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

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Learn how to find the area of the Triangle ABC. Important Geometry skills are also explained: circle theorem; Pythagorean theorem; triangle area formula; Two tangent theorem. Step-by-step tutorial by PreMath.com
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Комментарии : 34

@ramanivenkata3161 Месяц назад

Very well explained. I can say you are an expert in Analytical Geometry.

@PreMath Месяц назад

So nice of you Ramani dear 🌹 Thanks for the feedback ❤️

@giuseppemalaguti435 Месяц назад

(r+a)^2+(r+10)^2=(a+10)^2...r=1...1+2a+121=20a+100..22=18a ..a=11/9....Ablue=11(1+11/9)/2=110/9

@Mediterranean81 Месяц назад

area of triangle = rs= 1*s=s pythagorean theorem (x+1)^2+11^2=(10+x)^2 x^2+2x+1+121=100+200x+x^2 2x+22=200x 22=198x 1/9=x semi perimeter : (x+1+11+x+10)/2=(2x+22)/2=x+11 rs= 1(1/9+11) 11+99/9 = 110/9

@phungpham1725 Месяц назад

Alternative solution: Let A and r be the area of ABC and the radius of the circle. Just label AD=AE= m and DC= CF= n A= area of the square BEOF + area of AEOD + area of ODCF = sqr+ rm+rn And A= 1/2 AB.BC= 1/2 (r+m)(r+n)= 1/2 (sqr+rm+rn+mn)= 1/2(A+mn) --> A=mn Because r= 1 so, 1/2 ( m+1)(1+10) = 10 m --> m= 11/9 Area= mn= 110/9 sq cm

@user-qt8ks4gc2l Месяц назад

The area of a right triangle is the product of the two segments in which the hypotenuse is divided by the tangency point of the inscribed circle. In our case, area of ABC = AD * DC

@LuisdeBritoCamacho Месяц назад

Well Done!! ABC Area = DC * AD = 10 * 11/9 = 110/9 Square Units.

@himo3485 Месяц назад

r＊r＊π=π(cm^2) r=1 DC=DF=10 AD=AE=x (x+1)^2+(1+10)^2=(x+10)^2 x^2+2x+1+121=x^2+20x+100 18x=22 x=11/9 area of the Blue triangle : (11/9+1)＊11＊1/2=110/9cm^2

@marcgriselhubert3915 Месяц назад

EB = BF = 1 = radius of the circle. Let's now note AE = AD = x. In triangle ABC the Pythagorean theorem gives: (x +1)^2 + 11^2 = (x +10)^2, or x^2 + 2.x +1 + 121 = x^2 + 20.x +100, or 18.x = 22, or x = 22/18 = 11/9. Then AB = 20/9. The area of triangle ABC = (1/2).AB.BC = (1/2).(20/9).11 = 110/9

@jamestalbott4499 Месяц назад

Thank you!

@AmirgabYT2185 Месяц назад

S=110/9≈12,22 cm²

@prashant245100 Месяц назад

Nice

@raya.pawley3563 Месяц назад

Thank you

@santiagoarosam430 Месяц назад

CD=CF=10 ; AD=AE=t ; EB=BF=OD=OE=OF=Radio =r=1→ Área ABC =(t*1)+(1*1)+(10*1) = (1+10)*(t+1)/2→ t=11/9→ AB=20/9 ; BC=11→ Área ABC=(20/9)*11/2=110/9 =12,2222.......ud² Gracias y un saludo cordial.

@himadrikhanra7463 Месяц назад

R in=1 Delta/s=1 10-10,x-x,y-y 20+ 2x + 2y 2( 10 +x+y) Have a formula to find radius of excircle, unable to remember now ...there after we can get value of perimeter...after delta blue= 1 × semi perimeter...

@davidzagorski9756 Месяц назад

No need for the Pythagorean Theorem. Solve for BE as shown. Now you have K=rS, so the area of the triangle equals the radius of the inscribed circle multiplied by the semiperimeter. And take it from there.

@grink_coolhoznik Месяц назад

The area of the circle is pi*r^2. pi*r^2=pi; r^2=1; r=±1. The radius of the circle can only be positive, so r=1. AB, BC and AC are the tangents of the circle, so OE=OF=OD=r are the perpendiculars of the triangle's sides, where E, F and D are the points of intersection. Let's build a square EOFB. It's the square, because OE=OF=1 and the angle EBF=90 degrees. BO is the diagonal of the square; BO=sqrt(1+1)=sqrt(2). Let's build a line segment BD. BD is the altitude of the ABC for the hypotenuse AC. BD=BO+OD=sqrt(2)+1. The area of triangle is bh/2, where h is the altitude and b is the side that's perpendicular for the altitude. The area of ABC=(sqrt(2)+1)*10/2=5sqrt(2)+1 cm^2

@jimlocke9320 Месяц назад

Solution using the tangent double angle formula. At about 4:35, construct OC to form ΔCOF and add OD to form ΔCOD. Note that ΔCOF and ΔCOD are congruent by side - angle - side (OF = OD = radius,

@quigonkenny Месяц назад

Circle O: A = πr² π = πr² r² = 1 r = 1 FC and DC are tangents of circle O that intersect at C, so FC = DC = 10. AE and AD are tangents of circle O that intersect at A, so AE = AD = x. AB and BC are tangent to circle O at E and F respectively, so ∠BFO = ∠OEB = 90°. As ∠EBF = 90° as well and OE = OF = 1, then ∠FOE must equal 90° and BFOE is a square with side length 1. Triangle ∆ABC: AB² + BC² = CA² (x+1)² + 11² = (x+10)² x² + 2x + 1 + 121 = x² + 20x + 100 18x = 22 x = 22/18 = 11/9 Area = bh/2 = 11(20/9)/2 = 110/9 cm²

@egillandersson1780 Месяц назад

I find 11 ! With the sides of the triangle a=6+sqrt(14) and b=6-sqrt(14) : ab=22 (A=ab/2=11) and a^2+b^2=100=10^2

@egillandersson1780 Месяц назад

Ok ! I see my mistake : the length 10 is DC, not AC. The diagram is a bit confusing

@devondevon4366 Месяц назад

12.22 radius =1 Hence, the base =11 and the height = p+1 and the hypotenuse= 10 + p Let's employ Pythagorean Theorem ( p+1)^2 + 11^2 = (p+10)^2 p^2 + 2p +1 + 121= p^2 + 20p + 100 2p + 122= 20p + 100 22 = 18p 1.222 = p Hence, base = 12.2222 height = 2.2222 Area =12.22222

@wackojacko3962 Месяц назад

Ooooo! 😂

@baturnal89 Месяц назад

I always calculate the unidentified value {which in this case is 11/9} and then i realize that's not the whole answer, i was supposed to calculate the area of the triangle.

@misterenter-iz7rz Месяц назад

Rather easy, (a+10)^2=11^2+(a+1)^2, 18a=121+1-100=22, a=11/9, therefore th😂,e area is 1/2×11×20/9)=110/9.😊

@vivificateurveridique1420 Месяц назад

if BF=1 how FC=10? you shoud put in the exercice FC about 5, that seems more logical than 10.

@honestadministrator 20 дней назад

b * c /2 = ( b + c + √ ( b^2 + c^2)) /2 Herein (b + c + √ ( b^2 + c^2)) * ( b + c - √ ( b^2 + c^2)) = ( b + c) ^2 - ( b^2 + c^2) = 2 b c Hereby ( b + c) - √( b^2 + c^2) = 2 Given √( b^2 + c^2) = 10 Hereby b + c = 12 2 b c = ( b + c) ^2 - (b^2 + c^2) = 144 - 100 b c = 11 Area of ∆ A B C = b c /2 = 11/2

@prossvay8744 Месяц назад

Area of circle=π=πr^2 so r=1cm CD and CF are tangent So CD=CF=10cm BEOF is square BE=EO=OF=FB=1cm BC=BF+CF=1+10=11cm Let AB=x AE=AP=x-1 AC=x-1+10=x+9 x^2+11^2=(x+9)^2 x^2+121=x^2+18x+81 18x=121-81=40 x=40/18=20/9cm Area of the Blue triangle=1/2(20/9)(11)=110/9cm^2=12.22cm^2.❤❤❤

@unknownidentity2846 Месяц назад

Let's find the area: . .. ... .... ..... First of all we calculate the radius R of the inscribed circle: A(circle) = πR² πcm² = πR² 1cm² = R² ⇒ R = 1cm All three sides of the triangle are tangents to the circle. Therefore we can conclude: ∠ADO = ∠CDO = 90° ∠AEO = ∠BEO = 90° ∠BFO = ∠CFO = 90° AD = AE BE = BF CD = CF (= 10) The area of the triangle can be calculated from its perimeter and the radius of the inscribed circle: A(ABC) = P(ABC)*R/2 Since ABC is a right triangle, its area can also be calculated in the following way: A(ABC) = (1/2)*AB*BC Now we can conclude: P(ABC) = AB + AC + BC = AE + BE + AD + CD + BF + CF = AD + R + AD + CD + R + CD = 2*(AD + CD + R) P(ABC)*R/2 = (1/2)*AB*BC P(ABC)*R = AB*BC 2*(AD + CD + R)*R = (AE + BE)*(BF + CF) 2*(AD + CD + R)*R = (AD + R)*(R + CD) 2*(AD + 10cm + 1cm)*(1cm) = (AD + 1cm)*(1cm + 10cm) 2*(AD + 11cm) = (AD + 1cm)*11 2*AD + 22cm = 11*AD + 11cm 11cm = 9*AD ⇒ AD = (11/9)cm P(ABC) = 2*(AD + CD + R) = 2*(11/9 + 10 + 1)cm = 2*(11/9 + 11)cm = (220/9)cm A(ABC) = P(ABC)*R/2 = [(220/9)cm]*(1cm)/2 = (110/9)cm² AB = AE + BE = AD + R = (11/9 + 1)cm = (20/9)cm BC = BF + CF = R + CD = (10 + 1)cm = 11cm A(ABC) = (1/2)*AB*BC = (1/2)*[(20/9)cm]*(11cm) = (110/9)cm² ✓ Best regards from Germany

@LuisdeBritoCamacho Месяц назад

STEP-BY-STEP RESOLUTION : 1) Radius of Pink Circle is R = 1 2) CD = CF = 10 3) OC = sqrt(101) 4) BF = BE = 1 5) BC = BF + CF ; BC = 1 + 10 ; BC = 11 5) AE = AD = X 6) AC = 10 + X 7) AB = 1 + X 8) BC^2 = AC^2 - AB^2 9) 11^2 = (10 + X)^2 - (1 + X)^2 10) 121 = 100 + 20X + X^2 - 1 - 2X - X^2 11) 121 - 100 + 1 = 20X - 2X 12) 22 = 18X 13) X = 22/18 14) X = 11/9 15) AE = AD = 11/9 16) AB = 1 + 11/9 ; AB = 20/9 17) BC = 11 18) AREA = (20/9) * 11) / 2 19) AREA = 220/18 20) AREA = 110/9 Square Cm 21) ANSWER : The Area of Blue Triangle is equal to 110/9 Square Cm or approx. 12, 22(2) Square Cm.

@sergeyvinns931 Месяц назад

The redius the inscribed circie is 1, CD=CF=10, BC=CF+r=11, BD^2=BC^2-CD^2=21, BC=\/21, triangles BCD and BAD are similar, BD/AB=10/11, 10AB=11\/21, АВ=11\/21/10. Areq of the triqngle ABC = AB*BC/2 = 11*11\/21/10*2= =121\/21/20=27,7245829544...