Can you find the radius of the inscribed semicircle? | (Heron's Formula) | #math #maths | #geometry

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Can you find the radius of the inscribed semicircle? | (Heron's Formula) | #math #maths | #geometry

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[michaeldoerr]

I got that!!! Looks like we need more Heron's formula!!!

[Golololololo]

At first, looks difficult to solve; but when you start solving, it sounds easy!
Thanks

[dariosilva]

Wow, the picture can easily mislead you. You might easily assume AO = CO = 15/2. But the O is the center of the circle, not center of AC.

[MrPaulc]

Thank you. I did manage to get this one, but not before I went down a dead end first.

[giuseppemalaguti]

Briggs, cos(BAC/2)=√21*7/13*15=7/√65...cos(BCA/2)=√12/√15...r/sin(2arccos7/√65)+r/sin(2arccos√12/√15)=15...r=56/9

[ChuzzleFriends]

This is, again, an earlier lesson, but now, we just have to find the radius.
Draw radii DO & EO, they are perpendicular to sides BC & AB by the Circle Theorem.
Use Heron's Formula by finding the semi-perimeter of △ABC.
s = (a + b + c)/2
= 1/2 * (14 + 15 + 13)
= 1/2 * 42
= 21
A = √[s(s - a)(s - b)(s - c)]
= √[21 * (21 - 14) * (21 - 15) * (21 - 13)]
= √(21 * 7 * 6 * 8)
= √7056
= 84
Draw segment BO. It separates △ABC into two triangles, △AOB & △AOC. Find their areas.
A = (bh)/2
= 1/2 * 13 * r
= (13r)/2
A = 1/2 * 14 * r
= 7r
△ABC Area = △AOB Area + △AOC Area
84 = (13r)/2 + 7r
84 = (13r)/2 + (14r)/2
(13r + 14r)/2 = 84
(27r)/2 = 84
27r = 168
r = 168/27
So, the radius of the semicircle is 168/27 units (exact), or about 6.22 units (approximation).

[ОльгаСоломашенко-ьы]

По теореме косинусов находим cos(<A)=33/65, cos(< C)=3/5. sin(<A)=56/65, sin(<C)=4/5. r/( sin(<A))+r/(sin(<C))=15. r=56/9.

[prossvay]

S=(13+14+15)/2=21
area of triangle
Connect O to B to D and to E
Area of triangle ABC=area of triangle ABO + area of triangle BOC
84=1/2(13)(r)+1/2(14)(r)
So r=56/9 units =6.22 units.❤❤❤

[sundareshvenugopal]

(13 - x)^2 + r^2 = (14 - y)^2 + r^2 = 15^2.

x^2 + r^2 = y^2 + r^2.

[alexniklas]

Found:
AO=65/9; AE= 11/3.
According to the Pythagorean theorem
r = √(AO^2-AE^2)= 56/9= 6.22

[gelbkehlchen]

Solution:
Heron’s Formula says for the area A of a triangle:
A = √[s*(s-a)*(s-b)*(s-c)] with s = (a+b+c)/2 = (14+15+13)/2 = 21
(1) A = √[21*(21-14)*(21-15)*(21-13)] = √[21*7*6*8] = 84

One can draw a line from O to E, from O to B and from O to D. The the triangle ABC has also the area with r = OE = OD = radius of the semicircle:
(2) AE*r/2+EB*r/2+BD*r/2+DC*r/2 = (AE+EB+BD+DC)*r/2 = (AB+BC)*r/2 = (13+14)*r/2
= 13, 5*r
(1) = (2) ⟹ 84 = 13, 5*r |/13, 5 ⟹ r = 84/13, 5 = 168/27 = 56/9 = 6+2/9

[LuisPCoutinho]

Total Area = 84 sq un

13R/2 + 14R/2 = 27R/2

84 = 27R/2

168 = 27R

R = 168/27 Linear Units.

[santiagoarosam]

(AB+BC)r/2=Área ABC=27r/2 =(Área ABC según fórmula de Herón) → 27r/2=√(21*8*7*6)=84→ Radio r=2*84/27=168/27 =6, 2222....
Gracias y saludos.

[sergioaiex]

Solution:

Heron's Formula

2p = a + b + c
2p = 13 + 14 + 15
2p = 42
p = 21

A ∆ABC = √ [p .(p-a) (p-a) (p-c)]
A ∆ABC = √ (21 . 8 . 7 . 6)
A ∆ABC = √ (3 . 7 . 4 . 2 . 7 . 2 . 3)
A ∆ABC = √ (3² . 7² . 2² . 2²)
A ∆ABC = 3 . 7 . 2 . 2
A ∆ABC = 84

A ∆ABC = A ∆ABO + A ∆CBO

A = ½ b . h

A ∆ ABO = ½ . 13 . r
A ∆ ABO = 13r/2

A = ½ b . h
A ∆ CBO = ½ . 14 . r
A ∆ CBO = 7r

A ∆ABC = A ∆ABO + A ∆CBO
84 = 13r/2 + 7r (×2)
168 = 13r + 14r
27r = 168
r = 168/27 (÷3)

r = 56/9 units


Or

r = 6, 22 units

[unknownidentity]

Let's find the radius:
.
..
...
....


First of all we calculate the area of the triangle with the formula of Heron:

s = (AB + AC + BC)/2 = (13 + 15 + 14)/2 = 42/2 = 21
A = √[s*(s − AB)*(s − AC)*(s − BC)] = √[21*(21 − 13)*(21 − 15)*(21 − 14)] = √(21*8*6*7) = √(3²*7²*2⁴) = 3*7*4 = 84
A = (1/2)*BC*h(BC) ⇒ h(BC) = 2*A/BC = 2*84/14 = 12

Now let's assume that C is the center of the coordinate system and that BC is located on the x-axis. Then we obtain the following coordinates:

A: ( xA ; 12 )
B: ( 14 ; 0 )
C: ( 0 ; 0 )
D: ( xD ; 0 )
E: ( xE ; yE )
O: ( xO ; yO )

We get xA by applying the Pythagorean theorem:

AC² = (xA − xC)² + (yA − yC)²
15² = (xA − 0)² + (12 − 0)²
225 = xA² + 144
81 = xA²
⇒ xA = 9

The lines CA and AB can now be represented by the following functions:

CA: y = (yA − yC)*(x − xC)/(xA − xC) = (12 − 0)*(x − 0)/(9 − 0) = 12*x/9 = (4/3)*x
AB: y = (yA − yB)*(x − xB)/(xA − xB) = (12 − 0)*(x − 14)/(9 − 14) = 12*(x − 14)/(−5) = (−12/5)*x + 168/5

D is a point of tangency. Therefore the triangle CDO is a right triangle and with r being the radius of the semicircle we can conclude:

xO = xD ⇒ r = DO = yO − yD = yO = (4/3)*xO

E is also a point of tangency. Therefore the product of the slopes of the lines OE and AB is −1:

[(yE − yO)/(xE − xO)] * [(yB − yA)/(xB − xA)] = −1
[(yE − yO)/(xE − xO)] * [(0 − 12)/(14 − 9)] = −1
[(yE − yO)/(xE − xO)] * (−12/5) = −1
[(yE − yO)/(xE − xO)] = 5/12
yE − yO = (5/12)*(xE − xO)

r² = EO² = (xE − xO)² + (yE − yO)² = (xE − xO)² + (5/12)²*(xE − xO)² = (xE − xO)²*[1 + (5/12)²] = (xE − xO)²*(1 + 25/144) = (xE − xO)²*(169/144)
⇒ r = (13/12)*(xE − xO)
⇒ xE − xO = (12/13)*r = (12/13)*(4/3)*xO = (16/13)*xO
∧ yE − yO = (5/12)*(xE − xO) = (5/12)*(16/13)*xO = (20/39)*xO

xE = xO + (16/13)*xO = (29/13)*xO
yE = yO + (20/39)*xO = (4/3)*xO + (20/39)*xO = (52/39)*xO + (20/39)*xO = (72/39)*xO = (24/13)*xO

E is located on AB. Therefore we obtain:

yE = (−12/5)*xE + 168/5
(24/13)*xO = (−12/5)*(29/13)*xO + 168/5
120*xO = −348*xO + 2184
468*xO = 2184
⇒ xO = 2184/468 = 14/3
⇒ r = yO = (4/3)*xO = (4/3)*(14/3) = 56/9

Best regards from Germany

[wackojacko]

Heron is my Hero! ...wait a minute, that may be a contradictory statement. Anyhow, infinite extended lines of a triangle's sides have excircles that are tangent too a side and two extended lines and along with the inscribed circle of the triangle with infinite line extensions from the inscribed circle's center thru the excircles points of tangency of the triangle generate (MORE!) than just a various amount of terms in Geometry too consider. ...just sayin, ya need more than half a brain too figure that out. 🙂

[Golololololo]

At first, looks difficult to solve; but when you start solving, it sounds easy!
Thanks