Can you find the length AB? | (Two identical semicircles with radius 3) |

[@Aryan_1Official]

Love you sir ! You are awesome. Don't stop to make such video.❤😊😊😊

[@PreMath]

Thank you so much , dear😀 You are awesome. Keep it up ❤️

[@manishsurwade5803]

This is the only maths channel which make us do Non-routine maths❤❤ Keep us engaged😊 Thank you sir

[@PreMath]

It's my pleasure, dear❤️ Thanks for your feedback! Cheers! 😀 You are awesome. Keep it up 👍

[@marioalb9726]

Pytagorean theorem: (2R)² = R² + x² x² = 4R² - R² = 3 R² = 27 x = √27 cm Pytagorean theorem: (AB)² = R² + (2R + x)² (AB)² = 3² + (2.3 + √27)² AB = 11,591 cm ( Solved √ )

[@marioalb9726]

sin α = R / 2R = 1/2 α = 30° ; ½α = 15° sin ½α = 3 / x x = 3 / sin 15° x = 11,591 cm ( Solved √ )

[@Abby-hi4sf]

Just let you know, you are great teacher. I love the action plan. It helps to follow without wondering.

[@HappyFamilyOnline]

Great video 😊😊👍👍 Thanks for sharing

[@williamwingo4740]

Draw a vertical line through the contact point of the two semicircles; call it point R. This line will be perpendicular to PB and AD. Call the upper point S and the lower point T. Now draw radius QR to form right triangle QTR. QR = 3 and TR = 1.5, so the height is half the hypotenuse. Therefore this triangle is 30-60-90 and QT = (1.5)(sqrt(3)). So AT = 3 + (1.5)(sqrt(3)). Next consider right triangle ART. the base is 3 + (1.5)(sqrt(3)) and the altitude is 1.5; Invoking Pythagoras, we have AR^2 = (3 + (1.5)(sqrt(3)))^2 + (1.5)^2) = (5.5981)^2 + 2.25 = 31.3387 + 2.25 = 33.5887. So AP = sqrt(33.5887) = 5.7956. The desired length AB is twice this, or (2)(5.7956) = 11.59115. At first I went instinctively to the calculator; but it turns out it's not too difficult to just multiply it out. (3 + (1.5)(sqrt(3)))^2 + (1.5)^2) = (3 + (3sqrt(3)/2))^2 + (3/2)^2) = 9 + (2)(3)(3sqrt(3)/2) + (9)(3)(4) + 9/4 = 9 + 9sqrt(3) 27/4 + 9/4 = 9 + 36/4 + 9sqrt(3) = 18 + 9 sqrt(3). Take the square root and multiply by 2 and again we get 11.5911. Keep 'em coming! 🤠

[@PreMath]

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍

[@ramanivenkata3161]

Very well explained

[@PreMath]

Thanks for liking, Ramani dear You are awesome. Keep it up 👍

[@jimlocke9320]

Some additional observations: At 4:15, right ΔQCO has a hypotenuse twice as long as one of its sides (length CQ = 3, length OQ = 6), so it is a 30°-60°-90° special right triangle, therefore side CO = side CQ times √3 = 3√3. Now locate point R on PB so length PR = 3√3. Construct AR. ΔQCO and ΔAPR are congruent by side-angle-side (PA = CQ,

[@ybodoN]

There is a theorem that solves the problem of denesting two nested square roots: If _a_ and _c_ are rational numbers and _c_ is not the square of a rational number, there are two rational numbers _x_ and _y_ such that _√(a + √c) = √x + √y_ if an only if _a² − c_ is the square of a rational number _d._ If the nested radical is real, _x_ and _y_ are the two numbers _½ (a + d)_ and _½ (a − d)_ where _d = √(a² − c)_ is a rational number. Example: In _√(2 + √3), 3_ isn't the square of a rational number and _√(2² − 3)_ is a rational number. Then _d = √(2² − 3) = 1_ so _x = ½ (2 + 1) = 3/2_ and _y = ½ (2 − 1) = 1/2._ Thus _√(2 + √3) = √(3/2) + √(1/2) = √2 (√3 + 1) / 2 = (√6 + √2) / 2._

[@waheisel]

Good observation. So the denested solution is 6*(sqrt(3/2)+sqrt(1/2))

[@AmirgabYT2185]

AB=6√(2+√3)≈14,58

[@santiagoarosam430]

Distancia horizontal entre centros = c》c^2=(3+3)^2 -3^2》c=3sqrt 》AB^2=3^2 +(3+c+3)^2 》AB=6[sqrt(2+sqrt3)] Gracias y saludos.

[@mohabatkhanmalak1161]

Wonderfull math, love your short and quick solutions.💯

[@JSSTyger]

I get AB = 2sqrt[18+9sqrt(3)] The height of the point of tangency is r/2. We can draw a horizontal chord through one of the circles at that point and find that chord length to be r(sqrt(3)). The difference between this chord length and the diameter is 2r-rsqrt(3) = r[2-sqrt(3)]. Divide this by 2 to get the tiny horizontal distance between the intersection point and the horizontal diameter. Now construct a right triangle with hypotenuse "z", horizontal leg 2r-(r/2)[2-sqrt(3)], and vertical leg r/2. Use the pythagorean theorem to solve for z. Your final answer is 2z.

[@tombufford136]

At a quick glance the horizontal distance PB is calculated.Q to the touching tangent is the radius 3 and the vertical distance is 1.5. The horizontal distance is sqrt(9-2.25). PB = 2*(2.8+3) =. 11.57. Distance AB^2 = PB^2 + 3^2 = (11.57)^2 + 9 . AB = 11.95. This gives a close result , though I need to check for the discrepancy.

[@harikatragadda]

If the intersection point is E, drop a perpendicular EF on AB. If ∠OQB =θ then, Sinθ = 3/6 = 1/2 θ = 30° Hence, EF = 3Sin30= 3/2 QF = 3Cos30 = 3√3/2 AE² = AF² + EF² = 9(2+√3) AB = 2AE = 6√(2+√3)

[@PreMath]

Thanks for sharing! Cheers! You are awesome. Keep it up 👍

[@wackojacko3962]

@ 3:16 line AB has a perpendicular bisector which can be constructed with compass and ruler. The point of bisector is now collinear with points A & B. QO also has a perpendicular bisector which can be constructed the same way. Tricky part is finding length of AB. THIS was a cool problem! 🙂

[@PreMath]

Excellent! Thanks for your feedback! Cheers! 😀 You are awesome. Keep it up 👍

[@gayatrithanvi8901]

That is great question and good answer😃 I truly didn't find it Hats off to you sir

[@PreMath]

Glad you liked it Thanks for your feedback! Cheers! 😀 You are awesome. Keep it up 👍

[@niteshmalhotra6162]

Awesome!! Lovely session. I would request you to kindly post questions related to functions , inequality and no. Of integral solutions as well.

[@PreMath]

Keep watching... Thanks for your feedback! Cheers! 😀 You are awesome. Keep it up 👍

[@quigonkenny]

QO = 2r, as by rule the centers of two circles and their point of tangency are colinear. Triangle ∆QDO: OD² + QD² = OQ² QD² + r² = (2r)² QD² = 4r² - r² = 3r³ QD = √(3r²) = r√3 = 3√3 As PB and AD are parallel, as each line passes is coincident with the diameter of one of the congruent semicircles, and each semicircle is tangent to the other line, PA = r = 3. By observation, CP = r as well, as CP and PA are each tsngents of semicircle Q meeting at P. QA, QC, OB, and OD also equal r, as each is a radius of one of the congruent semicircles. As QC and OD are thus perpendicular to both PB and AD, OC = QD. PB = r + r√3 + r PB = (2+√3)r = 6 + 3√3 Triangle ∆APB: AP² + PB² = BA² 3² + (6+3√3)² = AB² AB² 9 + 36 + 36√3 + 27 AB² = 72 + 36√3 = 36(2+√3) AB = 6√[2+√3] √[2+√3] = √[2+√3(4/4)] = √[3/2 + 1/2 + 2√(3/4)] = √[(√(3/2)+√(1/2))²] = √(3/2) + √(1/2) = √(6/4) + √(2/4) = (1/2)(√6+√2) AB = 6(1/2)(√6+√2) AB = 3√6 + 3√2 ≈ 11.59

[@ybodoN]

Generalized: _AB = r (√2 + √6)_ where _r_ is the radius of the semicircles

[@PreMath]

Thanks for sharing! Cheers! You are awesome. Keep it up 👍

[@harryisaksson8923]

Yes, √(2 + √3) can be simplified as (√2 + √6)/2.

[@devondevon4366]

Easy 11.6 Draw a line from Q to O this is 6 ( the two radii) From O to D is 3 Use Pythagorean 6^2 - 3^2 sqrt 27 = 5.1961524 Add the two radii 5.1961524 + 6= 11.196 Pythagorean again using 11.196 and 3 yields 11.6 Answer.

[@mathbynisharsir5586]

Great 👍

[@PreMath]

Thanks for the visit You are awesome. Keep it up 👍

[@raya.pawley3563]

Thank you!

[@Birol731]

My solution is: If we draw a line through the point of the intersection (K) of these two circles, which combine their centers, QO CQ= r= 3 QO= 2r= 6 The angle COQ= α sin(α)= CQ/QO sin(α)= 3/6 = 1/2 α= arcsin(0,50) α= 30° This angle COQ= OQD The large angle in the circle is equal to AQK, AQK= 180°- α AQK= 150° According to the law of cosines, we can write: AK²= AQ²+QK²-2*AQ*QK*cos(AQK) AQ= QK= 3 AK²= 3²+3²-2*3*3*cos(150°) AK²= 9+9-18cos(150°) AK²= 18-18cos(150°) cos 150 = -cos 30 = -√3/2 ⇒ AK²= 18+18*√3/2 AK²= 18+9√3 AK= √18+9√3 = √9(2+√3) = 3√(2+√3) AB= 2*AK AB= 2*3√(2+√3) AB= 6√(2+√3) unit length is the answer 🤗

[@PreMath]

Thanks for sharing! Cheers! You are awesome. Keep it up 👍

[@Michael-nm1kq]

This is the method I used as well.

[@naitikchaurasia1925]

Cannot we simply add the two diameters, i.e 2r +2r 2*3+2*3 6+6 12 Which is approx equal to the ans 6√(2+√3) which is 11.6

[@harryisaksson8923]

The line AB does not go through the centers (Q and O) of the semicircles so AB is not two diameters.

[@giuseppemalaguti435]

AB^2=3^2+(3+sqrt(6^2-3^2)+3)^2

[@PreMath]

Thanks for sharing! Cheers! You are awesome. Keep it up 👍

[@giuseppemalaguti435]

​@@PreMaththanks

[@unknownidentity2846]

OK, let's do it: . .. ... .... ..... Pythagorean theorem applied for the right triangle COQ: CO² + CQ² = OQ² CO² + R² = (2R)² CO² + R² = 4R² CO² = 3R² ⇒ CO = √3R Pythagorean theorem applied for the right triangle ABP: AB² = AP² + BP² AB² = R² + (BO + CO + CP)² AB² = R² + (R + R + √3R)² AB² = R² + (2 + √3)²R² AB² = R² + (7 + 4√3)R² AB² = 4(2 + √3)R² ⇒ AB = √(2 + √3)*2R = 6√(2 + √3) Best regards from Germany

[@PreMath]

Great! Thanks for sharing! Cheers! You are awesome. Keep it up 👍

[@tombufford136]

Yes, I have found the discrepancy. I subtracted 10-2.75, when it should be 9-2.75. so rewriting my earlier comment .At a quick glance the horizontal distance PB is calculated.Q to the touching tangent is the radius 3 and the vertical distance is 1.5. The horizontal distance is sqrt(9-2.25). PB = 2*(2.6+3) =. 11.2. Distance AB^2 = PB^2 + 3^2 = (11.2)^2 + 9 . AB = 11.59.

[@misterenter-iz7rz]

Seems to be difficult,😮 it is difficult to determine OC, as PC=OB=3 clearly. We turn to another approach, upper and lower congruent right-angled triangles, a^2+b^2=36 ab=9, so a^b2+9^2/a^2=36, a^4-36 a^2+9^2, a=5.795555, b=2.552914, therefore the answer is 2x5.795555=11.59111.😊 I am wrong, OC is simply sqrt(36-9)=sqrt(27), therefore the answer is sqrt(9+(6+sqrt(27))^2)=11.59111 approximately. 😅

[@PreMath]

Thanks for sharing! Cheers! You are awesome. Keep it up 👍

[@NahidMiah-e6m]

Hey, pro. Thanks a lots for making this contain. 😊 i'm eagerly waiting for your response, i always comment for making these videos such : geometric of Olympic problem for class 9-10. Please feedback me. 😊 your explanations is so smooth. 😊

[@PreMath]

Glad you like them! Thanks for your feedback! Cheers! 😀 You are awesome. Keep it up 👍

[@eziopuiatti8801]

once we have found CO and then PB to be ca. 11.19 then we can easily calculate AB to be ca. 11.59 - why bother with all the rest ?🤔 why not simply calculate the Sq root of 27 ...

[@devondevon4366]

11.6

[@alexniklas8777]

АВ= 11,59

[@PreMath]

Thanks for sharing! Cheers! You are awesome. Keep it up 👍

[@alexniklas8777]

@@PreMath Thanks sir😊

[@eckhardfriauf]

Very nice. 👍👍👍

[@PreMath]

Thank you! Cheers! You are awesome. Keep it up 👍