and obtain 1 -sin(t) -(1/2).cos(t) = 0 or sin(t) = 1 - (1/2).cos(t) As (sin(t))^2 + (cos(t))^2 = 1, we have 1 -cos(t) + (5/4).(cos(t))^2 = 1 or (5/4).(cos(t))^2 - cos(t) = 0, which gives that cos(t) = 4/5 (as cos(t)<>0) That gives easily that sin(t) = 3/4, and so we have B(8; 6), VectorAB(3; 6), AB^2 = 45, AB = sqrt(45), VectorCB(8; -4), CB^2 = 80, CB = sqrt(80) Finally the green area of trangle ABC is (1/2).AB.CB) = (1/2).sqrt(45).sqrt(80) = (1/2).sqrt(3600) = (1/2).60 = 30.

Hot problem

1/ AC= 5sqrt5 (Pythagorean theorem) 2/ Draw the full circle: ABx5sqrt5= 15x5 ( chord theorem)-> AB = 3 sqrt5 -> BC= 4sqrt5 ( triangle ABC is a 3/4/5 triple) 3/ Area of the green triangle = 1/2x 3 sqrt5x4 sqrt5= 30 sq units😅😅😅

4:48 △AOF~△CBF(AA) AO:AF=1:sqrt5=CB:CF →CB=4sqrt5→AB=3sqrt5 △ABC=CBxAB/2=30😊

Once again Thales and Euclid have shown themselves trustworthy and I am confident too say that trust will never be betrayed. 🙂