Cosine rule: c²=a²+b²-2abcosα c= 19 cm Semi perimeter: s = ½(a+b+c) = 28cm Area of triangle: A² = s(s-a)(s-b)(s-c) A = 84√3 = 145,49 cm² Radius of circle: R = A/s = 3√3 = 5,196 cm Area of circle: A = πR² = 27π cm² ( Solved √ )
Make the left side a 30, 60, 90 triangle. Sides are 21(given), 21/2 (part of the base), and (21*sqrt(3))/2 (this is the triangle's height). Therefore, the triangle's area is 8*((21*sqrt(3))/2) Start simplifying: 4*(21*sqrt(3), so triangle area of 84*sqrt(3) un^2. Before calculating triangle area in terms of r, find length AC: The remaining part of the base is 16 - (21/2) = 11/2. (11/2)^2 + ((21*sqrt(3))/2)^2 = (AC)^2. 121/4 + 1323/4 = (AC)^2 1444/4 = (AC)^2 AC = sqrt(1444)/2 So (2*sqrt(361))/2. Conveniently, sqrt(361)=19, so AC = 19 Triangle area in terms of r = 28r. 28r = 84*sqrt(3) r = 3*sqrt(3) (3*sqrt(3))^2 * pi = 27pi approximates to 84.82 un^2
You don’t need to calculate the area of the triangle in this case. b = sqrt(21² − 10.5² + 5.5²) = sqrt(361) = 19 The perpendicular through O cuts BC in a point D, and BD = (16 + (21 − 19)) / 2 = 9 BDO is a 30-60-90 triangle, so: r = DO = 9 / sqrt(3) = 3 * sqrt(3)
I forgot about the triangle area from inradius formula, so I added the areas of triangles AOC, BOC and AOB (9.5r + 8r + 10.5r) to get 28r. so 84*sqrt(3) = 28r ==> r = 3*sqrt(3) ==> area of circle = pi * r^2 = 27pi.
Cosine law, AC = 19. Labeling 3 sets of equal corner tangents as a, b & c. Clockwise starting from corner A. Then a + b = 21, b + c = 19, c + a = 16. b = 19 - c , so a + b becomes a + 19 - c = 21. c = 16 - a , so a + 19 - c becomes a + 19 - 16 + a = 21. 2a - 3 = 21. a = 9. Joining B to centre of circle, then dropping perpendicular onto BC to make a triangle. Tan 30 = r / a = r / 9. r = 9 tan 30 = 9 x 1 /root 3. Area of circle = Pi x 9 x 9 x 1 / 3. 27 Pi.
At first I used the law of cosine to find missing side length. Then I used the theorem, that tangent points on a circle from a common external point have equal distances. The tangent points divide the side lengths, the values of the parts can be found by a system of three equations with three unknowns. Final step Pythagoras on the 30/60/90 right triangle in bottom left corner gives circle Radius and area. We have another 60° triangle with all sides being integers. Thanks 😊 for sharing this interesting 👍 geometry question, greetings and best 👌 wishes.
Área del triángulo =16*(21√3/2)/2 =84√3 ---> 16=r√3+(16-r√3) ; 21=r√3+(21-r√3) ; AC=21+16-2r√3=37-2r√3 ---> 84√3=(r/2)(16+21+37-2r√3)---> r=3√3 ---> Área del círculo =27π. Gracias y un saludo cordial.
I used law of cosines to get the missing side length (19), then used the formula for an inscribed circle, which is r^2=((28-19)*(28-21)*(28-16))/28, which gives a value of r^2=27. So the area of the circle is just 27*pi.
I found solving much easier if you attack the triangle with Pythagoras. Then you can find the circleradius two times the area of the perimeter triangle divided by the sum of perimeter.
1/From O drop OH, OI, and OK perpdndicular to BC, AB and AC respectively. Focus on the triangle BOH. The triangle BOH is a 30-60-90 special triangle, so BH = BI=rsqrt3. -> CH=CK= 16-rsqrt3 and AI=AK= 21-rsqrt3 2/Let p be the half of the perimeter of the triangle ABC: We have: p = rsqrt3+(16-rsqrt3)+(21-rsqrt3)= 37-rsqrt3 3/ By the cosine theorem we have: AC= 19-> p = 1/2(21+16+19)=28 --> 37-rsqrt3=28 -> rsqrt3= 9->r=3sqrt3 4/Area of the circle= 27 x pi sq units😅😅😅
AC^2 = BA^2 + BC^2 - 2.BA.BC.cos(60°) = 16^2 + 21^2 - 2.16.21.(1/2) = 256 + 441 - 336 = 361, so AC = sqrt(361) = 19. The perimeter of ABC is 16 + 21 + 19 = 56 and the half perimeter is 28. The Heron formula gives the area of ABC: Area = sqrt(28.(28 - 16).(28 - 21).(28 - 19)) = sqrt(28.12.7.9) = sqrt(21168) = 84.sqrt(3). The radius of the circle is: (2.area of ABC) / perimeter of ABC = (168.sqrt(3)) / 56 = 3.sqrt(3). Its area is ((3.sqrt(3))^2) .Pi = 27.Pi
Let's find the area: . .. ... .... ..... Let D (E,F) be the point of tangency on AB (AC,BC). According to the two tangent theorem we know: AD = AE BD = BF CE = CF With r being the radius of the inscribed circle we also know that OD=OE=OF=r. With BD=BF and OD=OF we can conclude that the right triangles OBD and OBF are congruent. Theref ore we know that ∠OBD=∠OBF=∠DBF/2=60°/2=30°. So OBD and OBF are 30°-60°-90° triangles and we can conclude: OB = 2r OD = OF = r BD = BF = √3*r AE = AD = AB − BD = 21 − √3*r CE = CF = BC − BF = 16 − √3*r AC = AE + CE = (21 − √3*r) + (16 − √3*r) = 37 − 2√3*r The area of the triangle ABC can be calculated in two different ways: A(ABC) = (1/2)*AB*BC*sin(∠ABC) A(ABC) = A(OAB) + A(OAC) + A(OBC) = (1/2)*AB*OD + (1/2)*AC*OE + (1/2)*BC*OF = (1/2)*(AB + AC + BC)*r (1/2)*AB*BC*sin(∠ABC) = (1/2)*(AB + AC + BC)*r AB*BC*sin(∠ABC) = (AB + AC + BC)*r 21*16*sin(60°) = [21 + (37 − 2√3*r) + 16]*r 21*16*√3/2 = (74 − 2√3*r)*r 168√3 = 74*r − 2√3*r² 2√3*r² − 74*r + 168√3 = 0 r² − (37/√3)*r + 84 = 0 r = 37/(2√3) ± √[37²/(2√3)² − 84] r = 37/(2√3) ± √(1369/12 − 84) r = 37/(2√3) ± √(1369/12 − 1008/12) r = 37/(2√3) ± √(361/12) r = 37/(2√3) ± 19/(2√3) We have two possible solutions: r = 37/(2√3) + 19/(2√3) = 56/(2√3) ⇒ AC = 37 − 2√3*r = 37 − 56 = −19 r = 37/(2√3) − 19/(2√3) = 18/(2√3) ⇒ AC = 37 − 2√3*r = 37 − 18 = 19 Finally we obtain the only useful solution: r = 18/(2√3) = 3√3 Now we are able to calculate the area of the inscribed circle: A = πr² = 27π Best regards from Germany
Solution: Let's label the sides: AB = c BC = a AC = b Applying Law of Cosines in ∆ ABC, we have: b² = a² + c² - 2 a c cos β b² = (16)² + (21)² - 2 (16) (21) cos 60° b² = 256 + 441 - 2 . 16 . 21 . ½ b² = 697 - 336 b² = 361 b = 19 Applying Heron's Formula to calculate ∆ ABC Area, we have: s = (a + b + c)/2 s = (16 + 19 + 21)/2 s = 28 ∆ ABC Area = √ [s (s - a) (s - b) (s - c)] ∆ ABC Area = √ [28 (28 - 16) (28 - 19) (28 - 21)] ∆ ABC Area = √ [28 (12) (9) (7)] ∆ ABC Area = √ (4 . 7 . 3 . 4 . 3 . 3 . 7) ∆ ABC Area = 4 . 7 . 3 √3 ∆ ABC Area = 84√3 Let's calculate the radius by Inradius Formula Inradius = Area/semiperimeter ∆ ABC Area = r . s 84√3 = r . 28 r = 84√3/28 r = 3√3 Final Step Green Circle Area = π r² Green Circle Area = π (3√3)² Green Circle Area = 27π square units ✅ Green Circle Area ≈ 84.8230 square units ✅
Your very good in analizing problems, but your solution is not applicable in work, very long, it takes too much time, can you not make shorter solution by applying trigonometry
What happened to my Comment? I made a Comment that was deleted and denounced. Why? I gave the right answer GCA = 27Pi Square Units. Whom did I offended?
Awesomeness sir ! Thanks as usual
Excellent! Glad you enjoyed it! 👍
You are very welcome!
Thanks for the feedback ❤️
Cosine rule:
c²=a²+b²-2abcosα
c= 19 cm
Semi perimeter:
s = ½(a+b+c) = 28cm
Area of triangle:
A² = s(s-a)(s-b)(s-c)
A = 84√3 = 145,49 cm²
Radius of circle:
R = A/s = 3√3 = 5,196 cm
Area of circle:
A = πR² = 27π cm² ( Solved √ )
@ 2:55 , At parties I use the Law of Cosines to calculate social distancing from circle jerks. 😊
Nice one Sir
Make the left side a 30, 60, 90 triangle.
Sides are 21(given), 21/2 (part of the base), and (21*sqrt(3))/2 (this is the triangle's height).
Therefore, the triangle's area is 8*((21*sqrt(3))/2)
Start simplifying:
4*(21*sqrt(3), so triangle area of 84*sqrt(3) un^2.
Before calculating triangle area in terms of r, find length AC:
The remaining part of the base is 16 - (21/2) = 11/2.
(11/2)^2 + ((21*sqrt(3))/2)^2 = (AC)^2.
121/4 + 1323/4 = (AC)^2
1444/4 = (AC)^2
AC = sqrt(1444)/2
So (2*sqrt(361))/2. Conveniently, sqrt(361)=19, so AC = 19
Triangle area in terms of r = 28r.
28r = 84*sqrt(3)
r = 3*sqrt(3)
(3*sqrt(3))^2 * pi = 27pi approximates to 84.82 un^2
You don’t need to calculate the area of the triangle in this case.
b = sqrt(21² − 10.5² + 5.5²) = sqrt(361) = 19
The perpendicular through O cuts BC in a point D, and
BD = (16 + (21 − 19)) / 2 = 9
BDO is a 30-60-90 triangle, so:
r = DO = 9 / sqrt(3) = 3 * sqrt(3)
Thank you!
use cosine rule to find side AC . Then using tangent rule you can write
21 -- sqrtof 3x r + 16 -- sqrt3x r = 19 and
find r and thus area of circle
I forgot about the triangle area from inradius formula, so I added the areas of triangles AOC, BOC and AOB (9.5r + 8r + 10.5r) to get 28r. so 84*sqrt(3) = 28r ==> r = 3*sqrt(3) ==> area of circle = pi * r^2 = 27pi.
Inradius formula A=r.s, is exactly the same that three times area of internal triangles
A=½bh+½bh+½bh =½p.r=s.r
You didn't forget !!
Cosine law, AC = 19.
Labeling 3 sets of equal corner tangents as a, b & c.
Clockwise starting from corner A.
Then a + b = 21, b + c = 19, c + a = 16.
b = 19 - c , so a + b becomes a + 19 - c = 21.
c = 16 - a , so a + 19 - c becomes a + 19 - 16 + a = 21.
2a - 3 = 21.
a = 9.
Joining B to centre of circle, then dropping perpendicular onto BC to make a triangle.
Tan 30 = r / a = r / 9.
r = 9 tan 30 = 9 x 1 /root 3.
Area of circle = Pi x 9 x 9 x 1 / 3.
27 Pi.
AD⊥BC BD=21/2 AD=21√3/2 DC=11/2
Triangle area = 16*21√3/2*1/2=84√3
(11/2)²+(21√3/2)²=AC² 121/4+1323/4=1444/4=361=AC² AC=19
16*r*1/2 + 19*r*1/2 + 21*r*1/2 = 84√3 56r = 168√3 r = 3√3
Green Circle area = 3√3*3√3*π = 27π
I followed along and got to the answer at the end. I feel good about that.
Fantastic, thank you teacher 👍🙏
At first I used the law of cosine to find missing side length. Then I used the theorem, that tangent points on a circle from a common external point have equal distances.
The tangent points divide the side lengths, the values of the parts can be found by a system of three equations with three unknowns.
Final step Pythagoras on the 30/60/90 right triangle in bottom left corner gives circle Radius and area.
We have another 60° triangle with all sides being integers.
Thanks 😊 for sharing this interesting 👍 geometry question, greetings and best 👌 wishes.
Área del triángulo =16*(21√3/2)/2 =84√3 ---> 16=r√3+(16-r√3) ; 21=r√3+(21-r√3) ; AC=21+16-2r√3=37-2r√3 ---> 84√3=(r/2)(16+21+37-2r√3)---> r=3√3 ---> Área del círculo =27π.
Gracias y un saludo cordial.
I used law of cosines to get the missing side length (19), then used the formula for an inscribed circle, which is r^2=((28-19)*(28-21)*(28-16))/28, which gives a value of r^2=27. So the area of the circle is just 27*pi.
cos(60/2)=√(p(p-c)/21*16)...c=19(Briggs)...r=2A/p=A/28=(1/56)21*16*sin60=3√3
I found solving much easier if you attack the triangle with Pythagoras. Then you can find the circleradius two times the area of the perimeter triangle divided by the sum of perimeter.
1/From O drop OH, OI, and OK perpdndicular to BC, AB and AC respectively.
Focus on the triangle BOH.
The triangle BOH is a 30-60-90 special triangle, so BH = BI=rsqrt3.
-> CH=CK= 16-rsqrt3
and AI=AK= 21-rsqrt3
2/Let p be the half of the perimeter of the triangle ABC:
We have:
p = rsqrt3+(16-rsqrt3)+(21-rsqrt3)= 37-rsqrt3
3/ By the cosine theorem we have: AC= 19-> p = 1/2(21+16+19)=28
--> 37-rsqrt3=28
-> rsqrt3= 9->r=3sqrt3
4/Area of the circle= 27 x pi sq units😅😅😅
S=27π≈84,86
AC^2 = BA^2 + BC^2 - 2.BA.BC.cos(60°) = 16^2 + 21^2 - 2.16.21.(1/2) = 256 + 441 - 336 = 361, so AC = sqrt(361) = 19.
The perimeter of ABC is 16 + 21 + 19 = 56 and the half perimeter is 28.
The Heron formula gives the area of ABC: Area = sqrt(28.(28 - 16).(28 - 21).(28 - 19)) = sqrt(28.12.7.9) = sqrt(21168) = 84.sqrt(3).
The radius of the circle is: (2.area of ABC) / perimeter of ABC = (168.sqrt(3)) / 56 = 3.sqrt(3). Its area is ((3.sqrt(3))^2) .Pi = 27.Pi
(right side length)^2 = 21^2 + 16^2 - 2(21)(16)cos60° = [(21 - r√3) + (16 - r√3)]^2
21^2 + 16^2 - (21)(16) = (21 + 16 - 2r√3)^2 = (21 + 16)^2 - 2(21 + 16)(2r√3) + 12r^2
12r^2 - (4√3)(21 + 16)r + (3)(21)(16) = 0
3r^2 - (√3)(37)r + (3)(3)(7)(4) = 0
(r - 3√3)(3r - (7)(4)√3) = 0
r = 3√3 or (28/3)√3) (not possible)
Circle area = 27π
Let's find the area:
.
..
...
....
.....
Let D (E,F) be the point of tangency on AB (AC,BC). According to the two tangent theorem we know:
AD = AE
BD = BF
CE = CF
With r being the radius of the inscribed circle we also know that OD=OE=OF=r. With BD=BF and OD=OF we can conclude that the right triangles OBD and OBF are congruent. Theref
ore we know that ∠OBD=∠OBF=∠DBF/2=60°/2=30°. So OBD and OBF are 30°-60°-90° triangles and we can conclude:
OB = 2r
OD = OF = r
BD = BF = √3*r
AE = AD = AB − BD = 21 − √3*r
CE = CF = BC − BF = 16 − √3*r
AC = AE + CE = (21 − √3*r) + (16 − √3*r) = 37 − 2√3*r
The area of the triangle ABC can be calculated in two different ways:
A(ABC) = (1/2)*AB*BC*sin(∠ABC)
A(ABC) = A(OAB) + A(OAC) + A(OBC) = (1/2)*AB*OD + (1/2)*AC*OE + (1/2)*BC*OF = (1/2)*(AB + AC + BC)*r
(1/2)*AB*BC*sin(∠ABC) = (1/2)*(AB + AC + BC)*r
AB*BC*sin(∠ABC) = (AB + AC + BC)*r
21*16*sin(60°) = [21 + (37 − 2√3*r) + 16]*r
21*16*√3/2 = (74 − 2√3*r)*r
168√3 = 74*r − 2√3*r²
2√3*r² − 74*r + 168√3 = 0
r² − (37/√3)*r + 84 = 0
r = 37/(2√3) ± √[37²/(2√3)² − 84]
r = 37/(2√3) ± √(1369/12 − 84)
r = 37/(2√3) ± √(1369/12 − 1008/12)
r = 37/(2√3) ± √(361/12)
r = 37/(2√3) ± 19/(2√3)
We have two possible solutions:
r = 37/(2√3) + 19/(2√3) = 56/(2√3) ⇒ AC = 37 − 2√3*r = 37 − 56 = −19
r = 37/(2√3) − 19/(2√3) = 18/(2√3) ⇒ AC = 37 − 2√3*r = 37 − 18 = 19
Finally we obtain the only useful solution:
r = 18/(2√3) = 3√3
Now we are able to calculate the area of the inscribed circle:
A = πr² = 27π
Best regards from Germany
Thanks a lot from the bottom of my heart ❤️
Solution:
Let's label the sides:
AB = c
BC = a
AC = b
Applying Law of Cosines in ∆ ABC, we have:
b² = a² + c² - 2 a c cos β
b² = (16)² + (21)² - 2 (16) (21) cos 60°
b² = 256 + 441 - 2 . 16 . 21 . ½
b² = 697 - 336
b² = 361
b = 19
Applying Heron's Formula to calculate ∆ ABC Area, we have:
s = (a + b + c)/2
s = (16 + 19 + 21)/2
s = 28
∆ ABC Area = √ [s (s - a) (s - b) (s - c)]
∆ ABC Area = √ [28 (28 - 16) (28 - 19) (28 - 21)]
∆ ABC Area = √ [28 (12) (9) (7)]
∆ ABC Area = √ (4 . 7 . 3 . 4 . 3 . 3 . 7)
∆ ABC Area = 4 . 7 . 3 √3
∆ ABC Area = 84√3
Let's calculate the radius by Inradius Formula
Inradius = Area/semiperimeter
∆ ABC Area = r . s
84√3 = r . 28
r = 84√3/28
r = 3√3
Final Step
Green Circle Area = π r²
Green Circle Area = π (3√3)²
Green Circle Area = 27π square units ✅
Green Circle Area ≈ 84.8230 square units ✅
Your very good in analizing problems, but your solution is not applicable in work, very long, it takes too much time, can you not make shorter solution by applying trigonometry
What happened to my Comment?
I made a Comment that was deleted and denounced. Why?
I gave the right answer GCA = 27Pi Square Units.
Whom did I offended?
There has got to be a mistake or miscommunication. My sincere apologies 🙏
You are the best and keep rocking👍