Find the Distance MN in this Right Triangle | Step-by-Step Tutorial

[@iMvJ27]

I just love Geometry ♥️ More please Dear SIR ☺️

[@PreMath]

Excellent! Glad to hear that! Thank you for your feedback! Cheers! You are awesome VJ. Keep it up 😀

[@yayat8801]

ولا

[@williamwingo4740]

As others have already noted, this one is unusual for using three different integer Pythagorean triangles. I recognized BCN as 7-24-25--probably because I've done so many of these by now--so I was able to skip the first step; but then ABC and AMN both turn out to be 3-4-5's, scaled up by 8 and 3 respectively. Didn't realize that until I'd done the calculations. Sneaky! Cheers. 🤠

[@fevengr9245]

The interesting thing about this problem to me is that in spite of the lines and angles going every which way, you still managed to select values that produced an integer result. Thanks!

[@williamwingo8952]

They're all "magic" Pythagorean triangles. Specifically, AMN and MNB are 3-4-5 (multiplied by 5); the big triangle ABC is 3-4-5 multiplied by 8; and BNC is 24-25-7.

[@marioperic9709]

Very nice task. Thank you professor for the good brain gym for the end of the week. well, it is easy: NB is the same as AN = sqrt(NC^2+BC^2)=25. which means AC is 32. the triangles AMN and ABS are similar. so: MN/AM=24/32 and MN/25=24/(2*MN). Multiplying these equivalence MN^2=225 so MN=15!

[@attilajanos2179]

That's what I did too! 🙂

[@Q_from_Star_Trek]

An interesting approach: A,C,B on the circle with the center on M. So AM=MB=MC=R. BMNC quadrilateral and

[@johnspathonis1078]

Another great video. Students are required to commit some things to memory such as ratios --- 3:4:5, 7:24:25 , Sqrt of 2, Sqrt of 3 etc. This problem employed triangles with side ratios of 3:4:5 and 7:24:25. Students will not learn the application of these ratios if you don't apply them in your problem solving. Cheers

[@silviatotaro9372]

Well done.I solved the problem exactly in the same way. I used the pythagorean triples to get the results. Hi from Italy!

[@PreMath]

Fantastic! Thank you for your feedback! Cheers! You are awesome Silvia. Keep it up 😀 Love and prayers from the USA!

[@Tmwyl]

Me too. I like doing these puzzles before I start work to give my brain a jump start!

[@eichkay8169]

AMN and ABC are similar triangles, and AB=2AM. From those we have the equations MN/AN=12/AM and AM/AN=(7+AN)/2AM. Apply Pythagorean theorem to ABC and we get the equation (7+AN)^2+24^2=(2AM)^2. We now have three unknowns and 3 equations. We can then use algebraic methods to solve for MN.

[@sandanadurair5862]

Another approach! Having established BN=AN=25 Extend AC to a point D such that AN=DN=25. Let MN = h Then BN = 2h since MN is parallel to MN and M and N are mid points of AB and AD. In the extended triangle BCD CD = 25-7=18 BC = 24 So BD^2 = BC^2+CD^2 4h^2 = 24^2+18^2 h = 15

[@gybx4094]

Very nice! A great exercise for the mind. I haven't done this in many decades since high school!

[@johnnath4137]

Circumcircle ABC is also a semi-circle, so MA = MB = MC = r (say); ΔAMN ≡ ΔBMN → AN = BN = 25 → AC = 32 → 2r = AB = 40 → r = 20 ; Δs AMN, ABC are similar → MN/AM = BC/AC = 24/32 → MN = r x 24/32 = 20 x 24/32 = 15.

[@seeithappen1]

Very good - brings back memories of my school time .....

[@mcdowelltw]

You're a wonderful teacher.

[@jaybling6687]

Without watching the video in full, first connect points N and B, which forms a right triangle NCB. Segment NB is the hypotenuse of NCB which has a length of 25 based on the 7-24-25 Pythagorean triple. Then we compare right triangles NMB and NMA and see they are similar since they share a height NM and segments MB and MA are the same length, therefore segments NB and NA are the same length as well so NA is also 25. That means segment AC is 32 (NC + NA = 7 + 25). Then we find the hypotenuse AC of right triangle ACB to be 40 based on the 3-4-5 Pythagorean triple multiplied by 8. Then MA = MB = AC/2 = 20. Finally with either right triangle NMA or NMB, we can find NM using the 3-4-5 Pythagorean triple multiplied by 5 to get a length of *15*.

[@PreMath]

Very well dome Jay Thank you for sharing! Cheers! You are awesome. Keep it up 😀

[@matheuslopes5287]

Solved a similar problem once, but with the triangle inscribed in a semicircle. In a few seconds, knew MN=15

[@devondevon4366]

AN = BN=25 BN=25 (using pythag.) hence AC=32 hence AB=40 AM=20 and since AN=25, then MN=15 since triangle AMN is a scaled up 3, 4, 5 right triangle. (scaled up by 5) Answer MN=15

[@nirupamasingh2948]

Your way of explanation marvellous. Thank youSir

[@PreMath]

You are most welcome Niru dear Thank you for your feedback! Cheers! You are awesome. Keep it up 😀 Love and prayers from the USA!

[@KundanKumar-pq5ec]

It was too easy problem and I solved it easily and quickly....

[@KundanKumar-pq5ec]

@MathsJokar this is a good channel for maths lover

[@MrPaulc222]

Great video, but I went a totally different route (don't I always? :) ). Pythagoras gives 25 as the hypotenuse of BNC --> so MN becomes cos(MN/25) hence 25cos(53.13) = 15. My calculator shows it as a smidgeon more than that but that's due to rounding errors on trigonometry values. However, your way looks cleaner.

[@parikshitparekh9609]

Easy :) Learnt previously only...

[@PreMath]

Super Parekh You are awesome. Keep it up 😀

[@williamwingo8952]

No peeking and no calculators, just multiple invocations of Pythagoras: Draw line NB. Length of NB is √(24^2 + 7^2) =√(576 + 49) = √625 = 25. Triangles ANM and NMB are congruent by side-angle-side, so AN = NB = 25. Now considering the entire triangle ABC: AC = 25 + 7 = 32; and the hypotenuse AB = √(32^2 + 24^2) = √(1024 + 576) = √1600 = √[(16)(100)] = (√16)(√100) = (4)(10) = 40; so AM = MB = 20. Then MN = √(25^2 − 20^2) = √(625 − 400) = √225 = √[(25)(9)] = (√25)(√9) = (5)(3) = 15. Thank you, ladies and gentlemen; I'm here all week. 🤠

[@SpikeTheNeuropsych]

Triangle Man would be proud, for you used nothing but triangles.

[@d.m.7096]

Another solution - Let AN = y, AM = x = MB ∆ANM ~ ∆ABC (by A-A-A theorem of similarity) Therefore, AN/AB = NM/BC = AM/AC Hence, y/2x = NM/24 = x/y+7 Hence, y^2 + 7y = 2x^2 ... (1) By Pythagoras Theorem for ∆ABC, (y+7)^2 + 576 = 4x^2 Hence, y^2 + 14y + 625 = 4x^2 ...... (2) By (1) and (2), y^2 +14y + 625 = 2y^2 + 14y Hence, y = 25 = AN Therefore, AB = 40 and AM = 20 By Pythagoras Theorem for ∆ANM, MN = 15

[@illyriumus2938]

Solved it the same way with the difference that ABC triangle is a 3-4-5 triangle and i found the AC to be 40 easily.

[@Taigan_HSE]

Alternatively, without constructing (NB) you can see that (AM)/(AN)=[(AN)+7]/2((AM). So 2(AM)^2=7(AN)+(AN)^2. Then, 4(AM)^2=[(AN)+7]^2+24^2. Two equations and two unknowns solve for (AM)= 20 and (AN)=25. Pythagorean gives us (MN)= 30

[@davidseed2939]

Side note. M is the centre of a circle radius 20, henceMC=20

[@dhrubajyotidaityari9240]

Complete the triangle ABD where B=90° .MN=½BD, BC²=AC.CD Let a=AN 24²=(a+7)(a-7), a=25 Now since AN=ND, CD=a-7=18 BD²=24²+18²=900=30² MN=15

[@nobody_knows675]

Very challenging problem. Took me like 15-20 min to find a solution

[@PreMath]

Great job Thank you for sharing! Cheers! You are awesome. Keep it up 😀 Love and prayers from the USA!

[@theoyanto]

Nice simple one for people like me, thanks again :-)

[@anatoliy3323]

Nice a task! I had solved it in the same way. Thank you so much, Sir. Lots of greetings from Ukraine>))

[@PreMath]

Excellent! Glad to hear that! Thank you for your feedback! Cheers! You are awesome Anatol dear. Keep it up 😀 Love and prayers from the USA!

[@rameshk631]

Be safe my friend.

[@242math]

got it, thanks for the challenge bro

[@gannonfitzgerald6485]

This one was very easy to figure out by oneself, good job though 👍

[@philipkudrna5643]

Before watching: The missing part of the other side of the right triangle is 25. (You can mirror it along the line MN to the other side and it forms the right triangle 7-24-25.) This leaves the „large“ right triangle with the dimensions 24^2+(7+25)^2=40^2. Half of the large Hypothenuse is therefore 20. By Pythagoras for the left triangle you get 25^2-20^2=15^2. So MN = 15.

[@thegeopoliticalanalyst1660]

Can you please share a diagram. A pic may be. Did not get this.

[@philipkudrna5643]

@@thegeopoliticalanalyst1660 just Watch the video. Our professor explains it. Just what I did, but step by step!

[@selvamd8861]

Marvellous... I liked it

[@erikguzik8204]

As always, another great video. Just wondering for Triangle AMN --- Could you just use the 3-4-5 Right triangle to get MN? (im sure someone has already mentioned this)

[@devondevon4366]

Since triangle ABC is also 3-4-5 (scaled up by 8), then MB = 20 (since it is the mid-point of line AB), hence triangle MBN is a 3-4-5 (scaled up by 5 since BN = 25, and MB=20 ), hence MN = 15. So the answer to your question is yes

[@Ninja_Legends]

I just want you too know I already found MN even before I saw this video

[@govindashit6524]

I solve it easily . Thanks.

[@JLvatron]

Hey, I solved it! Thank you!

[@PreMath]

You are very welcome. Thank you for your feedback! Cheers! You are awesome JL. Keep it up 😀 Love and prayers from the USA!

[@susennath6035]

Good explanation

[@PreMath]

Keep watching Thank you for your feedback! Cheers! You are awesome Susen. Keep it up 😀 Love and prayers from the USA!

[@thunderutkarsh9977]

Thank you sir 🙏🙏

[@extremegrief1525]

What about drawing a parallel to MN from C and using similar triangles? Would that work?

[@Reddogovereasy]

Why is a square positive when a square root of a number is positive or negative? Example 3 squared is 9 but the square root of 9 can be either positive or negative 3.

[@tijanimaths6006]

Goooood👌👌👌

[@vidyadharjoshi5714]

BN = Sqrt( 24sq + 7sq ) = 25. AM = MB So AN = BN. So AC = 7 + 25 = 32. So AB = Sqrt ( 32sq + 24sq ) = 40. So AM = 20. MN = Sqrt ( ANsq - AMsq ) = Sqrt ( 625 - 400 ) = 15

[@mustafizrahman2822]

15 units. Good question.

[@una_unknown]

wow I solved it from the thumbnail!

[@PreMath]

Excellent! Thank you for your feedback! Cheers! You are awesome Achraf. Keep it up 😀

[@walterbarboza1594]

Here NB=AN= 25 TRIANLE AMN AM =MN LET AM = x Then x squared+x squared = 25 squared. Solve fir x x= 25/root 2 = MN

[@mahalakshmiganapathy6455]

Super thank you

[@PreMath]

You are very welcome Mahalakshmi dear. Thank you for your feedback! Cheers! You are awesome. Keep it up 😀 Love and prayers from the USA!

[@mehmetdogancakr2878]

Thanks.

[@shashwatvats7786]

Solved easily

[@PreMath]

Great Vats Thank you for your feedback! Cheers! You are awesome. Keep it up 😀

[@pranavamali05]

Thnku

[@PreMath]

You are very welcome. You are awesome Pranav. Keep it up 😀 Love and prayers from the USA!

[@MusicHavenSG]

This one is really forcing you to fully use the Pythagoras' Theorem. That's why I love Trigo so much.

[@ДядяВолк-к9э]

3:4:5 special triangles in this figure

[@PreMath]

Right on! Thank you for your feedback! Cheers! You are awesome. Keep it up 😀

[@geoffroi-le-Hook]

and 7-24-25

[@Teamstudy4595]

MN = 15 units

[@kumudsaraswat1398]

I solve this problem when is see this Because I recently solved such kind of problem BTW I am from India

[@PreMath]

Excellent Kumud! Thank you for your feedback! Cheers! You are awesome. Keep it up 😀 Love and prayers from the USA!

[@kumudsaraswat1398]

@@PreMath thanks sir

[@JSSTyger]

OK I tried it all in my head and got 15.

[@EPaozi]

J'aime quand on reste dans les nombres naturels !!!

[@PreMath]

YES! Beaucoup plus simple à gérer ! Continue comme ça Paozi 😀

[@doremi.harizuki373]

Thaks

[@PreMath]

You are very welcome. Cheers! You are awesome. Keep it up 😀

[@doremi.harizuki373]

Please wish me a good luck I have a math exam soon😢🤜🤛

[@PreMath]

@@doremi.harizuki373 All the best Doremi dear. You are awesome 😀 Love and prayers from the USA!

[@vishwambharmahant9485]

MN =15

[@olivierhabineza7755]

The distance MN is of 15 units.

[@devondevon4366]

appears to =15 answer

[@فراسمعابره-ج5خ]

Mn=16.89

[@manualrepair]

👍

[@PreMath]

Thank you for your feedback! Cheers! You are awesome. Keep it up 😀

[@mathswan1607]

15

[@PreMath]

Great! You are very welcome. Thank you for your feedback! Cheers! You are awesome. Keep it up 😀

[@jhandle4196]

"Now drop and give me 20."

[@ВарвараДорофеева-з8у]

👍👍🤗

[@holyshit922]

Pythagorean identity is enough to solve it

[@adgf1x]

mn=12

[@imnitinnagar]

मजा आ गया गुरु तुम्हारे सवाल करके 🌹

[@manualrepair]

i hope that my feedback help you

[@manualrepair]

I said I didn't know your language well, but I know the language of mathematics and geometry

[@PreMath]

Excellent! Glad to hear that! Thank you for your feedback! Cheers! You are the best. Keep it up dear 😀 Love and prayers from the USA!

[@michaelkoch6863]

My answer : " Find the real mathematics " .....

[@rifatshan5362]

🙂

[@PreMath]

Thank you for your feedback! Cheers! You are awesome Rifat. Keep it up 😀 Love and prayers from the USA!

[@luiginodilenardo6820]

Guarda mio video Geometria numerica Area cerchio

[@richarddumont5389]

The drawing is deceptive

[@adilegarib4098]

like!

[@johnspathonis1078]

Another great video. Students are required to commit some things to memory such as ratios --- 3:4:5, 7:24:25 , Sqrt of 2, Sqrt of 3 etc. This problem employed triangles with side ratios of 3:4:5 and 7:24:25. Students will not learn the application of these ratios if you don't apply them in your problem solving. Cheers

[@AmirgabYT2185]

MN=15

[@jaibhuwan6189]

Easy :) Learnt previously only...

[@PreMath]

Good job Jai

[@mounirobeid6566]

15

[@PreMath]

Great! You are very welcome. Thank you for your feedback! Cheers! You are awesome Mounir. Keep it up 😀

[@abcoflife6420]

15

[@sarathsomaweerasilva]

15

[@Xyz-sd7ub]

15