I just used the leg ratios of a 30-60-90 & 45-45-90 triangle and trigonometry to work out the angles of ABC then the side MN to find the area 1/2(2srt2 x 6srt2) = 24/2 = 12
@ 4:07 my first observation is that the sides of a 30-60-90 triangle are always in proportion and that the length of the hypotenuse is twice the length of the shorter side and that the length of the longer side is 3^⅓ times the length of the shorter side . But it is always good to know alternate methods for finding side lengths. Absolutely love this problem because different methods are included. 🙂
Very nice! As an engineer I looked at it as a surveying problem, and after calculating the AB & BC, I used trigonometry to find the angle: N-A-M and proceeded to calculate the rest of the unknowns. I also solved it by assigning X-Y coordinates to the various points and proceeded to find the line equations, and from their intersections calculated the dimensions of the triangle CNM. In both cases I obtained the same answer.
From equilateral right ΔBCM with sides of length 6, hypotenuse CM has length 6√2. From equilateral right ΔMPN with sides found to have length 2, hypotenuse MN has length 2√2.So, right ΔCMN has sides of length 6√2 and 2√2. These sides are also its base and height. Area = (1/2)bh = (1/2)(6√2)(2√2) = 12 square units, as PreMath also found.
Once CB is calculated to be 6. Then CM = root 72 by Pythagoras. Angle BCM = 45. Tan BCA = 2. BCA = 63.435 degrees. Angle MCN = 63.435 - 45 = 18.435. Tan 18.435 = MN/CM. MN = CM x tan 18.435. Pink triangle area = 1/2 x CM x MN. 1/2 x root 72 x root 72 x tan18.435. 1/2 x 72 x tan 18.435. 36 x tan18.435. 12 ans.
I've found NM with trigonometry NM = 6√ 2*tan(alpha-45°) being tan (alpha) = 12/6=2 alpha is angle in C. tan(alpha-45°)=1/3 Otherwise we can build a triangle similar to AMN joining ABC with CMB rotated on the right side, so we have ACM with AM=6+12=18 (being angle MNA = 45° = CMB and A in common) its minor side CM being 6√2 as before. So we can compare the ratios base/right side of both triangle: For ACM base = 18 CM = 6√2 For AMN base = 6 MN = X 18/6 = 6√2/X x=2√2 = MN Area = 2√ 2*6√ 2*1/2 = 12
Well done!❤ Make a right triangle △ABC with legs in 2a/a proportion (longer leg 2a at the bottom) Midpoint AB = M, so AM = MB = a. Construct CM and the perpendicular to CM, intercepting hypotenuse AC at N. *The so constructed triangle △MCN has always area 1/3 of triangle △ABC!* Proof: MC = a√2 = base b of △MCN, h = MN = height of △MCN. Draw a perpendicular line through N to AM, making point P at AM. Then NP/AP = BC/AB = a/2a = 1/2 so 2·NP = AP due to similarity. Since NP = MP (isoceles triangle △PMN with 45°), we have AP + MP = 2·NP + NP = 3·NP = a. Then h = NP√2 = (a/3)√2. Area of triangle △MCN = (1/2)·b·h = (1/2)·a√2·(a/3)√2, which is a²/3 Area of triangle △ABC = (1/2)·2a·a = a² So area △MCN is 1/3 of area △ABC.
>MCB=α → tan α=1 >ACB=γ → tan γ= 2 >ACM=β γ = α + β tan γ = (tan α + tan β)/(1- tan α * tan β) 2= (1+ tan β)/(1- tan β) → tan β = 1/3 tan β = NM/MC MC = 6√2 1/3 = NM/ 6√2 → NM= 2√2 PΔ NMC = ½ * MC * NM = ½ *6√2*2√2= 6*2 = 12 ☺
I like your solution. I didn't use the exact same method, but ended up with 11.9965... which is 12 as near as matters. BCD as a 30-60-90 giving 6 for BC and 6*sqrt(2) for MC.
In triangle BCD Cos(30)=BD/BC=3√3/BC BC=3√3cos(30)=6 BC=BM Angle BCM=BMC=45 So triangle BCM is isosceles triangle CM^2=2BC^2=2(6^2) CM=6√2 units AB=12 units Tan(ACB)=12/6 Angle ACB=63.43 Hence: angle MCN=63.43-45=18.43 Tan(18.43)=MN/CM=MN/6√2 MN=6√2(0.333)=2.83 units Area of the purple triangle=1/2(2.83)(6√2)=12 square units. thanks ❤❤❤
First of all, the triangle BDC has only one interest in this problem: To calculate the length BC (or MB or MA) which is BD / cos (30°) = (3. sqrt(3)) / (sqrt(3) / 2) = 6. Now in the isoceles right triangle MBC we have MC = MB . sqrt(2) = 6. sqrt(2). To finish, then let's use an adapted orthonormal: M(0;0) B(6;0) A(-6;0). The equation of (MC) is y =x, and the equation of (MN) (which is perpendicular to (MC)) is y = -x. VectorAC (12.6) is colinear tu VectorU(2;1) and the equation of (AC) is (x + 6). (1) - (y). (2)= 0, or x -2y + 6 = 0. Then we have N (intersection of (MN) and (AC) when resolving the system y = -x and x -2y + 6 = 0. We find easily N(2;2) and then VectorMN(-2;2) and MN = sqrt( (-2)^2 + (2)^2) = sqrt(8) = 2.sqrt(2) Finally the area of the right triangle NMC is (1/2). MN . MC = (1/2). (2.sqrt(2)). (6.sqrt(2)) = 12.
Yay! I solved the problem. My first method was to calculate all the interior angles. MC = 6√2. Using the law of sines, MN=2√2. A = (1/2)*(2√2)*(6√2) = 12. My second method was to calculate area of triangle ABC and triangle AMN and triangle BMC and then the remaining area was the purple triangle. 36 - 18 - 6 = 12.
Il lato \\ è 3√3/cos30=6...il triangolo purple ha altezza h=√72..e base b,6/sin(arctg1/2+45)=b/sinarctg1/2(teorema dei seni)...b=4/√2...Ap=hb/2=√72*4/2√2=6*4/2=12
Measure of angle A is arc tan (1/2). We know angle nma is 45 degrees.So knowing the three angles of triangle ANM we can use the sine rule to find length of nm etc. Method on video much nicer.
Let's do it: . .. ... .... ..... The triangle BCD is a 30°-60°-90° triangle, so we can conclude that BC=2*CD. Now we apply the Pythagorean theorem: BD² + CD² = BC² (3√3)² + CD² = (2*CD)² 27 + CD² = 4*CD² 27 = 3*CD² 9 = CD² ⇒ CD = 3 ⇒ BC = 6 Now we apply the Pythagorean theorem again for the right triangle BCM: CM² = BC² + BM² CM² = 2*BC² ⇒ CM = (√2)*BC = 6√2 If we assume M to be the center of the coordinate system and A, M and B to be located on the x-axis, we can represent the lines AC and MN with the following functions: AC: y = (1/2)*[x + 6] MN: y = −x Now we can calculate the coordinates of N: −xN = (1/2)*[xN + 6] −xN = xN/2 + 3 −(3/2)*xN = 3 ⇒ xN = −2 ⇒ yN = +2 So the length MN and the area of the purple triangle turn out to be: MN² = (xN − xM)² + (yN − xM) = (−2)² + (+2)² = 2*2² ⇒ MN = 2√2 A(CMN) = (1/2)*CM*MN = (1/2)*6√2*2√2 = 12 Best regards from Germany
VERY nice use of superscripts, triangle glyphs, double arrows! I too like using 'em. Wrote a 'script enhancer for math' some years back that allows typing it in MUCH easier than finding the glyphs and patching them in. For example: This: "2/3 sqrt(3) * &mathx;^4 = θ" becomes "⅔√3 • 𝒙⁴ = θ" I am taking a bet that you use a similarly 'meta' way to enter your math formulae! Again, well one. GoatGuy
Sir, An innovative way to find out BC=6 because I thought that angle BCD=60degree So Sin60degree =BD/BC root 3/2=3. root 3/BC BC=6, But I prefered most the way as you have applied..... 🙏🙏🙏
Everyone finds the answer their own way! Excellent! How about this one… 30-60-90 △ has side lengths 1:2:√3 so in proportion BD / √3 = CD = 3 CD × 2 = BC = 6 BC = BM = AM = 6 MC = 6√2 ∠ CAB = arctan( BC / AB ) = arctan( 6 ÷ 12 ) = 26.565° ∠ NCM = 180° - (90° + ∠ CAB + ∠ MCB) = 18.435° NM = MC tan 18.435° = (6√2 × 0.3333333) = 2√2; Area △NMC = (½ × 6√2 × 2√2) = (½ × 6 × 2 × √2²) = 12 And we're done. Yay! GoatGuy
Resolution Proposal: CB = BM = AM = 6 tan (30) = CD / 3*Sqrt(3) CD = tan (30) * (3 * Sqrt(3)) = 3 lu The Area of Rectangle [ A B C A' ] = (6 * 12) = 72 su The Area of Triangle [ A B C ] = 36 su The Area of Triangle [ B C M ] = (6 * 6) / 2 = 36 / 2 = 18 su The Measure of the Segment CM = Sqrt (72) su ~ 8,5 lu The Area of Triangle [ A C M ] = (6 * 6) / 2 = 36 / 2 = 18 su The Angle BAC = Arctan (0,5) ~ 26,56 degrees So the Angle ACM (in degrees) = 90 - 45 - Arctan (0,5) = 45 - 26,56 ~ 18,44 degrees The Segment NM = tan (18,44) * 8,5 ~ 2,83 lu The Purple Triangle Area = (MN * CM) / 2 = (2,83 * 8,5) / 2 = 12 Answer : exactly 12 su.
I don't know if you have been made aware that TH-cam is POLLUTING your videos this nested ads. Nothing is more aggravating than have your attention rudely interrupted when trying to follow an involved explanation. That said how on earth could the ads be a successful revenue generator when all they accomplish is aggravation? HUGE THUMBS DOWN..
I m first pin me❤😊
Ok 😊
I liked this problem a lot. It is always a pleasure to follow your solutions!
Thanks ❤️
Very easy. And also, no need for trigo on the 30° angle in rt∆BDC. Just follow the 30-60-90 Theorem
Thanks ❤️
Amazing!
good solution.Thanks.
I just used the leg ratios of a 30-60-90 & 45-45-90 triangle and trigonometry to work out the angles of ABC then the side MN to find the area 1/2(2srt2 x 6srt2) = 24/2 = 12
@ 4:07 my first observation is that the sides of a 30-60-90 triangle are always in proportion and that the length of the hypotenuse is twice the length of the shorter side and that the length of the longer side is 3^⅓ times the length of the shorter side . But it is always good to know alternate methods for finding side lengths. Absolutely love this problem because different methods are included. 🙂
Thanks ❤️🌹
Very nice! As an engineer I looked at it as a surveying problem, and after calculating the AB & BC, I used trigonometry to find the angle: N-A-M and proceeded to calculate the rest of the unknowns. I also solved it by assigning X-Y coordinates to the various points and proceeded to find the line equations, and from their intersections calculated the dimensions of the triangle CNM. In both cases I obtained the same answer.
From equilateral right ΔBCM with sides of length 6, hypotenuse CM has length 6√2. From equilateral right ΔMPN with sides found to have length 2, hypotenuse MN has length 2√2.So, right ΔCMN has sides of length 6√2 and 2√2. These sides are also its base and height. Area = (1/2)bh = (1/2)(6√2)(2√2) = 12 square units, as PreMath also found.
I solved it the same way
Of course you meant isosceles, not equilateral 😉
Thanks ❤️🌹
Absolutely love your very efficient solution. Real nice! 😉
Once CB is calculated to be 6.
Then CM = root 72 by Pythagoras.
Angle BCM = 45.
Tan BCA = 2.
BCA = 63.435 degrees.
Angle MCN = 63.435 - 45 = 18.435.
Tan 18.435 = MN/CM.
MN = CM x tan 18.435.
Pink triangle area = 1/2 x CM x MN.
1/2 x root 72 x root 72 x tan18.435.
1/2 x 72 x tan 18.435.
36 x tan18.435.
12 ans.
Thanks ❤️
This is awesome, many thanks, Sir!
φ = 30° = CBD; BD = 3√3 → CD = 3 → CB = 6 = BM = AM → AB = 12 → CM = 6√2 → AC = √(144 + 36) = 6√5
CAB = θ → BCA = 3φ - θ = 3φ/2 + ϑ ↔ ϑ = MCN; sin(3φ/2) = cos(3φ/2) = √2/2
sin(θ) = √5/5 → cos(θ) = √(1 - sin^2(θ)) = 2√5/5 → sin(θ) = cos(3φ/2 + ϑ) →
cos(ϑ) = cos(3φ/2 - θ) = cos(3φ/2)cos(θ) + sin(3φ/2)sin(θ) = (√2/2)(2√5/5) + (√2/2)(√5/5) = 3√10/10 = 6√2/CN → CN = 4√5 → NM = √(80 - 72) = 2√2 → area ∆ CNM = (1/2)(2√2)(6√2) = 12
I've found NM with trigonometry NM = 6√ 2*tan(alpha-45°) being tan (alpha) = 12/6=2 alpha is angle in C. tan(alpha-45°)=1/3
Otherwise we can build a triangle similar to AMN joining ABC with CMB rotated on the right side, so we have ACM with AM=6+12=18 (being angle MNA = 45° = CMB and A in common)
its minor side CM being 6√2 as before. So we can compare the ratios base/right side of both triangle:
For ACM base = 18 CM = 6√2
For AMN base = 6 MN = X
18/6 = 6√2/X
x=2√2 = MN
Area = 2√ 2*6√ 2*1/2 = 12
Thanks ❤️
Excellent
Thank you so much 😀
Excellent presentation sir 🎉🎉🎉🎉🎉
Thanks ❤️
Thank you
You're welcome❤️🌹
Well done!❤
Make a right triangle △ABC with legs in 2a/a proportion (longer leg 2a at the bottom)
Midpoint AB = M, so AM = MB = a.
Construct CM and the perpendicular to CM, intercepting hypotenuse AC at N.
*The so constructed triangle △MCN has always area 1/3 of triangle △ABC!*
Proof:
MC = a√2 = base b of △MCN, h = MN = height of △MCN.
Draw a perpendicular line through N to AM, making point P at AM.
Then NP/AP = BC/AB = a/2a = 1/2 so 2·NP = AP due to similarity. Since NP = MP (isoceles triangle △PMN with 45°), we have AP + MP = 2·NP + NP = 3·NP = a.
Then h = NP√2 = (a/3)√2.
Area of triangle △MCN = (1/2)·b·h = (1/2)·a√2·(a/3)√2, which is a²/3
Area of triangle △ABC = (1/2)·2a·a = a²
So area △MCN is 1/3 of area △ABC.
Thanks ❤️🌹
>MCB=α → tan α=1
>ACB=γ → tan γ= 2
>ACM=β
γ = α + β tan γ = (tan α + tan β)/(1- tan α * tan β)
2= (1+ tan β)/(1- tan β) → tan β = 1/3
tan β = NM/MC MC = 6√2
1/3 = NM/ 6√2 → NM= 2√2
PΔ NMC = ½ * MC * NM = ½ *6√2*2√2= 6*2 = 12 ☺
I like your solution. I didn't use the exact same method, but ended up with 11.9965... which is 12 as near as matters.
BCD as a 30-60-90 giving 6 for BC and 6*sqrt(2) for MC.
Excellent❤
I have seen you before
Thanks 😊
In triangle BCD
Cos(30)=BD/BC=3√3/BC
BC=3√3cos(30)=6
BC=BM
Angle BCM=BMC=45
So triangle BCM is isosceles triangle
CM^2=2BC^2=2(6^2)
CM=6√2 units
AB=12 units
Tan(ACB)=12/6
Angle ACB=63.43
Hence: angle MCN=63.43-45=18.43
Tan(18.43)=MN/CM=MN/6√2
MN=6√2(0.333)=2.83 units
Area of the purple triangle=1/2(2.83)(6√2)=12 square units. thanks ❤❤❤
Thanks ❤️
First of all, the triangle BDC has only one interest in this problem:
To calculate the length BC (or MB or MA) which is BD / cos (30°) = (3. sqrt(3)) / (sqrt(3) / 2) = 6.
Now in the isoceles right triangle MBC we have MC = MB . sqrt(2) = 6. sqrt(2).
To finish, then let's use an adapted orthonormal: M(0;0) B(6;0) A(-6;0). The equation of (MC) is y =x, and the equation of (MN) (which is perpendicular to (MC)) is y = -x.
VectorAC (12.6) is colinear tu VectorU(2;1) and the equation of (AC) is (x + 6). (1) - (y). (2)= 0, or x -2y + 6 = 0.
Then we have N (intersection of (MN) and (AC) when resolving the system y = -x and x -2y + 6 = 0. We find easily N(2;2) and then VectorMN(-2;2)
and MN = sqrt( (-2)^2 + (2)^2) = sqrt(8) = 2.sqrt(2)
Finally the area of the right triangle NMC is (1/2). MN . MC = (1/2). (2.sqrt(2)). (6.sqrt(2)) = 12.
Thanks ❤️
3sqrt3=CBsqrt3/2》CB=6=AM=MB
Ángulo AMN=180-45-90=45°
AM=(6-h)+h ; pendiente de AC =6/12=1/2 ; pendiente de MN =1 》(6-h)(1/2)=1×h》h=2 》Área NMC=AMC-AMN =(6×6/2)-(6×2/2) =18-6=12
Gracias y saludos.
Thanks ❤️
Yay! I solved the problem. My first method was to calculate all the interior angles. MC = 6√2. Using the law of sines, MN=2√2. A = (1/2)*(2√2)*(6√2) = 12. My second method was to calculate area of triangle ABC and triangle AMN and triangle BMC and then the remaining area was the purple triangle. 36 - 18 - 6 = 12.
Thanks ❤️
Very nice and visual solution. My solution involves a lot of trigonometry.
Calculate BC=6 exactly like you did.
Calculate AB = 12 = 2*BC.
Tan (angle ACB) = AB/BC = 2.
Tan (angle MCB) = MB/BC = 1
Angle NCM = Angle ACB - MCB.
Tan (angle NCM) = (2 - 1)/(1 + 2*1) = 1/3. (Tan(x-y) = (tan(x) - tan(y))/(1 + tan(x)tan(y)).
Length MC = sqrt(MB*MB + BC*BC) = sqrt(72) = 6*sqrt(2).
Length MN = MC * tan(NCM) = 6*sqrt(2) * (1/3) = 2*sqrt(2).
Area purple triangle = (1/2) * MC * MN = (1/2)*6*sqrt(2)*2*sqrt(2) = 12.
Thanks ❤️
Il lato \\ è 3√3/cos30=6...il triangolo purple ha altezza h=√72..e base b,6/sin(arctg1/2+45)=b/sinarctg1/2(teorema dei seni)...b=4/√2...Ap=hb/2=√72*4/2√2=6*4/2=12
Thanks ❤️
Measure of angle A is arc tan (1/2). We know angle nma is 45 degrees.So knowing the three angles of triangle ANM we can use the sine rule to find length of nm etc. Method on video much nicer.
Thanks ❤️🌹
Let's do it:
.
..
...
....
.....
The triangle BCD is a 30°-60°-90° triangle, so we can conclude that BC=2*CD. Now we apply the Pythagorean theorem:
BD² + CD² = BC²
(3√3)² + CD² = (2*CD)²
27 + CD² = 4*CD²
27 = 3*CD²
9 = CD²
⇒ CD = 3
⇒ BC = 6
Now we apply the Pythagorean theorem again for the right triangle BCM:
CM² = BC² + BM²
CM² = 2*BC²
⇒ CM = (√2)*BC = 6√2
If we assume M to be the center of the coordinate system and A, M and B to be located on the x-axis, we can represent the lines AC and MN with the following functions:
AC: y = (1/2)*[x + 6]
MN: y = −x
Now we can calculate the coordinates of N:
−xN = (1/2)*[xN + 6]
−xN = xN/2 + 3
−(3/2)*xN = 3
⇒ xN = −2
⇒ yN = +2
So the length MN and the area of the purple triangle turn out to be:
MN² = (xN − xM)² + (yN − xM) = (−2)² + (+2)² = 2*2²
⇒ MN = 2√2
A(CMN) = (1/2)*CM*MN = (1/2)*6√2*2√2 = 12
Best regards from Germany
VERY nice use of superscripts, triangle glyphs, double arrows! I too like using 'em. Wrote a 'script enhancer for math' some years back that allows typing it in MUCH easier than finding the glyphs and patching them in. For example:
This: "2/3 sqrt(3) * &mathx;^4 = θ" becomes "⅔√3 • 𝒙⁴ = θ"
I am taking a bet that you use a similarly 'meta' way to enter your math formulae! Again, well one. GoatGuy
Thanks ❤️🇺🇸
Sir, An innovative way to find out BC=6 because I thought that angle BCD=60degree
So
Sin60degree =BD/BC
root 3/2=3. root 3/BC
BC=6, But I prefered most the way as you have applied..... 🙏🙏🙏
Thanks ❤️
Everyone finds the answer their own way! Excellent!
How about this one…
30-60-90 △ has side lengths 1:2:√3 so in proportion
BD / √3 = CD = 3
CD × 2 = BC = 6
BC = BM = AM = 6
MC = 6√2
∠ CAB = arctan( BC / AB ) = arctan( 6 ÷ 12 ) = 26.565°
∠ NCM = 180° - (90° + ∠ CAB + ∠ MCB) = 18.435°
NM = MC tan 18.435° = (6√2 × 0.3333333) = 2√2;
Area △NMC = (½ × 6√2 × 2√2) = (½ × 6 × 2 × √2²) = 12
And we're done.
Yay!
GoatGuy
GoatGuy👍
Thanks ❤️
Resolution Proposal:
CB = BM = AM = 6
tan (30) = CD / 3*Sqrt(3)
CD = tan (30) * (3 * Sqrt(3)) = 3 lu
The Area of Rectangle [ A B C A' ] = (6 * 12) = 72 su
The Area of Triangle [ A B C ] = 36 su
The Area of Triangle [ B C M ] = (6 * 6) / 2 = 36 / 2 = 18 su
The Measure of the Segment CM = Sqrt (72) su ~ 8,5 lu
The Area of Triangle [ A C M ] = (6 * 6) / 2 = 36 / 2 = 18 su
The Angle BAC = Arctan (0,5) ~ 26,56 degrees
So the Angle ACM (in degrees) = 90 - 45 - Arctan (0,5) = 45 - 26,56 ~ 18,44 degrees
The Segment NM = tan (18,44) * 8,5 ~ 2,83 lu
The Purple Triangle Area = (MN * CM) / 2 = (2,83 * 8,5) / 2 = 12
Answer : exactly 12 su.
Thanks ❤️
12
I don't know if you have been made aware that TH-cam is POLLUTING your videos this nested ads. Nothing is more aggravating than have your attention rudely interrupted when trying to follow an involved explanation. That said how on earth could the ads be a successful revenue generator when all they accomplish is aggravation? HUGE THUMBS DOWN..