I solvedd it with the second method. But when you have 12/16 = r/8, I always multiply diagonally. Same result, I know, but that's the easiest way for me. I love your way of explaining the math problems. Always a clear explanation, you are one of the best
My solution was similar to your second solution except I flipped OPC around the beta angle. BC is 16 and CP is 8 and both have the same angles. Therefore, OC is 10 and OP is 6. OP is the radius and I did half of the circle radius equation.
I solved this problem using the Pythagorean Theorem, two triangles and two variables and arrived at 32r = 192; r=6 . Area = Pi x 6^2 /2 or 18Pi square units.
AP = AB = 12 ( prop of tangent) AC = 12 + 8 = 20 by pythagorean theorem BC = SQROOT OF (20X20 -- 12X12) = 16 OC = 16 -- r where r is radius of semi circle by pyth theorem, (16 --r) sq = r sq + 8 x8 solve for r r = 6 hence semi circle area = pi36/2 = 18pi
By Two Tangent Theorem, PA=AB. Therefore CA = 8+12 = 20. As AB = 12 = 3(4) and CA = 20 = 5(4), this means that ∆ABC is a 4:1 ratio (3,4,5) Pythagorean Triple triangle and that BC = 4(4) = 16. As CA is tangent to Semicircle O at P, ∠CPO is 90°. As ∠BCA and ∠OCP are the same angle and ∠CPO and ∠ABC are both 90°, then ∠POC and ∠CAB are also the same angle and ∆ABC and ∆CPO are similar. BC/CP = AB/PO 16/8 = 12/r r = 12(1/2) = 6 A = (1/2)πr² = (π/2)6² = 36π/2 = 18π
Let r be the radius, then OC^2=64+r^2, therefore 400-144=(r+sqrt(64+r^2))^2, 256=16^2=(r+sqrt(64+r^2)), (16-r)^2=64+r^2, r=192/32=6, then the answer is 6x6/2 pi =18 pi.
B4 watching: The area is 18pi square units. The hypothenuse of the large triangle is 12+8=20. The bottom side is 16. (12^2+16^2=20^2 or 4 times the famous 3-4-5 right triangle). The radius r can be calculated as follows: r^2+8^2=(16-r)^2. This leads to r=6. The area of the semicircle is r^2*pi/2, which is 18pi.
So, once again, I did not chose the easiest approach… Quickly figured that the hypothenuse of the right triangle ABC is 20 (12+8). I have decomposed BC as 2*r+x and OC as r+x. Noticed that ABC and OPC are similar triangles so (r+x)/r=20/12=5/3 => 1+(x/r)=5/3 => x/r=2/3 => x=2/3*r. BC can thus be written as 2*r+(2/3)*r=(8/3)*r. Using the Pythagorean theorem : ((8/3)*r)^2 + 12^2 = 20^2 => (64/9)*r^2=256 => r^2=36 => r=6 (a length cannot be negative).
18 pi or 56.55 Line AP= 12 (Tangent circle theorem)/ Hence AC = 20 (12+ 18) . BC = 16 since this is a 3-4-5 right triangle scaled up by 4. Draw a perpendicular line from P to O, forming a right triangle. This is the radius, r . Line OC = 16- r (since line BO =r and BC=16), and PC =8 (given). Using Pythagorean theorem: r^2 + 8^2 = (16-r)^2 r^2 + 64 = 256+r^2 - 32r 64-256 = - 32r -192 = - 32 r 6=r so radius of the semi circle = 6. The area of the circle is pi 6^2 or 36 pi, hence the area of the semi circle is 18 pi or 56. 55 Answer
Let's use BC as the axis of symmetry to reflect the diagram and get a circle inscribed in a triangle with sides a = 24 and b = c = 20. We can then use the general formula for the inradius of an incircle in a triangle: r = √[(s − a)(s − b)(s − c)/s] where s = ½ (a + b + c).
Big Right triangle: Hypotenuse = 8 + 12 = 20cm h² = 20² -12² h = 16 cm Similarly of triangles: 16/20 = 8 / (h-r) h - r = 8 . 20 / 16 = 10 r = h - 10 = 6 cm Area = πr² /2 Area = 56,55 cm² ( Solved √ )
Nice n easy , thanks again 🤓👍🏻
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So nice of you.
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@@PreMathyour geometry problems help to keep me sane, they give me a welcome diversion.
@@theoyanto
Keep it up, my dear friend. Stay blessed.
I solvedd it with the second method. But when you have 12/16 = r/8, I always multiply diagonally.
Same result, I know, but that's the easiest way for me.
I love your way of explaining the math problems. Always a clear explanation, you are one of the best
Good presentation.
Thank you for showing the (AA ) similar triangles theorem use. Keep up the great work. You are great teacher
You're very welcome!
So nice of you.
Thank you! Cheers! 😀
My solution was similar to your second solution except I flipped OPC around the beta angle. BC is 16 and CP is 8 and both have the same angles. Therefore, OC is 10 and OP is 6. OP is the radius and I did half of the circle radius equation.
OPC is a similar triangle with ABC and therefore, a 3-4-5 triangle: If PC=8 (4x2), the other side must 3x2=6.
yes I saw that too. radius had to be 6
also once you figured out that the bottom was 16 than that further confirms 6 because that last side of the triangle using the radius had to be 10
This is what I did
Very nice explanations, thank you very much!
Glad you enjoyed it!
Wow, that semi circle thought he was some kind of tough guy!😂👍❤️
Well said!
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Tee-Hee!
Thanks for video.Good luck sir!!!!!!!!!!!!
So nice of you
thankyou v.much dear ❤
Most welcome 😊
So nice of you.
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Please give some challenging questions from algebra trignometry .
Menstruation is not included in our syllabus.
I solved this problem using the Pythagorean Theorem, two triangles and two variables and arrived at 32r = 192; r=6 . Area = Pi x 6^2 /2 or 18Pi square units.
Nice! tan(φ) = 12/16 = 3/4 = r/8 → r = 6 → yellow area = 18π
Very good!
I computed using special triangles and similar triangles to find the radius.
Super!
S=18π≈56,57
Keep it up
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Love it.
Excellent!
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Yay! I solved by the second method.
Bravo!
The 3rd method as derivative of the the 2nd:
tg(alpha) = 12/16;
tg(alpha)= r/8;
r/8 = 12/16;
r= 8*12/16= 6;
Ay = pi*r^2/2= pi*6^2/2=18*pi sq units.
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2nd method...now THAT'S GOD' SPEED! 😇
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Thanks
You are very welcome!
So nice of you.
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AP = AB = 12 ( prop of tangent)
AC = 12 + 8 = 20
by pythagorean theorem
BC = SQROOT OF (20X20 -- 12X12)
= 16
OC = 16 -- r where r is radius of semi circle
by pyth theorem,
(16 --r) sq = r sq + 8 x8
solve for r
r = 6
hence semi circle area = pi36/2
= 18pi
Thank you! Cheers! 😀
After getting
BC=16, we can proceed as-
∆OPC ~ ∆ABC (By AAA similarity)
Hence,
OP/PC = AB/BC
=> r/8 =12/16
=> r = 6
=> A =πr^2/2 = 16π
Thank you! If ABC is a 3-4-5 triangle (12, 16, 20), then so is OPC (6, 8, 10).
I think this is the fastest way, kind of shortcut!
Which test ?
By Two Tangent Theorem, PA=AB. Therefore CA = 8+12 = 20. As AB = 12 = 3(4) and CA = 20 = 5(4), this means that ∆ABC is a 4:1 ratio (3,4,5) Pythagorean Triple triangle and that BC = 4(4) = 16. As CA is tangent to Semicircle O at P, ∠CPO is 90°. As ∠BCA and ∠OCP are the same angle and ∠CPO and ∠ABC are both 90°, then ∠POC and ∠CAB are also the same angle and ∆ABC and ∆CPO are similar.
BC/CP = AB/PO
16/8 = 12/r
r = 12(1/2) = 6
A = (1/2)πr² = (π/2)6² = 36π/2 = 18π
Let r be the radius, then OC^2=64+r^2, therefore 400-144=(r+sqrt(64+r^2))^2, 256=16^2=(r+sqrt(64+r^2)), (16-r)^2=64+r^2, r=192/32=6, then the answer is 6x6/2 pi =18 pi.
B4 watching: The area is 18pi square units. The hypothenuse of the large triangle is 12+8=20. The bottom side is 16. (12^2+16^2=20^2 or 4 times the famous 3-4-5 right triangle). The radius r can be calculated as follows: r^2+8^2=(16-r)^2. This leads to r=6. The area of the semicircle is r^2*pi/2, which is 18pi.
Let OP = R. BC = 16. OC = 16 - R. In OPC OCsq = Rsq + 64. 256 - 32R + Rsq = Rsq + 64. 32R = 192. R = 6 Area = 18pi = 56.55
Excellent!
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So, once again, I did not chose the easiest approach… Quickly figured that the hypothenuse of the right triangle ABC is 20 (12+8). I have decomposed BC as 2*r+x and OC as r+x. Noticed that ABC and OPC are similar triangles so (r+x)/r=20/12=5/3 => 1+(x/r)=5/3 => x/r=2/3 => x=2/3*r. BC can thus be written as 2*r+(2/3)*r=(8/3)*r.
Using the Pythagorean theorem : ((8/3)*r)^2 + 12^2 = 20^2 => (64/9)*r^2=256 => r^2=36 => r=6 (a length cannot be negative).
Solving for x could be faster if you noticed that tangens of angle C is 12/16 so is 6/8 and also r/8 so r=6.
Per costruzione lipotenusa é 8+12=20...per Pitagora risulta 20^2=12^2+(r+rad(64+r^2))^2....r^2=36....Ay=36pi/2=18pi
18 pi or 56.55
Line AP= 12 (Tangent circle theorem)/ Hence AC = 20 (12+ 18) . BC = 16 since this is a 3-4-5 right triangle
scaled up by 4.
Draw a perpendicular line from P to O, forming a right triangle. This is the radius, r . Line OC = 16- r (since line BO =r
and BC=16), and PC =8 (given).
Using Pythagorean theorem: r^2 + 8^2 = (16-r)^2
r^2 + 64 = 256+r^2 - 32r
64-256 = - 32r
-192 = - 32 r
6=r
so radius of the semi circle = 6. The area of the circle is pi 6^2 or 36 pi, hence the area of the semi circle is
18 pi or 56. 55 Answer
There is one more method: if BC=16, then 8²= 16*(16-2r); 64=256-32r; 32r=192; r=6; then Area= π*6²/2 = 18π
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@@PreMath My pleasure! Cheers!🙏
Let's use BC as the axis of symmetry to reflect the diagram and get a circle inscribed in a triangle with sides a = 24 and b = c = 20.
We can then use the general formula for the inradius of an incircle in a triangle: r = √[(s − a)(s − b)(s − c)/s] where s = ½ (a + b + c).
Nice one
Big Right triangle:
Hypotenuse = 8 + 12 = 20cm
h² = 20² -12²
h = 16 cm
Similarly of triangles:
16/20 = 8 / (h-r)
h - r = 8 . 20 / 16 = 10
r = h - 10 = 6 cm
Area = πr² /2
Area = 56,55 cm² ( Solved √ )
∆POC is similar to ∆ BAC
AB / BC = r / PC
r = PC * AB / BC
= 8 * 12 / √( 20^2 - 12^2)
= 8 * 12 / 16 = 6 unit
I went for r^2 + 8^2 = (16-r)^2 for the radius, and went from there.
If the semicircle cuts BC at D, then BD. BC = BP^2
So BD = 8^2/16 = 4 so BD = 16-4 =12
So radius = 6
18π
x-c-lent
Generalized: the area of the yellow semicircle is _π a² b / (2 (2a + b))_ where _a_ = AB and _b_ = PC.
Thank you for your sharing,. I'm 74, I tried TH-cam, but It's slow to grow, help me proff, thank you.
18pi
.
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18 pi or 56.55
One method is a short step while the other is long one.
Excellent!
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@@PreMath by the way, may I ask if you are of Asian origin ( specifically from india)?. Don't mistake me for asking this question.
Only in 30 sec i got the answer
Area of yellow shaded semicircle=18π
r/8=12/16
r=6
I think r = 6 and area = 18π
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asnwer=18 is it hmm so than
Too early 😢
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18π