Can you find area of the Triangle ABC? | (Inscribed Circle) |
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- Опубликовано: 8 фев 2025
- 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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Very well explained.
I can say you are an expert in Analytical Geometry.
So nice of you Ramani dear 🌹
Thanks for the feedback ❤️
(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
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
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
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
Thank you!
Nice
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
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
Well Done!!
ABC Area = DC * AD = 10 * 11/9 = 110/9 Square Units.
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
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
Thank you
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...
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,
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
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.
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
S=110/9≈12,22 cm²
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
Ok ! I see my mistake : the length 10 is DC, not AC. The diagram is a bit confusing
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.❤❤❤
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
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.
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.
Ooooo! 😂
if BF=1 how FC=10? you shoud put in the exercice FC about 5, that seems more logical than 10.
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...