I solved it this way: I drew the scalene trapezoid APHK by drawing the perpendicular to AB that passes through the point C (above and below) and extending AB to the right until it meets the said perpendicular and therefore PH // to AB. I found PC by applying the cosine rule on PCB triangle (PC=a and angle in C = 120°), with cos 120° = -1/2 3² = 2² + a² - 2*2*a*(-1/2) a² + 2a - 5 = 0 a = √ 6 - 1 being PHC a right-angled triangle of the type 30°60°90°, it follows that PH = (√ 6 - 1)/2 HC = (3√ 2 - √ 3)/2 the height of the trapezium HK = HC + CK CK = √ 3 and BK = 1 (since BC = 2 and BCK is of the type 30°60°90) HK = (√ 3 + 3√ 2)/2 Finally we can find X with the Pythagorean theorem: X² = [(√ 3+3√ 2)]² + [ (2+1) - (√ 6 - 1)/2]² X² = 19 - 2√ 6
As ABCDEF is a regular hexagon, all internal angles are 120° and all sides = AB = 2. Let ∠BCP = θ. As PB is given as 3, BC = AB = 2, and θ = 120°, we can use the law of cosines to determine CP. Triangle ∆BCP: PB² = BC² + CP² - 2BC(CP)cosθ 3² = 2² + CP² - 2(2)CP(-1/2) 9 = 4 + CP² + 2CP CP² + 2CP - 5 = 0 CP = [-(2)±√((2)²-4(1)(-5))]/2(1) CP = -1 ± √(4+20)/2 CP = -1 ± √24/2 = -1 ± √6 CP = -1 + √6 | CP = -1 - √6 ❌ CP > 0 Draw AC. As FA and CD are parallel, then ∠FAC = ∠ACP = 90°. Let M be the midpoint of AC. As AB = BC, ∆ABC is an isosceles triangle and thus BM is a perpendicular bisector of ∆ABC, forming two congruent right triangles ∆BMA and ∆CMB. As ∠ABC = 120°, ∠ABM = ∠MBC = 60°, and thus ∆BMA and ∆CMB are special 30-60-90 right triangles. As AB = BC = 2, CM = MA = √3, so CA = 2√3. Triangle ∆ACP: AC² + CP² = PA² (2√3)² + (√6-1)² = x² 12 + (6-2√6+1) = x² 19 - 2√6 = x² x = √(19-2√6)
The answer is x = sqrt(19-2*sqrt(6)). This is an example of both using knowledge of the 30-60-90 triangle AND using auxiliary lines. And also combine that with interior angles of the hexagon. I better use that for practice!!!
For those who, like myself, aren't very clever with constructions: Extend AB to the right and drop a perpendicular from P. Label the intersection G. Extend DC until it meets the extended AB and label the intersection as point H. Note that ΔBCH is equilateral, so CH = BH = BC = 2. ΔPGH is a 30° - 60° - 90° right triangle, Let BG have length m, then GH has length 2 - m. From ratio of sides, PG has length (2 - m)(√3). Apply the Pythagorean theorem to ΔBPG: ( m² + ((2 - m)(√3))² = 3², which simplifies to 4m² - 12m + 3 = 0, roots m = (3 + √6)/2 and (3 - √6)/2. Since GH = 2 - m must be positive, we discard the first solution and m = (3 - √6)/2. Then, apply the Pythagorean theorem to ΔAPG. (AG)² + (PG)² = (AP)², (2 + m)² + ((2 - m)(√3))² = x². Expanding, 4 + 4m + m² + 12 - 12m + 3m² = x². Simplifying, 4m² - 8m + 16 = x² and m² - 2m + 4 = x²/4. We substitute m = (3 - √6)/2: ((3 - √6)/2)² - 2((3 - √6)/2) + 4 = (9 - 6√6 + 6)/4 - 2((3 - √6)/2) + 4 = (15 - 6√6)/4 - (12 + 4√6)/4 + 16/4. Simplifying, (19 - 2√6)/4. Set equal to x²/4: x²/4 = (19 - 2√6)/4, x² = 19 - 2√6, x = √(19 - 2√6), as Math Booster also found. Math Booster's clever constructions lead to much simpler algebra, but, with enough effort on my part, I reached the same solution.
If PB = 2√3 then x=4 cm If PB = 2 then x=2√3 cm If PB = 3 then a) Average: x= 3,732 +0,1/-0 b) Simple rule of 3, respect to "2"/"3" x = 3,821 +0/-0,1 c) Average of above: x = 3,776 ± 0,025
∠BCP = 120° so apply the law of cosines to △BCP to determine CP = √6 - 1 So PD = 2 - (√6 - 1) = 3 - √6 ∠ADP = 60° and AD = 4 so apply the law of cosines to △ADP to determine AP = √(19 - 2√6)
PC=a..PBC=β..PBA=γ..legge del coseno 9=a^2+4-4acos120..a=√6-1..a/sinβ=3/sin120..β=arcsin(√3(√6-1)/6)..γ=120-β...x^2=4+9-2*2*3cosγ/=16+√3√(5+2√6)-3√6=14,1010205...mah
A circle through the six vertices of the hexagon will have a radius of 2 . Angles in the in this circle standing on equal arcs will subtend equal angles . ( THIS WAS WRONG Angle ADB = Angle APB was assumed.) Angle ADB = 90º so sine of (ADB) =30º In triangle APB, sin( A) /3 = sin(B)/X = sin(P) /2 From here: sin(A) = 3/2 × sin(30º) = 3/4 , sin(A)/3 = 1/4 , X = 4 sin(B) sin (B) = sin(180º-A-P) = sin (A+P) = sin (A)cos (P) + sin(P)cos(A) = 3/4 × 3^½/2 +1/2 × (7^½)/4 so X = 3.921 approx This does not match the podcast so I have more mistakes to find: because my mistake might help people to avoid hasty steps!
α = BPC ; β = ABP
Sine rule:
sinα /2 = sin120° /3
α =35,2644° ; β =60°+α =95,2644°
Cosine rule:
x² = 2² + 3² - 2*2*3*cosβ
x = 3,755 cm ( Solved √)
Yours is very clean method and calculation.
I used cosine rule on angles BCP (to find a) and ADP to find x, AD being 2+2 and DP being 2-a.
I solved it this way:
I drew the scalene trapezoid APHK by drawing the perpendicular to AB that passes through the point C (above and below) and extending AB to the right until it meets the said perpendicular and therefore PH // to AB.
I found PC by applying the cosine rule on PCB triangle (PC=a and angle in C = 120°), with cos 120° = -1/2
3² = 2² + a² - 2*2*a*(-1/2)
a² + 2a - 5 = 0
a = √ 6 - 1
being PHC a right-angled triangle of the type 30°60°90°, it follows that
PH = (√ 6 - 1)/2
HC = (3√ 2 - √ 3)/2
the height of the trapezium HK = HC + CK
CK = √ 3 and BK = 1 (since BC = 2 and BCK is of the type 30°60°90)
HK = (√ 3 + 3√ 2)/2
Finally we can find X with the Pythagorean theorem:
X² = [(√ 3+3√ 2)]² + [ (2+1) - (√ 6 - 1)/2]²
X² = 19 - 2√ 6
As ABCDEF is a regular hexagon, all internal angles are 120° and all sides = AB = 2.
Let ∠BCP = θ. As PB is given as 3, BC = AB = 2, and θ = 120°, we can use the law of cosines to determine CP.
Triangle ∆BCP:
PB² = BC² + CP² - 2BC(CP)cosθ
3² = 2² + CP² - 2(2)CP(-1/2)
9 = 4 + CP² + 2CP
CP² + 2CP - 5 = 0
CP = [-(2)±√((2)²-4(1)(-5))]/2(1)
CP = -1 ± √(4+20)/2
CP = -1 ± √24/2 = -1 ± √6
CP = -1 + √6 | CP = -1 - √6 ❌ CP > 0
Draw AC. As FA and CD are parallel, then ∠FAC = ∠ACP = 90°. Let M be the midpoint of AC. As AB = BC, ∆ABC is an isosceles triangle and thus BM is a perpendicular bisector of ∆ABC, forming two congruent right triangles ∆BMA and ∆CMB. As ∠ABC = 120°, ∠ABM = ∠MBC = 60°, and thus ∆BMA and ∆CMB are special 30-60-90 right triangles. As AB = BC = 2, CM = MA = √3, so CA = 2√3.
Triangle ∆ACP:
AC² + CP² = PA²
(2√3)² + (√6-1)² = x²
12 + (6-2√6+1) = x²
19 - 2√6 = x²
x = √(19-2√6)
ua-cam.com/video/2D-55qRKCZI/v-deo.htmlsi=q2RAy6dGNwG9wQxO
Yours is ery clean method and calculation.
I used cosine rule on angles BCP (to find a) and ADP to find x, AD being 2+2 and DP being 2-a.
The answer is x = sqrt(19-2*sqrt(6)). This is an example of both using knowledge of the 30-60-90 triangle AND using auxiliary lines. And also combine that with interior angles of the hexagon. I better use that for practice!!!
In which grade are u at now?
For those who, like myself, aren't very clever with constructions: Extend AB to the right and drop a perpendicular from P. Label the intersection G. Extend DC until it meets the extended AB and label the intersection as point H. Note that ΔBCH is equilateral, so CH = BH = BC = 2. ΔPGH is a 30° - 60° - 90° right triangle, Let BG have length m, then GH has length 2 - m. From ratio of sides, PG has length (2 - m)(√3). Apply the Pythagorean theorem to ΔBPG: ( m² + ((2 - m)(√3))² = 3², which simplifies to 4m² - 12m + 3 = 0, roots m = (3 + √6)/2 and (3 - √6)/2. Since GH = 2 - m must be positive, we discard the first solution and m = (3 - √6)/2.
Then, apply the Pythagorean theorem to ΔAPG. (AG)² + (PG)² = (AP)², (2 + m)² + ((2 - m)(√3))² = x². Expanding, 4 + 4m + m² + 12 - 12m + 3m² = x². Simplifying, 4m² - 8m + 16 = x² and m² - 2m + 4 = x²/4. We substitute m = (3 - √6)/2: ((3 - √6)/2)² - 2((3 - √6)/2) + 4 = (9 - 6√6 + 6)/4 - 2((3 - √6)/2) + 4 = (15 - 6√6)/4 - (12 + 4√6)/4 + 16/4. Simplifying, (19 - 2√6)/4. Set equal to x²/4: x²/4 = (19 - 2√6)/4, x² = 19 - 2√6, x = √(19 - 2√6), as Math Booster also found.
Math Booster's clever constructions lead to much simpler algebra, but, with enough effort on my part, I reached the same solution.
ua-cam.com/video/2D-55qRKCZI/v-deo.htmlsi=q2RAy6dGNwG9wQxO
If PB = 2√3 then x=4 cm
If PB = 2 then x=2√3 cm
If PB = 3 then
a) Average:
x= 3,732 +0,1/-0
b) Simple rule of 3, respect to "2"/"3"
x = 3,821 +0/-0,1
c) Average of above:
x = 3,776 ± 0,025
α = BPC ; β = ABP
Sine rule:
sinα /2 = sin120° /3
α =35,2644° ; β =60°+α =95,2644°
Cosine rule:
x² = 2² + 3² - 2*2*3*cosβ
x = 3,755 cm ( Solved √)
∠BCP = 120° so apply the law of cosines to △BCP to determine CP = √6 - 1
So PD = 2 - (√6 - 1) = 3 - √6
∠ADP = 60° and AD = 4 so apply the law of cosines to △ADP to determine AP = √(19 - 2√6)
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PC=a..PBC=β..PBA=γ..legge del coseno 9=a^2+4-4acos120..a=√6-1..a/sinβ=3/sin120..β=arcsin(√3(√6-1)/6)..γ=120-β...x^2=4+9-2*2*3cosγ/=16+√3√(5+2√6)-3√6=14,1010205...mah
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BD^2 =12
BP^2=BC^2+CP^2-2BC CP COS120
BP=-1+racine6
PD=2-PC
=3-racine6
X^2=AD^2+DP^2 - 2AD DP cos60
=19-2racine6
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A circle through the six vertices of the hexagon will have a radius of 2 .
Angles in the in this circle standing on equal arcs will subtend equal angles . ( THIS WAS WRONG Angle ADB = Angle APB was assumed.)
Angle ADB = 90º so sine of (ADB) =30º
In triangle APB, sin( A) /3 = sin(B)/X = sin(P) /2 From here: sin(A) = 3/2 × sin(30º) = 3/4 , sin(A)/3 = 1/4 , X = 4 sin(B)
sin (B) = sin(180º-A-P) = sin (A+P) = sin (A)cos (P) + sin(P)cos(A) = 3/4 × 3^½/2 +1/2 × (7^½)/4 so X = 3.921 approx
This does not match the podcast so I have more mistakes to find: because my mistake might help people to avoid hasty steps!
Point P is on the hexagon, but it is not on the circle. That is why my answer was incorrect.
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Guys..you are very smart. Good
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Pleas compair the.sum of.diameter.and perimeter
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What dis math du yuo naw math uot area
Extremely annoying accent
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