x+y = 15 y+z = 14 z+x = 13 And we have system of three linear equations Two tangent segments are equal in length We can prove it if we draw two radiuses (radius is perpendicular to the tangent line)and use Pythagorean theorem We may also need to connect vertex and circle center
Use Heron's law to get the area and then the various angles. Total area is of triangle is 84. Use incircle rule to find radius of circle. Inradius = area/semiperimter. 84/21 = radius of 4. Make that inradius the sine of angle at point A which is 29.74 degrees. Then go Pythagorean. This produces a hypotenuse between point A and the center of the circle along a length of 8.07. So X, which is line AE becomes 7. X =7
Let's find x: . .. ... .... ..... AB, AC and BC are tangents to the inscribed circle. According to the two tangent theorem we know that AD=AE, BE=BF and CD=CF. So we can conclude: AD = AE = x BE = BF = AB − AE = 15 − x CD = CF = AC − AD = 13 − x BC = BF + CF 14 = (15 − x) + (13 − x) 0 = 14 − 2*x ⇒ x = 7 Best regards from Germany
Let the circle center be O, for sake of naming. By the Two Tangent Theorem, as DA and AE are line segments that are tangent to circle O and intersect at A, DA = AE = x. Similarly, as EB = AB-AE = 15-x and CD = CA-DA = 13-x, BF = EB = 15-x and FC = CD = 13-x. BC = BF + FC 14 = (15-x) + (13-x) 14 = 28 - 2x 2x = 28 - 14 = 14 x = 14/2 = 7 units
The answer is x = 7. By golly I am glad that I have reviewed the Two-Tangent Theorem. I wonder if your channel has a playlist of videos that makes use of Two-Tangent Theorem.
The solution for x is straight forward. It turns out the radius of the circle is a nice number as well, but I don’t know how to solve for it. This might be another good puzzle.
My way of solution ▶ Let's consider the right triangles ΔAEO and ΔAOD, for the trinagle ΔAEO ∠ OAE = α ∠ AEO = 90° ∠ EOA = β for the trinagle ΔAOD ∠ DAO = α ∠ ODA = 90° ∠ AOD = β both trinagles are similar: ΔAEO ~ ΔAOD while [AO] is the hypotenuse of the both right triangles, and [OD] =[OE] = r ⇒ [AE]= [DA]= x In the same way: [EB]= [BF]= y [FC]=[CD]= z ⇒ x+y= 15 y+z= 14 x+z= 13 ⇒ x+y= 15 x= 15-y ⇒ x+z= 13 15-y+z= 13 y-z= 2 ⇒ y-z= 2 y+z= 14 ⇒ 2y= 16 y= 8 ⇒ x= 7 z= 6 [AE]= [DA]= x= 7 length units
Thanks.but why I got 6.6 as answer. By Heron formula, area is 84. By connecting C to the tangent point in base 15, there is the height . 84 =15h/2===>h=11.2. So X^2+11.2^2=13^2===> X=6.6
Hi, here, centre(O) of circle to AB let's say OE makes 90⁰ with AB.BUT, O , C, E need not be collinear. They are collinear only incase of equilateral or isosceles triangle. By observing the diagram, we can say this triangle is neither of the above mentioned one. Also, the picture is not to scale.
xis =6. Method of Proof:- Let the External points are. A. B. And. C. Tangents are drawn to a circleof radius say R. The points where they touch the circle be D. E. And F. For the tangent of length 13 cms it 's point is D.Similarly for AC of length the touching point be E and the otherof BC ofit touches at F.BC need not be a tangent but the line drawn. AO the center being O extended to meet this BC line at F. Now we have 14-Y(Assume AD)=15-X. Next 14-Y=15-X. Then X-Y=1. But you have AB+AC-2Y=27-2Y.We have. BD as 13-Y-Y=1.@ 13-2Y=1. So Y =6.We have from AB BD is 14-Y. So X is 7 cms. Hence the answer.
انشااللة في ميزان حسناتك استاذ
14 = (13-x)+(15-x)
2x = 28-14 = 14
x = 7 cm ( Solved √ )
x+y = 15
y+z = 14
z+x = 13
And we have system of three linear equations
Two tangent segments are equal in length
We can prove it if we draw two radiuses (radius is perpendicular to the tangent line)and use Pythagorean theorem
We may also need to connect vertex and circle center
After seeing the Two-Tangent Theorem in previous videos, the problem became a piece of cake
Thanks Sir
Very nice and useful
Good luck with glades
❤❤❤❤
Use Heron's law to get the area and then the various angles. Total area is of triangle is 84. Use incircle rule to find radius of circle. Inradius = area/semiperimter. 84/21 = radius of 4. Make that inradius the sine of angle at point A which is 29.74 degrees. Then go Pythagorean. This produces a hypotenuse between point A and the center of the circle along a length of 8.07. So X, which is line AE becomes 7. X =7
(15 - x) + (13 - x) = 14
28 - 2x = 14
2x = 28 - 14
2x = 14
x = 7
Let's find x:
.
..
...
....
.....
AB, AC and BC are tangents to the inscribed circle. According to the two tangent theorem we know that AD=AE, BE=BF and CD=CF. So we can conclude:
AD = AE = x
BE = BF = AB − AE = 15 − x
CD = CF = AC − AD = 13 − x
BC = BF + CF
14 = (15 − x) + (13 − x)
0 = 14 − 2*x
⇒ x = 7
Best regards from Germany
Simple and elegant solution!
A different take on another puzzle.
AD = x
CD = 13-x
CB = 14, which splits into 13-xand 15-x.
(13-x) + (15-x) = 14
28-2x = 14
x=7
Let the circle center be O, for sake of naming. By the Two Tangent Theorem, as DA and AE are line segments that are tangent to circle O and intersect at A, DA = AE = x. Similarly, as EB = AB-AE = 15-x and CD = CA-DA = 13-x, BF = EB = 15-x and FC = CD = 13-x.
BC = BF + FC
14 = (15-x) + (13-x)
14 = 28 - 2x
2x = 28 - 14 = 14
x = 14/2 = 7 units
Great use of circle theorem and algebra
This was one of the easier ones.
Thank you 😊
x = ½(13+14+15)-14
x = 7 cm ( Solved √ )
AB=15=x+b→ b=15-x ; AC=13=x+c→ c=13-x ; BC=14=b+c=15-x+13-x→ x=7.
Gracias y un saludo
Fine. That was easy.
The answer is x = 7. By golly I am glad that I have reviewed the Two-Tangent Theorem. I wonder if your channel has a playlist of videos that makes use of Two-Tangent Theorem.
Thank you!
AD=AE=x CD=CF=13-x BE=BF=15-x 13-x+15-x=14 2x=14 x=7
Eb =15-x and same for fb which makes cf =14-(15-1)=x-1. Therefore x-1+x =13 therefore x = 7
AB=AE+BE
15=x+BE
So BE=15-x=BF
AE=AD=x
AC=AD+CD
13=x+CD
So CD=13-x
CD=CF=13-x
BC=BF+FC
14=13-x+15-x
So x=7 units.❤❤❤
2 tangent theorem and simul equation
The solution for x is straight forward. It turns out the radius of the circle is a nice number as well, but I don’t know how to solve for it. This might be another good puzzle.
х=7. AD=x, DC=13-x, CF=13-x, BE=15-x, BF=15-x, CF+BF=BC, 13-x+15-x=14, 28-2x=14, 2x=14< x=14/2, x=7.
thought by myself b4 watching
x + y = 15
x + z = 13
y + z = 14
x + y + z = 21
x = x + y + z - (y + z).= 21 - 14 = 7
x×y=15, y+z=14, z+x=13, so x=(x+y+z)-(y+z)=(13+14+15)/2-14=7.😊
My way of solution ▶
Let's consider the right triangles ΔAEO and ΔAOD, for the trinagle ΔAEO
∠ OAE = α
∠ AEO = 90°
∠ EOA = β
for the trinagle ΔAOD
∠ DAO = α
∠ ODA = 90°
∠ AOD = β
both trinagles are similar: ΔAEO ~ ΔAOD
while [AO] is the hypotenuse of the both right triangles, and
[OD] =[OE] = r
⇒
[AE]= [DA]= x
In the same way:
[EB]= [BF]= y
[FC]=[CD]= z
⇒
x+y= 15
y+z= 14
x+z= 13
⇒
x+y= 15
x= 15-y
⇒
x+z= 13
15-y+z= 13
y-z= 2
⇒
y-z= 2
y+z= 14
⇒
2y= 16
y= 8
⇒
x= 7
z= 6
[AE]= [DA]= x= 7 length units
X =7 was easy if all that is true. Question is can we have such a triangle with a circle inscribed? How do we prove that?
When you say two lines are congruent, shouldn't you say that they are equal length instead? Congruent just applies to shapes, angles etc
I agree with you
Isn't it that X is half of 15 which is 7.5 units?. E is tangent with the bottom of the circle. This is not a problem. Straight away u know the answer.
x=7
Extra credit: find r :-) It's a little more fun than finding x
Thanks.but why I got 6.6 as answer. By Heron formula, area is 84. By connecting C to the tangent point in base 15, there is the height . 84 =15h/2===>h=11.2. So X^2+11.2^2=13^2===> X=6.6
Connecting C to the tangent point in the 15 length segment doesn’t form a right angle. That segment doesn’t go through the circle center.
You are considering Base of the height on the contact point of the tangent,which is wrong !
@@brucezas1I connected the height from center of the circle to the tangent of 15 base. Line on the tangent of cricle makes angle of 90°
@@nicolewhite5296if we mark the center as "O" we have 3 triangles COA, COB, AOB IN WHICH all have the height of "r"
Hi, here, centre(O) of circle to AB let's say OE makes 90⁰ with AB.BUT, O , C, E need not be collinear. They are collinear only incase of equilateral or isosceles triangle. By observing the diagram, we can say this triangle is neither of the above mentioned one. Also, the picture is not to scale.
X+x+1=15...x=7
Next, calculate the diameter of the circle?🤔
Cosine formula to find angle A.
Join AO.
Tan (A / 2) = r / 7.
Xis 7 cms. Not 6 as mentioned first.
STEP-BY-STEP RESOLUTION PROPOSAL :
01) Triangle [ABC] Area = 84 sq un. Using Heron Formula.
02) Triangle [ABC] Perimeter = 13 + 14 + 15 = 14 * 3 = 42 lin un
03) AD = AE = X lin un
04) CD = CF = (13 - X) lin un
05) BE = BF = (15 - X) lin un
06) AD + AE + CD + CF + BE + BF = 42
07) X + X + 2 * (13 - X) + 2 * (15 - X) = 42
08) 2X + 26 - 2X + 30 - 2X = 42
09) - 2X = 42 - 56
10) -2X = - 14
11) X = -14 / -2
12) X = 7
Therefore,
OUR BEST ANSWER IS :
X equal 7 Linear Units.
Not 14-Y but 13-Y. So that X is 7.
xis =6.
Method of Proof:- Let the External points are. A. B. And. C. Tangents are drawn to a circleof radius say R. The points where they touch the circle be D. E. And F. For the tangent of length 13 cms it 's point is D.Similarly for AC of length the touching point be E and the otherof BC ofit touches at F.BC need not be a tangent but the line drawn. AO the center being O extended to meet this BC line at F.
Now we have 14-Y(Assume AD)=15-X. Next 14-Y=15-X. Then X-Y=1. But you have AB+AC-2Y=27-2Y.We have. BD as 13-Y-Y=1.@ 13-2Y=1. So Y =6.We have from AB BD is 14-Y. So X is 7 cms. Hence the answer.
14-(15-X)=13-X ; 14-15+X=13-X ; 14+2X=28 ; 2X=14 ; X=7
15-x=x+1
X=7