A Very Nice Geometry Problem | You should be able to solve this!

A Very Nice Geometry Problem | You should be able to solve this! | 3 Different Methods

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Күн бұрын

Пікірлер: 119

@thummalurusrinivasareddy1078 28 күн бұрын

Circumradius R =abc/4 area of triangle a=5 ,b=5 c =6 , area of triangle = square root of s(s-a) (s-b) (s-c) and s=a+b+c÷2. s= 5+5+6÷2. s= 16÷2=8. Area of the triangle = square root of 8×[8-5] ×[8-5]×[8-6] = square root of. 8×3×3×2 =square root of 144 =12 circumradius R= abc÷ 4 area of the triangle R= 5×5×6÷4×12 =150÷48=25÷8= 3.125

@Logging_Out 8 күн бұрын

This is the right answer ✅

@pooransingh1882 3 ай бұрын

The method of intersecting chords will also yield the radius more quickly(4*x=3*3).This yields a diameter of 6.25.,r=3.125

@Ellehsdee Ай бұрын

I'm new to this, but it seems that it should be (4x = 5*5), which solves to x = 6.25. Or am I missing something? Again, I am new to this, so please correct me if I am wrong.

@pooransingh1882 Ай бұрын

@Ellehsdee The x value is the section of the diameter below the chord.This value is calculated to be 2.25 from the formula for intersecting chords (Euclid) The length of the diameter is calculated to be 4+2.25=6.25. From this, the radius is calculated to be 6.25/2 =3.125.

@Ellehsdee 29 күн бұрын

@pooransingh1882 Ahh, thank you so much for the charity on this.

@Logging_Out 8 күн бұрын

Right ✅✔️

@monroeclewis1973 3 ай бұрын

4th way: Find h=4 by the Pythagorean theorem as you have. Then, draw a line from the center of the circle on h to intersect line AC at the midpoint, Q, at a right angle. Triangle AOQ is similar to the larger right triangle ACP. Therefore, PC/AP=OQ/2.5. 3/4=OQ/2.5. OQ = 15/8. Now use the Pythagorean theorem to find the radius, OA: (15/8)^2 + (5/2)^2 = OA^2. OA or R = 25/8.

@imetroangola17 3 ай бұрын

_Outro método:_ A área ABC é facilmente encontrada fazendo [ABC]=2×(4×3)/2=12 [ABC]= abc/4R, onde a,b e c são os lados do triângulo ABC e R é o raio da circunferência. Assim, 12=5×5×6/4R 2=5×5/4R 8R=5×5 *R=25/8*

@hongningsuen1348 3 ай бұрын

The most direct method using no construction is to use Heron's formula to calculate area of the triangle from the 3 sides and then use circumradius formula to find R = abc/4xarea.

@Monsieur_Cauchemar-YT 3 ай бұрын

You don't need Heron's formula - it's an isosceles triangle. The line from the apex to the base intersects the base at 90 degrees at the midpoint of BC at say, D. So BD = CD = 6/2 = 3. You then have a Pythagorean 3-4-5 triangle, with AD = 4. The area is then BD*AD = 3*4 = 12.

@hongningsuen1348 3 ай бұрын

@@Monsieur_Cauchemar-YT My point is not using any construction including perpendicular bisector theorem for isosceles triangle.

@Monsieur_Cauchemar-YT 3 ай бұрын

@@hongningsuen1348 Fair enough, I just personally think Heron's formula is a lot more work.

@WahranRai 3 ай бұрын

@@Monsieur_Cauchemar-YT No, because the Heron is general for any triangle ! His formula is correct

@Monsieur_Cauchemar-YT 3 ай бұрын

@@WahranRai I know it does. I just said in the case of this particular isosceles triangle that it's a lot more work. Lol.

@devondevon4366 3 ай бұрын

25/8 Draw a perpendicular line from BC to form two 3-4-5 right triangles, ABP and ACP. BP = 3 and AP=4 label O on line AP as the circle's center. O will be located on AP, given the isosceles triangle, ABC. From O, draw a line to B, then draw another to P to form a triangle, BOP OB = r OP= 4 -r (since AP=4), and BP =3 Using Pythag : r^2 = 3^2 + ( 4-r)^2 r^2 = 9 + 16 + r^2 - 8r 8r = 25 r = 25/8 Answer

@thangavelm.m8003 3 ай бұрын

Nicely explained sir., thanks.

@devondevon4366 3 ай бұрын

@@thangavelm.m8003 Glad to be of help

@jamesfaizi7771 3 ай бұрын

Just draw diameter from point A intersecting BC in 2 segments of 3 Th height of isosceles triangle is obviously 4 (due to two 345 triangles or 30,60,90 degree angles) Now use intersecting rule 3x3=9 9 ÷ 4= 2.25 2.25 + 4 = 6.25 diameter 6.25 ÷ 2 = 3.125 Radius

@WahranRai 3 ай бұрын

My method, use of formulas : R = a*b*c/ 4*A and (Heron) A = sqrt(p*(p-a)*(p-b)*(p-c) with half perimeter p=0.5*(a+b+c) and A = triangle area p = 0.5(5+5+6) = 8 ---> A = sqrt(8*3*3*2) = 12 ----> R = 5*5*6 / 4*12 = 25/8

@rakeshseth1461 2 ай бұрын

😮T😮

@chinkythapa1250 2 ай бұрын

One of the Finest Method No other Calculation is required after this Heron's formula is enough for all types of Triangles to calculate Area.....

@lukicborislav7591 Ай бұрын

A=SQR(S(S-a)(S-b)(S-c))=(abc)/4R; S=(a+b+c)/2=8; A=SQR(8x2x3x3)=12; A=(5x5x6)/4R=12; R=25/8

@anandharamang3289 3 ай бұрын

Without using heron formula or any construction other than vertical height. After finding height, we can use a simple formula as below. Radius= product of sides of triangle other than base divided by 2 times of height. So, Radius= 5*5/(2*4)= 25/8.

@renewd 15 күн бұрын

I used 2nd method. I was not aware of 3rd method. Very interesting use of law of sines to equal 2R of the circumcircle.

@cjburian1 2 ай бұрын

This was a fun one, thanks. Seemingly perplexing problem actually able to be solved in a person's head!

@makermaker11 3 ай бұрын

We divide triangle in middle to 2 right triangles and they will be both similiar because it is icoseles so 3-4-5 specaial triangle so height is 4 then we use inside circle cors where from radius to cord it divides cord to 2 which makes base 3 and side with 5 to 5/2 then we measure height for the base triangles which will be |/r²-9 then we add both r and |/r²-9 to find that both equal 4

@santiagoarosam430 3 ай бұрын

D es el punto medio de BC y "s" es la flecha del arco BC---> Potencia de D respecto a la circunferencia =BD*DC=AD*s---> 3*3=4s---> s=9/4---> 2r=4+(9/4)=25/4--->r=25/8. Gracias y saludos.

@josephsalinas6725 3 ай бұрын

Quarto método: teorema de Faure. Quinto método: S=abc/4R. Aprendi aqui, professor !

@lusalalusala2966 3 ай бұрын

After finding the length of the height from A, I would introduce cartesian coodinates with the origin to be B(0,0) and BC the x-axis, so C has coordinates C(6,0) and A(3,4). Now you have to find the radius though 3 points B(0,0), C(6.0) and A(3,4) with the center O(a,b). Because the triangle is isocele, the point O is on the vertical line through A, so a=3. (4-b)^2=R^2=9+b^2, so b=7/8. Finally R=4-7/8=25/8.

@劉益安-h8s 3 ай бұрын

設半徑長為R,畫∠BAC角平分線交BC於D,離A點R長度取得圓心O位置,則AD長為4,OD長為4-R , CD長為3→三角形ODC R^2= (4-R)^2+3^2→ R=25/8

@michaeldoerr5810 3 ай бұрын

The radius is 25/8. This is definitely an example of Pythagorean triples. Foe the first method, this involves the perpendicular bisector theorem and the intersecting chords theorem. For the second method, this uses the perpendicular bisector theorem and shwos the right scalene triangle that is contained in the equilateral triangle. I like the second method and I think that I need to start visualizing the equivaleny intersecting chids theorem on equilateral triangles. And I think that THIS is anothe property of Pythagorean triples. Also please look up my comment for yesterday's video.

@alexniklas8777 3 ай бұрын

4th method, the easiest 1) From vertex A we lower the perpendicular AP to BC. 2) From the middle of AC to AP, construct a segment NO perpendicular to AC. AN= NC= 5/2= 2.5. AO= R. 3) 5/4=R/2.5; R=5×2.5/4= 3.125

@ناصريناصر-س4ب 3 ай бұрын

In triangle ABC drawn in a circle of radius R where AH is the height related to side BC we have the relationship: AB*AC=AH*2R and from it 5*5=4*2R so R=25/8

@rajgopalraogade4351 Ай бұрын

Insted of using sine formula we can solve it in an easy way, viz , Extended BP touches the circumference (say) at Q. Then BP = 2C, Hence, (BP)^2 = (PQ)(AP) Or 3^2 =. 4(PQ) Or. PQ = 9/4 = 2.25 Or AQ = 4+2.25 = 6.25 which is 2r Hence r = 6.25/2 or 3.125

@zawatsky 25 күн бұрын

Ещё ∠ВОС=2∠ВАС (по правилу градусной меры дуги, на которую оба опираются). Так можно доказать, что ΔCOP~ΔACP, а значит - тоже египетский. Коэффициент подобия ¾, т. о. РС=¾*4, ОР=¾*3, ОС=R=¾*5=15/4=3³/₁₅=3¹/₅.

@Amt-p2h 25 күн бұрын

You deserve 10⁶ likes, thanks for ur valuable videos ❤

@lasalleman6792 3 ай бұрын

Find the angles. Simple Pythagorean. Then use this formula: circumradius = a/ 2sin(A) Where a , is a side of the triangle, and A, is the angle opposite of side a. Circumradius R = 3.125

@maluhadli468 3 ай бұрын

What is the value of the angle A how to find explainit

@lasalleman6792 3 ай бұрын

@@maluhadli468 It's a simple Isosceles triangle. You can split the existing triangle into two equal right triangles and simply use the Pythogorean rule to find the angles. The angles at B and C are both 53.13 degrees. Angle A is 73. 74 degrees. The height from line BC is 4. So you basically have two 3,4,5 right triangles joined together. They share the same height, which is 4. Measured from the base line BC

@Amt-p2h 25 күн бұрын

4th method ABC is the length of the largest arc in the circle BC is the length of the chord of the largest arc in the circle AP is the height of the triangle Its length = 4 Radius = [(the square of the length of the chord ÷ 8 the height of the triangle) + the length of the height of the triangle ÷ 2] R = [(6²÷8×4)+4÷2] =[(36÷32)+2]=3.125

@王剛-m7n 3 ай бұрын

If we use the formula r²=a²b²c²÷[(a+b-c)(b+c-a)(c+b-a)(a+b+c)], then it is so brutal

@is7728 3 ай бұрын

Me thinking about the same thing

@KipIngram 3 ай бұрын

Split the triangle vertically to get a 3/4/5 right triangle. The angle at C thus has sine of 4/5 = 0.8. By the law of sines we then have 5/0.8 = D, the diameter. Therefore the radius is 2.5/0.8 = 25/8. Q.E.D.

@adgf1x 13 күн бұрын

Ar. of inscribed triangle=12sq.Radius (R)=5×5×6/4×12=25/8 unit=3.125 unit.ans (by abc/4.delta)

@murdock5537 3 ай бұрын

4(2r - 4) = 9 → r = 25/8; or: ABC = BCA = δ → AOC = 2δ → sin⁡(δ) = 4/5 → cos⁡(δ) = 3/5 → cos⁡(2δ) = cos^2(δ) - sin^2(δ) = -7/25 → 25 = 2r^2(1 - cos⁡(2δ)) → r = 25/8

@RAHUL_CARS 2 ай бұрын

By using herons formula Half perimeter of ∆ABC is 5+5+6/2=8 :.area=√s(s-a)(s-b)(s-c) =√8(8-5)(8-5)(8-6) =√8*3*3*2 =3√2*2*2*2*2 ১ =12√2 1/2*6*x=12√2 3x=12√2 x=4√2 By Circle triangle theorem Radius=4√2*2\3 =8√2/3 =8*1.141/3 =3.77(aprox)

@bpark10001 Ай бұрын

There is simpler method than all 3 approaches shown. Proceed as in method 1 to get AP = 4. Construct line CQ. Triangle ACQ is right triangle & is similar to triangle ACP having 3:4:5 side ratio with scaling factor = 5/4. So AQ = (5/4)(5) = 25/4. Radius is half that = 25/8.

@KrylovYurii Ай бұрын

Both ACP and AQC are right triangles with common angle A. In this reason they are similar ==> |AQ|/|AC| = |AC|/|AP| ==> |AQ| = |AC|*|AC|/|AP| = 5 * 5 / 4 = 25/4 R = |AQ|/2 = (25/4)/2 = 25/8

@RAHUL_CARS 2 ай бұрын

7:17 Objection Why you useing pq as radius?

@cesarelai 3 ай бұрын

A theorem says radius = (ABxACxBC)/(4 x Area). Or: in triang. (ABQ) you can easily find PQ by Euclides 2nd Theorem: PQ = BP^2/AP

@alexniklas8777 3 ай бұрын

The easiest way R=abc/(4S); S= 6×4/2= 12 R= 5×5×6/(4×12)= 150/48= 3.125

@MrPaulc222 2 ай бұрын

I went for 4(2r - 4) = 9 8r - 16 = 9 8r = 25 r = 25/8 or 3.125 if preferred.

@SkinnerRobot 2 ай бұрын

Excellent!

@adgf1x 23 күн бұрын

ar.of inscribed circle=12 sq.radius of circle=5×5×6/4.12=25/8=3.125 unit.ans

@AlbaRosa-ui3xs 5 күн бұрын

Calcolato A P, si applica il teorema di Euclide al triangolo AQB (rettangolo perché iscritto in una semicircconerenza) e si calcola AQ...

@gnanadesikansenthilnathan6750 9 күн бұрын

Can we apply the concept of circumcentre

@RobertRoth-oj6zz Ай бұрын

The way the teachers explain it to you complicates it even more

@felixmgallegos3630 3 ай бұрын

(2"R-4)*4=3*3 por el Teorema de las cuerdas

@sujoymukherjee1930 2 ай бұрын

How do you conclude the perpendicular passes through centre?

@pojuellavid 2 ай бұрын

По определению. Перпендикуляр к хорде, проведенный через ее середину является диаметром

@kateknowles8055 3 ай бұрын

Midpoint of BC: calling it M . Extending AM to P where AP is diameter. BC =6 ,given. 6/2=3 BM.MC =3.3.=9 Two chords bisecting : product of their parts are equal. AM.MP =9 AM.AM= 5.5-3.3 =16 so AM=4. So MP=9/4 AP is diameter = 4 +9/4 equals 2.radius. 2r=6+1/4 Radius is 3+1/8. Now I'll check this.

@globalterroil3208 3 ай бұрын

I feel fully math boosted now!

@wasimahmad-t6c 2 ай бұрын

5×5=25-3×3=16squroth=4)(3×3=9÷4=2.5+4=6.5÷2=3.25×3.25=105626×3.14159268=33.183

@longduong7373 2 ай бұрын

r x cos(1/2 A) = 2.5 5 x cos(1/2 A) = 4 so we have: r/5 = 2.5/4 r= 25/8

@adgf1x 5 күн бұрын

radius=5×5×6/4.12=25/8=3.125unit.ans

@RealQinnMalloryu4 3 ай бұрын

(5)^2 =25 (5)^2 =25 (6)^2=36 {25+25+36}=86 360°ABC/86=4.16ABC 4.4^4 2^2.2^2^2^2 1^1.1^1^1^2 1^2 (ABC ➖ 2ABC+1).

@laurentblanchet1803 3 ай бұрын

cos A/2 = h÷5 (h hauteur du triangle isocèle de sommet A et de coté 5) cos A/2 = 2,5÷R (R rayon du cercle de cente O et de rayon recherché et 2,5 demi-base du triangle isocèle de sommet O et de base 5) h coté de l'angle droit du triangle rectangle d’hypoténuse 5 et de second coté de l' angle droit qui vaut 3 (demi-base du triangle isocèle de sommet A) Donc h=4 (Pythagore) d'où : 4÷5 = 2,5÷R R = (5 × 2,5)÷4 R = 3,125

@mthapa5 2 ай бұрын

It is very useful.

@CharlesChen-el4ot 3 ай бұрын

r^2 = 9 + (4 - r)^2 r^2= 9+ 16 - 8 r + r^2 8r = 25 r = 25/8= 3 + ( 1/8)

@richardmayerthaler3645 2 ай бұрын

Why so complicate? If you have QP = 9/4 and AP = 4 .... so you have the diameter 9/4 + 4 .... r = diameter/2

@wasimahmad-t6c Ай бұрын

94=2.25+4=6.25÷2=3.125×3.125×3.14159268=30.6796ful area

@nizarahammad4500 3 ай бұрын

Excellent

@violetadeliu734 2 ай бұрын

Gjate gjithr jetes time kam pasur respekt te vecante per perdonat qe kishin arsye te shendoshe ne matematike. Por me falni per nje verejtje ne vizatimin e krsaj figure. Te paktn me sy seg BC ta benit me te gjate. Kete e arrinit ta vendosni me afer pike O. Pak a shume edhe nje figure e ndertuar mire te ndihmon ne zgjidhje. Flm

@OguOne 2 күн бұрын

OM L AB => Tr. OMB=MOA MA = 2 1/2 => MA/R = 4/5 FROM ABP => R=25/8

@saeedhabibi547 3 ай бұрын

Very good

@viveksinghchauhanbikingsin1425 Ай бұрын

2/3 of perpendicular which is 4 ans is 8/3

@contelefurca 26 күн бұрын

minute 5.50- is there not an error ?

@sorourhashemi3249 2 ай бұрын

Thanks easy

@SOLOPOT_GH 3 ай бұрын

Why not adding 4 and 9/4 and divide by 2

@次野先生 Ай бұрын

R=5×5×6/(4×12)

@wasimahmad-t6c 2 ай бұрын

3.125×3.125×3.14159268=30.6796 full area

@subhashchandramishra8067 3 ай бұрын

Second method is easier than other methods

@ivaldopinhofilho7654 3 ай бұрын

S= abc/ 4r ,mais simples

@tuttouno 24 күн бұрын

AP2 = AP x PQ; 9=4xPQ; PQ=9/4 ...

@rgcriu2530 3 ай бұрын

👍👍👍👍👍👍👍👍👍👍👍👍

@qasima.5180 Ай бұрын

R=abc/4s.

@user-cq8xh5vl1p 7 күн бұрын

Wow what is this accent from which country?

@mohammadwatan1992 Ай бұрын

God taught mathematics and geometry to humans, so who can understand God's mathematics.

@rezatakdast 2 ай бұрын

3.13

@MathsSirg 3 ай бұрын

❤

@Ari-pw6nu Ай бұрын

Mericans no unusand maff.

@kolayskyboy2674 Ай бұрын

Faking difficult, so complex faking maths😅

@2012tulio 3 ай бұрын

3.12

@devondevon4366 3 ай бұрын

25/8

@ManojkantSamal 3 ай бұрын

25/8=3.25

@dipankardasgupta5125 2 ай бұрын

4 cm

@adgf1x 3 ай бұрын

r=25/8

@moviesforyou6794 2 ай бұрын

It's 3.125

@heriqnazinjixulyan9028 Ай бұрын

Ավելի էֆեկտիվ կլիներ օգտագործել լարին ուղղահայաց տրամագծի հատկությունը երբ հաշվել էր բարձրությունը

@prashantvishwa 2 ай бұрын

Pata nahi yaar main to sqrt(3^2+x^2)=4-x karke solve kar lunga maths student bhi nahi hun

@theresaotoole9141 Ай бұрын

The right angle 3,4,5 triangle is immediately obvious, while knowing that BP x PC (3 x 3 = 9) divided by 4 (= 2.25) gives you the diameter of 4 + 2.25 = 6.25. The radius is then 6.25/2 = 3.125 (or 25/8). Done in 20 seconds. Too easy.