Find all missing sides of the triangle | Two Methods | (Step-by-step explanation) |

[@parthtomar6987]

Nice solution sir

[@PreMath]

Thanks and welcome ❤️ You are awesome. Keep it up 👍

[@AndreasPfizenmaier-y7w]

Which Solution is meant?

[@chanpangchin9744]

The radius of the circle is 50cm. Use Pythagoras triplets, 7,24,25 BD=48 Likewise, 3,4,5 gives AB=60 And 3,4,5 gives BC=80 Solved it in 2 minutes. All mental calculations.

[@arnavkange1487]

m is called the geometric mean and the theorem of geometric mean states that if a line perpendicular from a vertex of a triangle to the hypotenuse ,then line is geometric mean of the divided parts of hypotenuse I.e.A2=B.C

[@PreMath]

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

[@andydaniels6363]

This is the same equation that is obtained from considering similar triangles and indeed this problem is one of the motivations for the definition of the geometric mean.

[@HappyFamilyOnline]

Great presentation 👍 Thanks for sharing 😊

[@soli9mana-soli4953]

This problem is perfect for Euclid's theorems 1) BD² = 36*64 = 6²*8² = (6*8)² = (48)² ( II Euclid's theorem) 2) AB² = 36*100 = 6²*10² = (6*10)² = (60)² ( I Euclid's theorem) 3) BC² = 64*100 = 8²*10² = (8*10)² = (80)² ( I Euclid's theorem)

[@xof-woodworkinghobbyist]

I just used the Pythagorean theorem for the 3 triangles ABC, ABD, BDC. That gives 3 equations with 3 unknowns... easy to solve. Thanks for helping our brains to exercise!

[@PreMath]

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

[@Copernicusfreud]

That is how I solved the problem.

[@gulzarsingh1411]

I also did like this only

[@chanpangchin9744]

I thank my sifus Mr Loke Siu Kow, Mr Oon Kee Soon from San Min School, Telok Intan plus Mr Lim Chee Lin and Mr Nga Been Sing from National Junior College, Singapore.

[@tombufford136]

At a quick glance, A 3,4,5 triangle then 6,8 10 triangle might work (multiply by 10). The two triangles ABD and BDC are similar to ABC. Then 36/AB = 64/AB and AB / 100 =36/AB and 36/BD = AB /BC and BD/64 = AB /BC. Then AB^2 = 3600 = Then AB = 60 Then 60, 80, 100 triangle, BC = 80. Then BD^2 = 3600 - 36^2 and BD = 48.

[@Copernicusfreud]

Yay! I solved the problem.

[@PreMath]

Awesome ❤️

[@ybodoN]

∠ABC is 90° ⇒ AC is a diameter of the circumcircle of △ABC (converse of Thales's theorem) ⇒ BD = √(36 ⋅ 64) = 48 (chord theorem) gcd(36, 48) = 12 ⇒ 36 / 12 = 3 and 48 / 12 = 4 ⇒ AB is the hypotenuse of the (3, 4, 5) Pythagorean triangle ABD ⇒ AB = 5 ⋅ 12 = 60 gcd(48, 64) = 16 ⇒ 48 / 16 = 3 and 64 / 16 = 4 ⇒ BC is the hypotenuse of the (3, 4, 5) Pythagorean triangle BCD ⇒ BC = 5 ⋅ 16 = 80

[@PreMath]

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

[@phungpham1725]

Use the right triangle altitude theorem sq BD = 36 x64 --> BD= 4x8=48 Be cause AD=36 and BD = 48 so AB= 5x12 =60 (3-4-5 triple) Similarly, BC= 80

[@devondevon4366]

48, 60, 80 I assumed it was a 3-4-5 triangle scaled up by 20 since the hypotenuse =100 (64 + 36, divide 100 by 20 = 5, multiply the other two sides ( 3 and 4) by 20 = 60, 80. Set BC = 80 and AB 64. Divide 80 by 5 = 16 (since the triangles are similar); hence, divide 64/16 (to test and see if it is 4 If I did not get 4 here, I would need to find another method). Finally, multiply 3* 16 = 48 (Since I assumed in the beginning that it was a 3-4-5 scaled up by 20 since the hypotenuse 100 is divisible by 5)

[@じーちゃんねる-v4n]

think x1/4 angleA=angleDBC=θ AC=25 BC=25sinθ=16/sinθ sinθ=4/5 cosθ=3/5 AB=15 BD=12 BC=20 real ones= x4

[@PreMath]

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

[@rick57hart]

The big tringle is also similar to the both smaller ones. After we know that the ratio of the kathedes is 3 : 4 we also know that the ratio is 3: 4 :5. 36 + 64 = 100 here represented by 5. To get there we have to multiply 5 by 20. 3 times 20 = 60. 4 times 20 = 80. I don't even needed pythagoras🤔😇

[@honestadministrator]

∆ ABD, ∆ ACB are all similar AB / AC = AD /AB AB^2 = AD. AC = AD ( AD + DC) BC^2 = AC^2 - AB^2 = DC ( AD + DC)

[@Ibrahimfamilyvlog2097l]

Good 👍🌹❤

[@PreMath]

Thanks dear ❤️ You are awesome. Keep it up 👍

[@michaelkouzmin281]

Just one more solution x^2+y^2=(64+36)^2 (1) m^2+64^2=x^2 (2) m^2+36^2=y^2; (3) let us add (2)+(3): 2m^2+64^2+36^2=x^2+y^2 (4) subst (1) into (4): 2m^2+64^2+36^2=(64+36)^2 => m^2= 64*36 => m= 8*6=48; subst m^2 into (2) x^2=64*36 + 64^2=64(64+36)=64*100 => x=8*10=80; subst m^2 into (3) y^2= 64*36+36^2=36(64+36)=36*100 => y=6*10=60.

[@PreMath]

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

[@AndreasPfizenmaier-y7w]

36*64=2304=m^2,hence m=48, Rest by Pythagoras

[@pramodsingh7569]

Thanks

[@PreMath]

You are very welcome! ❤️ You are awesome. Keep it up 👍

[@williamwingo4740]

No peeking, no calculator: all work done by hand:🤞 Using your lettering convention and invoking Pythagoras, x^2 + y^2 = (36 + 64)^2 = (100)^2 = 10,000; call this equation [1]: Also by Pythagoras, using your lower-case b for line BD: b^2 + 36^2 = y^2; equation [2]: and from the other direction, b^2 + 64^2 = x^2; equation [3]: We now have three independent equations relating x, y, and b. Add equations [2] and [3]: b^2 + 36^2 + b^2 + 64^2 = x^2 + y^2; rearranging and multiplying out the square terms, this becomes: 2 (b^2) + 1296 + 4096 = x^2 + y^2; carrying out the addition: 2 (b^2) + 5392 = x^2 + y^2; but from equation [1], x^2 + y^2 = 10,000;:substituting that value in and rearranging: 2 (b^2) = 10,000 -- 5392 = 4608; dividing both sides by 2: b^2 = 2304. Now 2304 is a little less than 2500, so b must be a little less than 50. Hmmmm..... Since 2304 ends in a 4, let's try 48, whose square must also end in a 4: (48)(48) = 2304--right the first time! So: b = 48. Now if b = 48 and y = 36, then triangle ABD is 3-4-5 multiplied by 12; so y must equal (5)(12) = 60. And similarly, triangle BDC is 3-4-5 multiplied by 16, so x = (5)(16) = 80. Voila! Also FWIW, the entire triangle ABC is 3-4-5 multiplied by 20. Cave Canem! 🤠

[@misterenter-iz7rz]

h^2=36×64, h=48, then AB=60, BC=80.😊

[@PreMath]

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

[@prossvay8744]

BD=48, AB=60,BC=80

[@PreMath]

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

[@ИванПоташов-о8ю]

АB=√AC*AD=√36*100=60; BC=√AC*CD=√64*100=80;

[@PreMath]

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

[@eduardofiolmujica2996]

There is a more elegant solution. Start by AD×DC = m^2 (I don't remember the name of this theorem)

[@rvbln]

Euclid’s theorem of altitude or geometric mean theorem

[@PreMath]

Thank you! Cheers! 😀

[@wackojacko3962]

😉

[@PreMath]

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

[@michaelgarrow3239]

Can you do cold fusion next?!! 😀

[@PreMath]

Hello dear, if you an example, please send it to [e-mail] Cheers 😀

[@alster724]

Easy

[@AmirgabYT2185]

60; 80; 48

[@giuseppemalaguti435]

60,48,80

[@PreMath]

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

[@MathsMadeSimple101]

What do you call the study of mathematics?

[@medusa_404]

Robert, but his friends call him Bob