Calculate side length X of the right triangle | Circumference and Perimeter are known |

[@chelliahRaveedrarajah]

Sir, your explanation is very nice. Thank you sir for the explanation for this neat question.😂🎉😂

[@PreMath]

Most welcome, dear Thanks for your feedback! Cheers! 😀 You are awesome. Keep it up 👍

[@D.P9]

Easy approach ------- In right angle triangle Semi perimeter + inradius = Hypotenuse From this, we can write Hypotenuse = x = 1008 - 160 = 848 unit ✓✓

[@PreMath]

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

[@D.P9]

@@PreMath thank you sir I must say your all questions are amazing and most fruitful too. Your questions enhance my thinking skill , keep posting such a wonderful question sir

[@georgebliss964]

2pi x r =320 pi. r = 160. Labelling the three pairs of equal tangents at each corner of the triangle as a,a then b,b then c,c. Then 2a + 2b + 2c = perimeter = 2016. So a + b + c =1008. We know b = r from the square at the right angled corner of the triangle. Thus a + 160 + c = 1008. a + c = 848. x = a + c.

[@hemantdikshit3008]

The sum of a + b should be 848 it can be any triangle

[@quigonkenny]

​@@hemantdikshit3008Pretty sure those tangents were named as such to match their respective vertices of the triangle (a and a intersect at A, etc.). That's why b = r and a+c = x.

[@اممدنحمظ]

تمرين جميل جيد . رسم واضح مرتب . شرح واضح مرتب. شكرا جزيلا لكم والله يحفظكم ويرعاكم ويحميكم جميعا . تحياتنا لكم من غزة فلسطين .

[@PreMath]

Thanks dear❤️

[@danlayne9436]

I did it very similarly. Since for any right triangle with an inscribed circle, a + b - c=2r. We calculate this as a + b - c=320. And we know a + b + c=2016 Since the two equations share a + b, c must be the average of two solutions... (2016-320)/2 = 848

[@angeluomo]

This is the approach I took. I just added the two equations together and got 2c=1696, so the value of the hypoteneuse was simply half that.

[@TomasPböckerlyftningschack]

A slightly different (but I think simpler way, avoiding some minus signs): The tangents BE= BF= 160. AE = AD, call them a. CD=CF, call them b. Then 2x160 + 2a + 2b = 2016 2a+2b= 1696 and of course X=a+b so X = a+b=848

[@Micboss1000]

Yes, this solution is a fair bit simpler. I found it the same way.

[@murdock5537]

Nice! Many thanks,Sir! 2πr = 320π → r = 160 → c = a - 160 + b - 160 = (a + b) - 320 = 2016 - (a + b) → 2(a + b) = 2336 → a + b = 1168 → c = 2016 - (a + b) = 848

[@PreMath]

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

[@devondevon4366]

x= 848 Answer The other two sides are 448 and 720 At first, I thought it was the Pythagorean triplet 3-4-5 scaled up by 168 since this would lead to a perimeter of 2016, but it didn't add up when I factored in the radius. It is a Pythagorean triplet but it is not the 3-4-5 triplet. It is 28-45 -53 triplet scaled up by 16 First, calculate the radius: Circumference = 2 pi radius 320 pi = 2 pi r 320 = 2 r 160 = radius Let the height of the triangle =y and base z Hence x+ y + z = 2016 equation 1 Construct a square from the circle's center with sides = the circle's radius which = 160 Draw a perpendicular line from the point of tangency on line x to the circle's center, from line y to the circle's center and from line z to the circle's center x = z- 160 + y- 160 (tangent circle theorem) hence x = z+y - 320 hence z+y -320 + y + z = 2016 ( substitute the value of x into equation 1) 2z+ 2y - 320= 2016 2z + 2y = 2016 + 320 2z + 2y = 2336 2(z +y ) = 2336 z + y = 2336/2 = 1,168 hence x = ( 2016-1168) =848 Answer I noticed that the factor of 848 includes 53 which is the hypotenuse in the triplet 28,45, 53 848 = 53 * 16 720 = 45 *16 448 = 28*16 ----- 2016 So the other two sides of the triangle are 720 and 448 Checking 448^2 + 720^2 = 848^2 200,704 + 518,400 = 719,104 719, 104 = 719,104

[@mvrpatnaik9085]

Explanation is quite good. Thanks.

[@quigonkenny]

Circle O: C = 2πr 320π = 2πr r = 320π/2π = 160 By Two Tangent Theorem, DA = AE, EB = BF, and FC = CD. By observation, EB = BF = 160. Let DA = AE = y and FC = CD = z. Note that y + z = x. Triangle ∆ABC: P = AB + BC + CA 2016 = y + 160 + z + 160 + y + z 2y + 2z = 2016 - 320 = 1696 y + z = x = 848

[@soli9mana-soli4953]

2π*r = 320 π (circumference) then r = 160 [1] a + b + x = 2016 (perimeter) [2] x = (a-r) + (b-r) = a - 160 + b - 160 = a + b - 320 combining a + b + a + b - 320 = 2016 we get a + b = 1168 being x = 2016 - (a+b) from the [1] we get x = 848

[@ramanivenkata3161]

Well explained

[@PreMath]

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

[@michaelgarrow3239]

Mr. Math; Can you explain why the circumstance of a circle is 2pi•r? Wouldn’t it be simpler to do D•pi? Or 2R•pi? Thank you.

[@KAvi_YA666]

Thanks for video.Good luck sir!!!!!!!!!!!!

[@devondevon4366]

This is a 28, 45, 53 Pythagorean triplet scaled up by 16. Hence the sides are 848, 720, 448 Yes, a 3-4-5 triangle (another triplet) scaled up by 168, would also have a perimeter of 2016, the same as the one above, and in this case 'x' would be 840, height 672, and base 504 (840+ 672+504 =2016] but this is not it. 840 certainly would be one of the choices in a multiple-choice examination So the answer is 848

[@PreMath]

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

[@bigm383]

Neat problem. Thanks Professor!❤😀

[@PreMath]

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

[@unknownidentity2846]

Circumference of the circle: C = 2πR = 320π ⇒ R = 320π/(2π) = 160 Perimeter of the triangle: P = AB + AC + BC = 2016 (AE + BE) + (AD + CD) + (BF + CF) = 2016 From the two tangent theorem we can conclude: AE = AD BE = BF CD = CF Additionally, we have BE = BF = R: (AE + BE) + (AD + CD) + (BF + CF) = 2016 2*AD + 2*R + 2*CD = 2016 AD + R + CD = 1008 ⇒ x = AC = AD + CD = 1008 − R = 1008 − 160 = 848 Best regards from Germany

[@giuseppemalaguti435]

dopo vari calcoli risulta c1c2=2016r....c1+c2=r+1008....( r=160 ,c1^2+c2^2=x^2)....(r+1008)^2-4032r=x^2...x=848

[@PreMath]

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

[@williamwingo4740]

As usual, a slightly different and mostly algebraic approach: I got EB = BF = 160 from the circumference of the circle, as you did. Now let AE = y and FC = z. From the symmetry of the diagram, x = y + z; Invoking Pythagoras, (y + 160)^2 + (z + 160)^2 = (y + z)^2; expanding everything, y^2 + 320x + 160^2 + z^2 + 320z + 160^2 = y^2 + 2yz + z^2; the y^2 and z^2 terms on each side subtract out. Collecting terms and simplifying: 320(y + z) + 2(160^2) = 2z; dividing everything by 2 and simplifying some more: 160(y + z + 160) = yz; label this equation 1. Now from the perimeter of the triangle given, 2y + 2z + 320 = 2016; dividing by 2: y + z + 160 = 1008; solving for z in terms of y: z = (1008 - 160 - y) = 848 - y. label this equation 2 and plug this value for z back into equation 1: 160(y + (848 - y) + 160) = y(848 - y) = 848 y - y^2; simplify some more: 160y + (160)(848) - 160 y + 160^2) = 848y - y^2; the 160y term subtracts out and we have: (160)(848 + 160) = 848y - y^2; converting to the standard quadratic form: y^2 - 848y + (160)(1008) = 0; applying the quadratic formula: y = (848 +/- √(848^2 - (4)(160)(1008)))/2; so [I confess I used a calculator for this]: y = (848 +/- √73984)2 = (848 +/- 272)/2 = 424 +/- 136 = either 560 or 288; using equation 2 to solve for z: z = either 848 - 560 = 288; or 848 - 288 = 560; so either y = 560 and z = 288; or vice-versa. And either way, x = y + z = 848. Check: perimeter of the triangle = 2y + 2z + 320 = 2(848) + 320 = 1696 + 320 = 2016. It's also worth noting that the triangle ABC is integer Pythagorean: 28-45-53 x 16. I didn't have that one, but it's now in my table. Belated and probably posthumous thanks to Mr. Clements and Miss Whaley (Birmingham, Alabama, 1962-3); and Cheers. 🤠

[@marioalb9726]

Radius of Circle: P = 2πr r = P / 2π = 320π / 2π r = 160 cm Perímeter of triangle:: P = 2. x₁+ 2.x₂ + 2r P = 2 x + 2r x = (P -2r) / 2 x = (2016 - 320)/2 x = 848 cm ( Solved √ )

[@michaelkouzmin281]

320/2=160

[@marioalb9726]

​@@michaelkouzmin281 Thanks

[@mathbynisharsir5586]

An excellent presentation sir

[@PreMath]

Many thanks, dear You are awesome. Keep it up 👍

[@mathmurthy993]

If area of incircle is aπ & the perimeter of given right ∆ is b then its hypotenuse is (b-2a)/2

[@PreMath]

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

[@ybodoN]

A square sign is missing between a and π or a square root sign is missing between 2 and a... 🧐

[@engralsaffar]

I did it much simpler 2pi r= 320 pi r=160 units AD=AE=a DC=FC=b EB=FB=r 2a+2b+2r=2016 a+b+r=1008 a+b=x=1008-160=848 units

[@PreMath]

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

[@danielsgrunge]

Fun one!

[@PreMath]

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

[@parthtomar6987]

Nice way to solve plz can you explain calculus

[@PreMath]

Thanks for your feedback! Cheers! 😀 I've already uploaded videos on Calculus. Please check out the playlist. You are awesome. Keep it up 👍

[@Copernicusfreud]

Yay! I solved the problem.

[@PreMath]

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

[@soniamariadasilveira7003]

Gosto muito de suas explicações, obrigada!

[@PreMath]

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

[@laxmikatta1774]

I thought of linear equations in two variables but it was another way

[@erwinkurniadi1850]

Radius = 160 AB = 160 + A BC = 160 + B X = A + B perimeter = 2016 AB + BC + X = 2016 160 + A + 160 + B + A + B = 2016 320 + 2A + 2B = 2016 2A + 2B = 2016 - 320 A + B = (2016 -320)/2 = 848 X = A + B = 848

[@mathukuttyvarghese2738]

Sir if a,b&c of a right triañgle,it is in ratio 3:4:5. So side c=5X2016/12=840. Other sides 504& 672.

[@D.P9]

Sir it can be solved in one another way Here let hypotenuse h = x d = 320π/π = 320 d = P +b - h = 320 P + b + h = 2016 h = x = (2016-320)/2 = 848

[@PreMath]

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

[@santiagoarosam430]

2b+2h+320=2016》b+h=848=X Gracias y saludos cordiales.

[@PreMath]

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

[@kennethstevenson976]

I had the answer quickly and didn't know it. (a+b) + (a+160) + (b+160) = 2016 ; 2a + 2b + 320 = 2016 ; a + b + 160 = 1008 ; a+ b = 848 ; X = a + b Therefore X = 848.

[@AmirgabYT2185]

x=692

[@ybodoN]

Bonus question: what are the lengths of the other two sides of the right triangle ABC? 🤔

[@Abby-hi4sf]

Let AE = a , and FC = b , we know EB= BF = R= 160 , We know AE = AD, also FC = DC (a + 160)^2 + (b + 160) ^2 = (a + b)^2 a^2 + 360a + 25600 + b^2 + 360b + 25600 = a^2 + 2ab + b^2 ----------- here both a^2 and b^2 cancel each other 2ab = 360( a+ b) + 2(25600) ------- we devide all by 2, and we know (a+b) = 848 from PreMath answer ab= 160 ( 848) + 25600 ab = 161280 1) a + b = 848 ---->>> b= 848 - a 2) ab = 161 280 Then you get quadriatic eqn. by substiuting b= 848 on eqn 2 a (848 - a) = 161280 a^2 - 848a + 161280. sloving give us a and b to be 288 AND 560 add R= 160 to each a & b AB = 448, BC = 720

[@ybodoN]

@@Abby-hi4sf Well done! 👍

[@ybodoN]

💡 _In every Pythagorean triangle, the radius of the incircle is a natural number._ Since x turns out to be a natural number, ABC must be a Pythagorean triangle. The GCD of x and the radius of the circle is 16, leading to the triple (28, 45, 53). Thus, the triangle ABC has sides (28, 45, 53) × 16, i.e. 448, 720 and 848 units.

[@bb55555555]

didn't need the Pythagorean theorem at all on this one. I thought for sure he would use it.

[@quigonkenny]

Same thing happened to me when I worked on it. Started in on the equation for the tangents and the perimeter, fully expecting to have to tie it in to Pythagoras, and realized "Hey, that's x". I kinda want to see if I can determine the other two sides of the triangle now...

[@manepetros2181]

Good luck everything is so Easy 😂😂😂and if someone knows a little Math and trigonometry could solve this Problem,,,😂😂😂😂 Thanks 🙏🙏🙏🙏🙏,,,for explain,,, God bless you 🙏🙏🙏❤️😂

[@giovannicis4982]

2x=2016-320

[@sebasiegrist9341]

I deducted 2r from perimeter, and found out I was left with 2 pairs of equal segments. Then X= 1 pair, so I divided by 2.

[@CharlesB147]

Also, after further solving down, the triangle is a very little used Pythagorean triple - a (28, 45, 53), each side of which is scaled by a factor of 16 to produce the side lengths of 448, 720, and 848.

[@hemantdikshit3008]

The sum of a + b should be 848 it can be any triangle which set is 5 the condition of sum of the sides. 28 into 16 and 45 into 16 is also one of the solutions

[@PreMath]

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

[@ronaldmontgomery8446]

Something is dreadfully wrong with this figure and or dimensions. I drew this on CAD. A right triangle has a circle tangent to the 3 sides and the circle has a diameter of 320, then it can not have a side of 848. Will somebody else please draw this figure on Cad, and tell me I am wrong.

[@mrabhinav223]

Love from india ❤❤❤❤❤❤

[@batavuskoga]

I made one stupid mistake, so my solution was not correct. For the rest my calculation was correct. My mistake : 320π=2πr --r=80, should be 160 of course 😩

[@PreMath]

No worries. We are all lifelong learners. That's what makes our life exciting and meaningful! You are awesome. Keep it up 👍

[@pintomed]

x = a + b (a + b) + (160 + b) + (160 + a) = 2016 a + b = 848 looks neater in my opinion . Thank you

[@fryespinoza]

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