After finding the lenght of the radius, we know that the triangles CSQ and BPQ are congruent, because they have a right angle, a congruent acute angle and a congruent side( SQ=PQ)----> CB= 4square root of 3. And Area blu= 12square root of 3 - 6pi.
I agree as well. The expression (1/2)*(4√3)*(4√3)*(√3/2) of the area of the triangle ABC (in cm²) is correct, but that expression turns out to be 12√3. I guess this is good news: It means that PreMath is a human being.🙂 Best regards from Germany
Right on the tail end, your area calculation should result in 12sqrt(3), not 24sqrt(3). The rest up to there is good, you just forgot to divide by 2 twice. 😅 So the final answer should be 12sqrt(3)-6pi. Thank you for the video as always, though.
Area of Equilateral Triangle (A) = (√3/4)a² So all the is needed is the side length of the equilateral triangle to calculate its area an subtract 6pi. Draw the line PQ, since the radius is 2 that line must have a length of 4. This line now is one side of the right triangle PQB, where the other 2 sides are parts of the sides of the given triangle. By the information provided we know, that this right triangle we just constructed must have angles of 90, 60 and 30 degrees. - We know (maybe from previous Videos on this channel) that the side lengths of an 30-60-90 Triangle have a ratio of 1:√3:2 Since PQ (the second longest side) has a length of 4, QB equals 4/√3 and PQ = 8/√3 Thus the side length = 12/√3 Therefor Area of Equilateral Triangle (A) = (√3/4)*(12/√3)² = (√3/4)*(144/3) =(√3/4)*48 = 20,785 (rounded to the 3rd decimal) => 20,785 - 6pi is the blue area. (roughly 1.935) So we do not need to fuss around with sin
The radii are 2. PSQ is equilateral with sides of 4, so a height of 2*sqrt(3). Height measured from a point T which is the midpoint of PQ and where the tangents meet. The inside of the main triangle may be split into a series of 30,60, 90 triangles. APS has mid-length side of 4, so PA is 4/sqrt(3) = (4*sqrt(3))/3 and AS = (8*sqrt(3))/3. SC = PA = (4*sqrt(3))/3 Main triangle side lengths are (8*sqrt(3))/3 + (4*sqrt(3))/3 = (12*sqrt(3))/3 = 4*sqrt(3) h is square root of (48 - 12) = sqrt(36) = 6. Area is 6 * ((6*sqrt(3))/3) = (36*sqrt(3))/3 = 12*sqrt(3) un^2 Blue area is 12*sqrt(3) - 6pi Approx (12*1.732) - (6*3.1416) = 1.935 (rounded). Just looked at the video and was wondering where I went wrong. Then I checked the comments below and felt some relief.
Let's find the area: . .. ... .... ..... We start with the calculation of the radius r of the yellow semicircles: A = πr²/2 (2π)cm² = πr²/2 4cm² = r² ⇒ r = 2cm Since every pair of semicircles has exactly one point of intersection, the distances of their centers are all the same: PQ = PS = QS = r + r = 4cm It is obvious that PQ (PS, QS) is perpendicular to BC (AB, AC). Since all interior angles of the equilateral triangle ABC are 60°, the triangles APS, BPQ and CQS are congruent 30°-60°-90° triangles. So we can conclude: PQ = PS = QS = 4cm ⇒ AP = BQ = CS = 4cm/√3 ∧ AS = BP = CQ = 8cm/√3 Therefore the side length and the area of the equilateral triangle ABC turn out to be: s(ABC) = AP + BP = AS + CS = BQ + CQ = 4cm/√3 + 8cm/√3 = 12cm/√3 A(ABC) = (√3/4)*[s(ABC)]² = (√3/4)*(12cm/√3)² = (√3/4)*(144/3)cm² = (12√3)cm² Now we can calculate the size of the blue area: A(blue) = A(ABC) − 3*A(semicircle) = (12√3 − 6π)cm² ≈ 1.935cm² Best regards from Germany
It is obvious that PQ (PS, QS) is perpendicular to BC (AB, AC) but a proof is needed.The proof of angle PQB = 90 is made by constructing triangle PQE which is a right-angled triangle as QE is centre Q to point of tangency E. In triangle PQE, sin(angle EPQ) = 2/4 hence angle EPQ = 30 and then angle PQB = 180 - 60 - 30 = 90. Similarly angles APS and CSQ are proven to be 90.
@@hongningsuen1348 Thanks for your feedback and you are absolutely right. I only used symmetry considerations to derive that PQ (PS, QS) is perpendicular to BC (AB, AC). Indeed this is not enough. After obtaining the side lengths PQ=4 and EQ=2 and the angle ∠PEQ=90° it can be concluded that EPQ and therefore also BPQ are 30°-60°-90° triangles. That should have been shown first before using symmetry arguments. Best regards from Germany
I first calculated the area of the blue sector in the center (area of triangle PQS minus 3 circle sectors of 60°). The total area of the other blue parts is 2 times the area of the center blue part: 3 (1/2) + 3 (1/6) = 2. So multiply the area of the center blue sector by 3 and you have the answer.
SQP is equilateral triangle so reverse calculation to find PQ,PE,(QB=AP),EB and then use sqrt3/4 *( AB*AB) to find ABC triangle and minus 6pi will be answer
1) Radius (R) of Yellow Semicircles: 2) Pi*R^2 = 4*Pi ; R^2 = 4 ; R = 2 cm 3) Area of Equilateral Triangle [PQS] (A) with Side Length = 2*R = 4 cm; using Heron's Formula: 4) A ~ 6,93 Square Cm 5) Area of 3 Equal Right Triangles : [CSQ] ; [APS] ; [BPQ]. The Triangles are (30º ; 60º ; 90º) 6) BQ = (4 * tan(30º)) cm ~ 2,31 cm 7) Area of each Triangle (AT) 8) 2*AT = (4 * [4 * tan(30º)]) ; 2*AT = 16*tan(30º) ; AT = 8*tan(30º) ~ 4,62 Square Cm 9) Area of all 3 Right Triangles = AT [CSQ] + AT [APS] + AT [BPQ] = 24*tan(30º) ~ 13,86 Square Cm 10) Area of the Great Triangle (GT) = 24*tan(30º) + 6,93 ; GT ~ 13,86 + 6,63 ~ 20,79 Square Cm 11) Blue Region Area (BRA) = 20,79 - 6*Pi ~ 20,79 - 18,85 ~ 1,94 Square Cm. 12) Answer : The Blue Region Area is equal to 1,94 Square Centimeters.
Blue area = area of ∆ABC minus three yellow semicircle areas. - semicircle areas: given (= 6π) - ∆ABC area : unknown - Need: ∆ABC side length x AB = BC = CA = x Semicircle S: A = πr²/2 2π = πr²/2 r² = (2/π)2π = 4 r = √4 = 2 Let M be the poont of tangency between semicircle P and CA, N be the point of tangency between semicircle Q and AB, and T be rhe point of tangency between semicircles P and Q. As ∆ABC is an equilateral triangle, ∠ABC = ∠BCA = ∠CAB = 60°. Draw MP, QN, and QP. As the centers of two tangent circles and their point of tangency are colinear, and as all three semicircles are fongruent, point T is the midpoint of QP. PM = QN = PT = QT = r = 2. As ∠MAP = ∠NBQ = 60°, ∠PMA = ∠QNB = 90°, and PM = QN, ∆PMA and ∆QNB are congruent 30-60-90 special tight triangles. In a 30-60-90 special right triangle, if the short leg (opposite 30°) is length k, the long leg (opposite 60°) is length √3k and the hypotenuse (opposite 90°) is 2k. Thus PM = QN = √3k = 2. √3k = 2 k = 2/√3 NB = k = 2/√3 PA = 2k = 4/√3 As QN = r, QP = 2r, and ∠PNQ = 90°, ∆PNQ is also a 30-60-90 special right triangle, and PN = √3r = 2√3. AB = PA + PN + NB s = 4/√3 + 2√3 + 2/√3 s = √3(4/3+2+2/3) s = 4√3 Blue area = ∆ABC - 6π A = bh/2 - 6π A = (4√3)(2√3(√3))/2 - 6π A = (4√3)(6)/2 - 6π A = 12√3 - 6π cm² ≈ 1.935 cm²
It should have been stated in the given information that each semicircle was tangent to one of the sides. Else we need to prove it. It is not obvious from the diagram.
Once you've learned that it's an equilateral triangle its area is (length of the sides squared times √3) divided by 4, which is the same as 1/2absin60º therefore the area is 12√3 NOT 24√3!
The introduction of point E is redundant. We can figure out AP by dividing the radius by cos30 and likewise we get PB by dividing 2 times the radius by cos30. After that you can get the side length of the triangle by adding AP and PB and the rest follows. Keep up the good work.
E is not redundant. In your calculation, you assume angle PQB = 90 such that triangle PQB is a right-angled triangle. The proof of angle PQB = 90 is made by constructing triangle PQE which is a right-angled triangle as QE is centre to point of tangency. In triangle PQE, sin(angle EPQ) = 2/4 hence angle EPQ = 30 and then angle PQB = 180 - 60 - 30 = 90.
Drav a line PS, SQ, PQ. Area trangle PSQ = \/p(p-2r)^3. p=3r, r=\/2, area PSQ=4\/3, area ASP=area CQS=area BPQ =8\/3 Area blue trangle = 4\/3+8\/3=12\/3, area of the blue shaded region = 12\/3-6(3,14)=1,9356!
I found AB in another way *The triangle SPQ is equilateral triangle which its side=r=2 *angel SPB=90°(because AB is tangent of the circle which SP is diameter. Angle(QPB)=Angel(SPB)-Angel (SPQ) 30°=60°-90°= Similar angle(BQP)=90° In the right angled triangle (BQP) PB=y, QB=x, y=2x x²+4²=y² x²+16=(2x)² 4x²-x²=16 3x²=16 x=4sqrt(3)/3 AB=x+y=x+2x=3x=4sqrt(3)
Correction: The answer is supposed to be 12√3−6π. My sincere apologies for that!
I was mad trying to find my mistake....kkkkkk Very goos professor!!
Everyone makes mistakes. Keep on submitting good work, though, I really enjoy doing these problems.
@@CharlesB147
Thanks dear ❤🙏
Thank you! I was wondering, because the blue area cannot be bigger than the area of the three semicircles.
@@lillaschwarte8760
Thanks for your understanding. ❤🙏
Area of triangle ABC is actually 12sqrt3, not 24sqrt3, the mistake is only the calculation
Yes, I noticed that too.
After finding the lenght of the radius, we know that the triangles CSQ and BPQ are congruent, because they have a right angle, a congruent acute angle and a congruent side( SQ=PQ)----> CB= 4square root of 3. And Area blu= 12square root of 3 - 6pi.
I agree. PreMath made an error somewhere
I agree as well. The expression (1/2)*(4√3)*(4√3)*(√3/2) of the area of the triangle ABC (in cm²) is correct, but that expression turns out to be 12√3.
I guess this is good news: It means that PreMath is a human being.🙂
Best regards from Germany
You are absolutely unforgotten I always see you both making myself free and challenged to your question.
The area of the ABC triangle is actually 12sqrt(3) not 24sqrt(3). This makes the Blue Area = 12sqrt(3) - 6Pi =1.935
i noticed that mistake too
I'm glad I finally read your comment about the correction. I checked and rechecked my math. Thank you.
Right on the tail end, your area calculation should result in 12sqrt(3), not 24sqrt(3). The rest up to there is good, you just forgot to divide by 2 twice. 😅 So the final answer should be 12sqrt(3)-6pi. Thank you for the video as always, though.
Area of Equilateral Triangle (A) = (√3/4)a²
So all the is needed is the side length of the equilateral triangle to calculate its area an subtract 6pi.
Draw the line PQ, since the radius is 2 that line must have a length of 4. This line now is one side of the right triangle PQB, where the other 2 sides are parts of the sides of the given triangle. By the information provided we know, that this right triangle we just constructed must have angles of 90, 60 and 30 degrees. - We know (maybe from previous Videos on this channel) that the side lengths of an 30-60-90 Triangle have a ratio of 1:√3:2
Since PQ (the second longest side) has a length of 4, QB equals 4/√3 and PQ = 8/√3
Thus the side length = 12/√3
Therefor Area of Equilateral Triangle (A) = (√3/4)*(12/√3)²
= (√3/4)*(144/3)
=(√3/4)*48 = 20,785 (rounded to the 3rd decimal)
=> 20,785 - 6pi is the blue area. (roughly 1.935)
So we do not need to fuss around with sin
The radii are 2.
PSQ is equilateral with sides of 4, so a height of 2*sqrt(3). Height measured from a point T which is the midpoint of PQ and where the tangents meet.
The inside of the main triangle may be split into a series of 30,60, 90 triangles.
APS has mid-length side of 4, so PA is 4/sqrt(3) = (4*sqrt(3))/3 and AS = (8*sqrt(3))/3.
SC = PA = (4*sqrt(3))/3
Main triangle side lengths are (8*sqrt(3))/3 + (4*sqrt(3))/3 = (12*sqrt(3))/3 = 4*sqrt(3)
h is square root of (48 - 12) = sqrt(36) = 6.
Area is 6 * ((6*sqrt(3))/3) = (36*sqrt(3))/3 = 12*sqrt(3) un^2
Blue area is 12*sqrt(3) - 6pi
Approx (12*1.732) - (6*3.1416) = 1.935 (rounded).
Just looked at the video and was wondering where I went wrong.
Then I checked the comments below and felt some relief.
Let's find the area:
.
..
...
....
.....
We start with the calculation of the radius r of the yellow semicircles:
A = πr²/2
(2π)cm² = πr²/2
4cm² = r²
⇒ r = 2cm
Since every pair of semicircles has exactly one point of intersection, the distances of their centers are all the same:
PQ = PS = QS = r + r = 4cm
It is obvious that PQ (PS, QS) is perpendicular to BC (AB, AC). Since all interior angles of the equilateral triangle ABC are 60°, the triangles APS, BPQ and CQS are congruent 30°-60°-90° triangles. So we can conclude:
PQ = PS = QS = 4cm
⇒ AP = BQ = CS = 4cm/√3
∧ AS = BP = CQ = 8cm/√3
Therefore the side length and the area of the equilateral triangle ABC turn out to be:
s(ABC) = AP + BP = AS + CS = BQ + CQ = 4cm/√3 + 8cm/√3 = 12cm/√3
A(ABC) = (√3/4)*[s(ABC)]² = (√3/4)*(12cm/√3)² = (√3/4)*(144/3)cm² = (12√3)cm²
Now we can calculate the size of the blue area:
A(blue) = A(ABC) − 3*A(semicircle) = (12√3 − 6π)cm² ≈ 1.935cm²
Best regards from Germany
It is obvious that PQ (PS, QS) is perpendicular to BC (AB, AC) but a proof is needed.The proof of angle PQB = 90 is made by constructing triangle PQE which is a right-angled triangle as QE is centre Q to point of tangency E. In triangle PQE, sin(angle EPQ) = 2/4 hence angle EPQ = 30 and then angle PQB = 180 - 60 - 30 = 90. Similarly angles APS and CSQ are proven to be 90.
@@hongningsuen1348 Thanks for your feedback and you are absolutely right. I only used symmetry considerations to derive that PQ (PS, QS) is perpendicular to BC (AB, AC). Indeed this is not enough. After obtaining the side lengths PQ=4 and EQ=2 and the angle ∠PEQ=90° it can be concluded that EPQ and therefore also BPQ are 30°-60°-90° triangles. That should have been shown first before using symmetry arguments.
Best regards from Germany
6(2√3-π)≈1,92
I first calculated the area of the blue sector in the center (area of triangle PQS minus 3 circle sectors of 60°). The total area of the other blue parts is 2 times the area of the center blue part: 3 (1/2) + 3 (1/6) = 2. So multiply the area of the center blue sector by 3 and you have the answer.
King of math "Premath" 👑👑
SQP is equilateral triangle so reverse calculation to find PQ,PE,(QB=AP),EB and then use sqrt3/4 *( AB*AB)
to find ABC triangle and minus 6pi will be answer
Wow
1) Radius (R) of Yellow Semicircles:
2) Pi*R^2 = 4*Pi ; R^2 = 4 ; R = 2 cm
3) Area of Equilateral Triangle [PQS] (A) with Side Length = 2*R = 4 cm; using Heron's Formula:
4) A ~ 6,93 Square Cm
5) Area of 3 Equal Right Triangles : [CSQ] ; [APS] ; [BPQ]. The Triangles are (30º ; 60º ; 90º)
6) BQ = (4 * tan(30º)) cm ~ 2,31 cm
7) Area of each Triangle (AT)
8) 2*AT = (4 * [4 * tan(30º)]) ; 2*AT = 16*tan(30º) ; AT = 8*tan(30º) ~ 4,62 Square Cm
9) Area of all 3 Right Triangles = AT [CSQ] + AT [APS] + AT [BPQ] = 24*tan(30º) ~ 13,86 Square Cm
10) Area of the Great Triangle (GT) = 24*tan(30º) + 6,93 ; GT ~ 13,86 + 6,63 ~ 20,79 Square Cm
11) Blue Region Area (BRA) = 20,79 - 6*Pi ~ 20,79 - 18,85 ~ 1,94 Square Cm.
12) Answer : The Blue Region Area is equal to 1,94 Square Centimeters.
Blue area = area of ∆ABC minus three yellow semicircle areas.
- semicircle areas: given (= 6π)
- ∆ABC area : unknown
- Need: ∆ABC side length x
AB = BC = CA = x
Semicircle S:
A = πr²/2
2π = πr²/2
r² = (2/π)2π = 4
r = √4 = 2
Let M be the poont of tangency between semicircle P and CA, N be the point of tangency between semicircle Q and AB, and T be rhe point of tangency between semicircles P and Q.
As ∆ABC is an equilateral triangle, ∠ABC = ∠BCA = ∠CAB = 60°. Draw MP, QN, and QP. As the centers of two tangent circles and their point of tangency are colinear, and as all three semicircles are fongruent, point T is the midpoint of QP. PM = QN = PT = QT = r = 2. As ∠MAP = ∠NBQ = 60°, ∠PMA = ∠QNB = 90°, and PM = QN, ∆PMA and ∆QNB are congruent 30-60-90 special tight triangles.
In a 30-60-90 special right triangle, if the short leg (opposite 30°) is length k, the long leg (opposite 60°) is length √3k and the hypotenuse (opposite 90°) is 2k. Thus PM = QN = √3k = 2.
√3k = 2
k = 2/√3
NB = k = 2/√3
PA = 2k = 4/√3
As QN = r, QP = 2r, and ∠PNQ = 90°, ∆PNQ is also a 30-60-90 special right triangle, and PN = √3r = 2√3.
AB = PA + PN + NB
s = 4/√3 + 2√3 + 2/√3
s = √3(4/3+2+2/3)
s = 4√3
Blue area = ∆ABC - 6π
A = bh/2 - 6π
A = (4√3)(2√3(√3))/2 - 6π
A = (4√3)(6)/2 - 6π
A = 12√3 - 6π cm² ≈ 1.935 cm²
All right.
It should have been stated in the given information that each semicircle was tangent to one of the sides. Else we need to prove it. It is not obvious from the diagram.
Once you've learned that it's an equilateral triangle its area is (length of the sides squared times √3) divided by 4, which is the same as 1/2absin60º therefore the area is 12√3 NOT 24√3!
I took a different, exploiting the two tangent theorem and special triangles. But, I got there in the end. Thank you!
The introduction of point E is redundant. We can figure out AP by dividing the radius by cos30 and likewise we get PB by dividing 2 times the radius by cos30. After that you can get the side length of the triangle by adding AP and PB and the rest follows. Keep up the good work.
E is not redundant. In your calculation, you assume angle PQB = 90 such that triangle PQB is a right-angled triangle. The proof of angle PQB = 90 is made by constructing triangle PQE which is a right-angled triangle as QE is centre to point of tangency. In triangle PQE, sin(angle EPQ) = 2/4 hence angle EPQ = 30 and then angle PQB = 180 - 60 - 30 = 90.
Please check triangle EBQ, its an isosceles triangle since eq and qb is a radius of semi circle ,so it can't be right angle triangle
Drav a line PS, SQ, PQ. Area trangle PSQ = \/p(p-2r)^3. p=3r, r=\/2, area PSQ=4\/3, area ASP=area CQS=area BPQ =8\/3
Area blue trangle = 4\/3+8\/3=12\/3, area of the blue shaded region = 12\/3-6(3,14)=1,9356!
6(2√3 - π)
Area of triangle ABC will be 12underroot3not 24 underroot3
I found AB in another way
*The triangle SPQ is equilateral triangle which its side=r=2
*angel SPB=90°(because AB is tangent of the circle which SP is diameter.
Angle(QPB)=Angel(SPB)-Angel (SPQ)
30°=60°-90°=
Similar angle(BQP)=90°
In the right angled triangle (BQP)
PB=y, QB=x, y=2x
x²+4²=y²
x²+16=(2x)²
4x²-x²=16
3x²=16
x=4sqrt(3)/3
AB=x+y=x+2x=3x=4sqrt(3)
I did 3 equilaterals triangles of side 2/sin60 per semicircle...so the final answer is 18/sin60 - 6pi
ABC area is 12*Sqr(3) ...not 24*
The correct area is 12√3 - 6π, I think
12√3-6π
Love it!!!!!!
12SQUAR3 WHY 24SQUAR3 ?
delete this video and make it again. save people the time and aggravation.