Can you find area of the Triangle ABC? | (Inscribed Circle area is 225pi) |

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Learn how to find the area of the Triangle ABC. Important Geometry skills are also explained: circle theorem; Heron's formula; triangle area formula; Two tangent theorem. Step-by-step tutorial by PreMath.com
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Can you find area of the Triangle ABC? | (Inscribed Circle area is 225pi) | #math #maths | #geometry
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Комментарии : 44

@mehdismaeili3743 3 месяца назад

Excellent .

@PreMath 3 месяца назад

Excellent! Glad to hear that! Thanks for the feedback ❤️

@montynorth3009 3 месяца назад

Tan EAO = 15/25 = 0.6. EAO = 30.96 degrees. Tan EBO = 15/26 = 0.577. EBO = 29.98 degrees. Sum of 3 half angles in triangle = 90 degrees. Therefore FCO = 90 - 30.96 - 29.98 = 29.06. Tan FCO = 15/FC. FC = 15/tan 29.06 = 27. Total area = sum of 6 small triangles. (25 x 15 ) + (26 x 15) + (27 x 15). (3 congruent pairs) 1170.

@awandrew11 3 месяца назад

Use similar triangles CEB and COF give an area of 1147?

@LuisdeBritoCamacho 3 месяца назад

I solved this Problem with a little help from my friend, Analytic Geometry: 01) Coordinates of Center of the Circle : (25 ; 15) with Radius equal to 15. 02) Coordinates of Point A (0 ; 0). 03) Coordinates of Point B (51 ; 0). 04) Two Straight Lines Tangent to Circle. Straight Line AC and Straight Line BC. 05) The interception of these two Straight Lines. 06) Solution Coordinates : (24,471 ; 45,882). 07) Area of Triangle ABC = (45,882 * 51) / 2. 08) A = 2.339,982 / 2. 09) A =1.169,991 Square Units. 10) ANSWER : Area of Triangle ABC equal to 1.170 Square Units. That's all for today.

@bobbyheffley4955 3 месяца назад

Placing one point of the triangle at the origin simplifies calculations.

@PreMath 3 месяца назад

Excellent! Thanks for sharing ❤️

@quigonkenny 3 месяца назад

Circle O: Aᴏ = πr² 225π = πr² r² = 225 r = √225 = 15 As DA and AE are tangents of circle O that intersect at A, AE = DA = 25. As BF and EB are tangents of circle O that intersect at B, EB = BF = 26. As DA = AE = 25, OD = OE = 15, and OA is common, ∆ODA and ∆AEO are congruent triangles by side-side-side. Triangle ∆AED: EO² + AE² = OA² 15² + 25² = OA² OA² = 225 + 625 = 850 OA = √850 = 5√34 Let ∠DAO = ∠OAE = α. ∠DAE = ∠DAO+∠OAE = α+α = 2α. cos(2α) = 2cos²(α) - 1 cos(2α) = 2(25/5√34)² - 1 cos(2α) = 2(25/34) - 1 cos(2α) = 25/17 - 1 = 8/17 sin(2α) = 2sin(α)cos(α) sin(2α) = 2(15/5√34)(25/5√34) sin(2α) = 2(15/34) = 15/17 As FC and CD are tangents to circle O that intersect at C, FC = CD = x. By the law of cosines: BC² = AB² + CA² - 2AB(CA)cos(2α) (x+26)² = 51² + (x+25)² - 2(51)(x+25)(8/17) x² + 52x + 676 = 2601 + x² + 50x + 625 - 48x - 1200 52x + 676 = 2026 + 2x 50x = 1350 x = 27 Triangle ∆ABC: Aᴛ = AB(CA)sin(2α)/2 Aᴛ = 51(52)(15/17)/2 Aᴛ = 3(26)(15) = 45(26) = 1170 sq units

@marcgriselhubert3915 3 месяца назад

Something different: The radius of the circle is15 (evident) Let a = angleOAE and b = angle OBE In triangle OAE: tan(a) = OE/AE = 15/25 = 3/5 In triangle OBE: tan(b) =OE/BE = 15/26 angleBAC = 2.a and tan(2.a) =(6/5)/(1 - 9/25) =15/8 angle ABC = 2.b and tan(2.b) = (30/26)/(1 -225/676) = 780/451 Let H the orthogonal projection of A on (AB) and h = CH In triangle AHC: tan(2.a) = h/AH, so AH = h/tan(2.a), so AH = (8/15).h In triangle BHC: tan(2.b) = h/BH, so BH = h/tan(2.b) = (451/780).h As AH + BH = AB = 51, we have 51 = ((8/15) + (451/780)).h = (867/780).h, So h = (780.51)/867 = 780/17. Now the area of ABC is (1/2).AB.h .So it is: (1/2).(51).(780/17) = 1170.

@PreMath 3 месяца назад

Excellent! Thanks for sharing ❤️

@unknownidentity2846 3 месяца назад

Let's find the area: . .. ... .... ..... First of all we calculate the radius R of the inscribed circle: A = πR² 225π = πR² 225 = R² ⇒ R = 15 According to the two tangent theorem we know that AE=AD=25, BE=BF=26 and CD=CF=x. The area of the triangle can be calculated either from its perimeter P and the radius R of its inscribed circle or according to the formula of Heron: AB = AE + BE = 25 + 26 = 51 AC = AD + CD = 25 + x BC = BF + CF = 26 + x P = AB + AC + BC = 51 + (25 + x) + (26 + x) = 102 + 2*x ⇒ P/2 = 51 + x R*P/2 = √[(P/2)(P/2 − AB)(P/2 − AC)(P/2 − BC)] R²*(P/2)² = (P/2)(P/2 − AB)(P/2 − AC)(P/2 − BC) R²*P/2 = (P/2 − AB)(P/2 − AC)(P/2 − BC) 15²*(51 + x) = [(51 + x) − 51][(51 + x) − (25 + x)][(51 + x) − (26 + x)] 225*(51 + x) = x*26*25 9*(51 + x) = 26*x 459 + 9*x = 26*x 459 = 17*x ⇒ x = 27 Now we are able to calculate the area of the triangle: P = 102 + 2*x = 102 + 2*27 = 156 A(ABC) = R*P/2 = (1/2)*15*156 = 1170 Let's check the result: AB = 51 AC = 25 + 27 = 52 BC = 26 + 27 = 53 P = 51 + 52 + 53 = 156 ⇒ P/2 = 78 A(ABC) = √[78*(78 − 51)*(78 − 52)*(78 − 53)] = √(78*27*26*25) = √(3*26*3*9*26*25) = 9*26*5 = 1170 ✓ Best regards from Germany

@unknownidentity2846 3 месяца назад

Solved exactly like PreMath. Does that mean I am a math genius, at least for today? OK, just kidding.🙂

@robertlynch7520 3 месяца назад

I guess I'm the only fool that tried the tan( 2 theta ) formula to solve this. Once the radius is determined to be 15 units, then the slope of the left side of the triangle is tan 2θ = 2 tan θ / (1 - tan² θ) is the identity Working with the 'any triangle with a circle inscribed', having a radius O and an adjacent A tan θ = O/A = 15/25 ... thus tan 2θ = 2 OA / (A² - O²). ... which susses out asa tan 2θ = 750/400 = 15/8 For the right hand part, its exactly the same but with 26 in stead of 25 tan 2Φ = 780/451 Then since the tangents are ALSO the slopes of the lines, we have 2 line functions which can be equated: 15/8 x = -780/451 x + 88.204 x = 88.204 / 3.6045 x = 24.471. ... and plugging back in to get y, the height y = 45.882 Well, now we have the baseline (25 + 26 = 51) and a height (45.882). Area of triangle is 0.5 * 51 * 45.882 = 1170. exactly See? none of that special triangles memorization. Just good old algebra and a single trig identity. GoatGuy

@PreMath 3 месяца назад

Excellent! Thanks for sharing ❤️

@devondevon4366 3 месяца назад

1,170.15 The radius of the circle = 15 (sqrt 225) Draw a perpendicular line from D to O (the circle's center) and from the center to E . Draw a line from O to A, forming two congruent right triangles ADO, and ADE Since length DO =15 and AD 25, then AO = 5 sqrt 34 (Pythaogrean Theorem) Angle AOD = Sine 90 * 25/ sqrt 34 (Law of sine) = 0.8575 = 59.036 degrees Hence, AOE, also = 59.036 degrees Draw a perpendicular line from O to F and from O to E . Draw a line from O to B, forming two congruent right triangles , BFO, and BEO. I am basically doing the same thing I did on the left side. Again, using the law of sines (not shown), angle BOF = 60.018 degrees Hence , BOE also = 60.018 degrees Add these all together 59.036 + 59.036 + 60.018 + 60.018 = 238.108 Hence, angle AOF (the top of the triangle) = 121.892 degrees ( 360 -238.108) Draw a perpendicular line from DO and OF and a line from O to C to form two congruent triangles since angle AOF = 121.892 degrees (see just above), then one of the angles of the congruent triangle = 60.946 (or 1/2 of 121.892) Using the Law of Sine, the length of DC = 27.01 (recall there are two of them) Let's find the area of these 6 triangles or three pairs of congruent triangles 27.01 * 15 = 405.15 (no need to divide by 2 since both congruent triangles have the same area) the other two congruent triangles' dimensions were (see above) 15 * 26 = 390 (again,n since there are two, there is no need to use 1/2 base * height) The other congruent triangle dimensions were 15 * 25 = 375 (again, since there are two, there is no need to use 1/2 base * height) Add these together 405. 15 + 390 + 375 =1,170.15 Answer

@PreMath 3 месяца назад

Excellent! Thanks for sharing ❤️

@popoolachristopher5457 3 месяца назад

I don't think the analysis is correct...an inscribed circle in equilateral triangle touching the circle at different distance to the tip of the circle...pls let look into that....I need more clarification

@SkinnerRobot 3 месяца назад

It's not an equilateral triangle.

@misterenter-iz7rz 3 месяца назад

Using trigonometric method to determine the last side, it is 27, then the answer is 15×78=1170.😊

@AmirgabYT2185 3 месяца назад

It looks like an equilateral triangle 😂😂😂

@hongningsuen1348 3 месяца назад

Method using calculator: 1. Let CG be the height of triangle ABC with G on AB. Then area of triangle ABC = (1/2)(25+26)(CG)= (51/2)(CG) 2. Radius of incircle = 15 (from given area of incircle) 3. Angle OAD = arctan (15/25) and angle OBF = arctan 15/26 4. Angle CAG = (2 x angle OAD) and angle CBG = (2 x angle OBF) (Vertex to incentre are angle bisectors.) 5. (CG)/tanCAG + (CG)/tanCBG = 25 + 26 CG = 51/(1/tanCAG + 1/tanCBG) 6. Area of triangle ABC = 51^2/2(0.5333+0.5782) = 1170.0404

@prossvay8744 3 месяца назад

Area of circle=225π πr^2=225π r^2=225 So r=15 AD=AE=25 BE=BF=26 CD=CF=x S=(25+x+26+x+51)/2=51+x Area of the triangle=√(51+x)(51+x-25-x)(51+x-26-x)(51+x-51)=5√1326x+26x^2) And area of the triangle=15/2(25+x+26+x+51)=5√1326x+26x^2 So x=27 Hence area of the triangle ABC=√(51+27)(25)(26)(27)=1170 square units.❤❤❤

@AdemolaAderibigbe-j8s 3 месяца назад

Let radius of circle be R. then Pi*(R^2) = 225*Pi so R = 15cm. We draw radii OE, OD and OF perpendicular to AB, AC and CB respectively. Also tangents from an eternal point to a circle are equal in length so AD = AE = 25, EB = FB = 26 and DC = CF = X. Thus triangles AEO and ADO are congruent, triangles EOB and OFB are congruent and triangles ODC and OCF are congruent. Let angle AOE = angle AOD = a, angle FOB = angle EOB = b and angle DOC = angle COF = c. Then Tan(a )= 25/15 and a =59.0362 degrees. Tan(b) = 26/15 and b = 60.01836 degrees. Then all angles at the center of the circle sum to 360 degrees so 2*c = (360 - 2*a -2*b) and thus c = 60.9454 degrees and Tan(c) = X/R = X/15 and X = 15*Tan(60.9454) = 27. Area of triangle ABC = (2*Area of triangle AOE + 2*Area of triangle FOB +2* Area of triangle COD) = (25*R + 26*R + 27*R) = 1170 cm^2.

@leonidtsilker8330 3 месяца назад

Вычисление x не требует ни одного уравнения. tg(COF) = tg((90-(BOF+AOD)) = ctg(BOF+AOD) = 1/tg(BOF+AOD) = (1-tg(BOF)*tg(AOD))/(tg(BOF)+tg(AOD)) = (1-(15/26)*(15/25))/((15/26)*(15/25))) = 5/9. Отсюда x = OF/tg(COF) = 15/(5/9) = 27.

@Vakyanshpro 3 месяца назад

Make vids on integration

@DaveKube-cx4sn 3 месяца назад

You are the best. Even a little better then MindYourDecision and it is very good channel. You invent these problems alone or it is taken over from somewhere else? You are really very good mathematician.👍👏

@PreMath 3 месяца назад

You are too generous! I'm just an ordinary human being. We are all lifelong learners. That's what makes our life exciting and meaningful! Thanks for the feedback ❤️ Stay blessed😀

@mvrpatnaik9085 2 месяца назад

Excellent Approach.

@jimlocke9320 3 месяца назад

Several viewers have noted that the problem may be solved by trigonometry. Here's my trigonometric version, which may be the same as another viewer's. Construct OA, OB, OC, OD, OE and OF, resulting in 6 triangles, which are congruent in pairs, ΔAOE and ΔAOD, ΔBOE and ΔBOF, ΔCOD and ΔCOF. We have found that the radius of the circle is 15, so OD = OE = OF. We could take the tangents of

@MrPaulc222 3 месяца назад

The two immediate observations are that r =15 and AB = 51 (due to 2-tangents). CD = CF = x OFB is 390/2 = 195 ODA is 375/2 AOB is 765/2 Area excluding CDOF = 195 + 187.5 + 382.5 = 765, which indicates that OFB + ODA = ADB This leaves the awkward quadrilateral CDOF OD = OF = 15 CD = CF and ODC = OFC Call AC, x and BC, x+1 (Scrap some of the above as I am changing tack). I wanted to avoid trigonometry, but can't see how else to do this. Ultimately, I want

@giuseppemalaguti435 3 месяца назад

AB=25+26=51..CB=26+15tg(arctg15/26+arctg15/25)=26+27=53...Atr=(1/2)53*51*sin(2arctg15/26)=53*51*390/901=1170

@mosquitobight 3 месяца назад

I find the fact that the length of all three sides of the triangle are integers even more impressive than the solution to the problem.

@luigipirandello5919 3 месяца назад

Professor, poderia fazer um vídeo demonstrando a fórmula de Heron, por favor? Estou curioso para saber de onde vem essa fórmula.

@jamestalbott4499 3 месяца назад

Thank you!

@PreMath 3 месяца назад

You are very welcome! Thanks for the feedback ❤️

@Holyxxrd 3 месяца назад

5:31 what is the name of theory of triangle area=rs

@saketkumar4080 3 месяца назад

It's 10th grade concept

@rorydaulton6858 3 месяца назад

I've never seen a name for that equation, but I suppose we could call it the inradius area formula. You can see it in Wikipedia: see either the "Area of a Triangle" page 'Other area formula's' section or the "Incircle and Excircles" page 'Relation to the area of a triangle' section.

@unknownidentity2846 3 месяца назад

AB, AC and BC are tangents to the circle. Therefore we known that OE (OD, OF) is perpendicular to AB (AC, BC). So we can conclude: A(ABC) = A(ABO) + A(ACO) + A(BCO) = (1/2)*AB*OE + (1/2)*AC*OD + (1/2)*BC*OF With r=OE=OD=OF we obtain: A(ABC) = (1/2)*AB*r + (1/2)*AC*r + (1/2)*BC*r = r*(AB + AC + BC)/2 With s=(AB+AC+BC)/2 being half of the triangles perimeter we get A(ABC)=r*s. Best regards from Germany