Let's Think outside the Box! | Calculate the Chord AB | (Step-by-step explanation) |
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- Learn how to calculate the Chord AB length. Area of the Blue square is 2. Important Geometry skills are also explained: area of the square formula; Pythagorean Theorem; similar triangles; Thales' theorem. Step-by-step tutorial by PreMath.com
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Let's Think outside the Box! | Calculate the Chord AB | (Step-by-step explanation) | #math #maths
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Intersecting Chords theorem at F.
√6*FB = (2+√2)*(2-√2)
FB = 2/√6
AB = √6 + 2/√6 = 8/√6
I solved it this way too! : )
Since the blue square is 2 cm², its sides are √2 cm and its diagonals are 2 cm. Diagonal OE also is the radius of the semicircle.
Radius OA = 2 cm and side OF = √2 cm ⇒ AF = √6 cm. By the chord theorem, AF · BF = FE² ⇒ BF = √6 / 3 cm ⇒ AB = 4√6/3 cm.
Technically the screen is a rectangular box, so we can’t think outside the box…
Brilliant! Thanks!!
Perfect
MashaAllah. Excellent work 👍. Very informative video
Thanks for liking ❤️ 🌹
Good job sar❤👍
By extending ef to meet the semicircle at f’ by proving the triangles feo and off’ congruent, it follows that ef= ff’ so af by fb = ef to be squared etc
At a quick glance, Reflect the Blue square about the line OF, to form a second square OFGH. The squares have side lengths sqrt(2). Draw a line HE parallel with AB. HE^2 = sqrt(2)^2 + (2)^2 = 2 + 4 = 6 and HE = sqrt(6). AF = HE = sqrt(6). AO = sqrt(6 - 2). AO = 2 Then Bisect AB from O to Q to form triangle OQF. Using similar triangles AFO and AQF. Then AF/sqrt(2) = sqrt(2)/QF Then QF = 2 / AF Then QF=2/ sqrt(6) and AQ = AF - QF = sqrt(6) - (2/sqrt(6) )= 1.633 and AB = 2 * AQ =3.26. Chord AB = 3.26
you have a good sense of humor
By observation, the diagonal OE of the square is a radius of the semicircle. We'll also need to know the side length as well.
Square ODEF:
A = d²/2
2 = r²/2
r² = 4
r = 2
A = s²
2 = s²
s = √2
Triangle ∆AOF:
OF² + OA² = FA²
(√2)² + 2² = FA²
FA² = 2 + 4 = 6
FA = √6
Draw radius OG so that it bisects AB at H. As ∠OHA = 90° because any radius that bisects a chord is perpendicular to that chord, and ∆OHA and ∆AOF share internal angle ∠A, the two triangles are similar. Let HA = x. AB = 2HA = 2x.
Triangle ∆OHA:
HA/AO = AO/FA
x/2 = 2/√6
x = 4/√6
AB = 2x = 2(4/√6) = 8/√6
AB = 8√6/6 = (4√6)/3 cm ≈ 3.266 cm
A more trigonometric approach (all angles in degrees):
The side of the blue square is sqrt(2) and the diagonal, also the radius of the big semicircle, is 2; so angle _alpha_ = arctan(sqrt(2)/2) = arctan(0.7071) = 35.254.
I suspected that angle B might be 90 degrees and was glad to have Thales confirm it. That means angle _beta_ = (180 -- (90 + 35.254)) = 54.736.
Then the desired length AB = (4)(sin(54.736)) = (4)(0.8165) = 3.266 cm.
Alea jacta est! 🤠
Thank you
Easy one
AO=OC=Radio r →→ OF=√2 → Potencia de F respecto a la circunferencia = FE²=(√2)²=2 =(r-√2)(r+√2)=r²-2 → r=2 → AF=√[(√2)²+2²] =√6 → Potencia de F =2 =FB*AF=FB√6→ FB=(√6)/3 → AB=AF+FB =(4/3)√6
Gracias y saludos.
It's good to know the *intersecting chords theorem* for this.
The square has area 2, therefore side length sqrt(2). The radius of circle is the diagonal of the square which is sqrt(2)·sqrt(2) = 2.
Double the upper square side EF to the left intersection with the halfcircle. This chord then has length 2·sqrt(2).
The longer part of the wanted chord AF is given by Pythagorean theorem sqrt(2² + sqrt(2)²) = sqrt(6).
Then use the intersecting chords theorem to get to shorter part and add up to the longer part.
Side of square,
and radius of circle:
A = S² = ½R² = 2 cm²
S = √2 cm
R = 2 cm
Pytagorean theorem:
AF² = S² + R² = 2 + 2²
AF = √6 cm
Intersecting Chords Theorem
AF. FB = (R+S).(R-S)
FB = (2+√2).(2-√2)/√6
FB = 2/√6 = 0,8165 cm
AB= AF +,FB = √6+2/√6
AB = 8/√6 = 3,266 cm ( Solved √ )
Extend EF horizontally and OF vertically and label the side of the square and the radius of the circle as a and R.
We have a= sqrt2.
Use chord theorem: AFxFB = sqa=2
and (R-sqrt2)(R+sqrt2) = sqa=2---->.sqR -2= 2---> R = 2
Use Pythagorean theorem AF= sqrt6-----> FB=2/sqrt6------> AB=AF+FB= sqrt6 + 2/sqrt6 = 8/sqrt6= 4sqrt6/3 cm
AB=(4√6)/3≈3,27 cm
I turned triangle AOB into an isosceles triangle with angles 35.26°, 35.26°, and 109.47°. Since r = 2 we can use the law of sines. sin(35.26°)/2 = sin(109.47°)/AB
i edge to your videos
you can use the formula √2*side of square to calculate the diagonal of square ........
Side of square,
and radius of circle:
A = S² = ½R² = 2 cm²
S = √2 cm
R = 2 cm
Similarity of triangles:
AB / 2R = R / √(R²+S²)
AB = 2R² / √(R²+S²)
AB = 8/√6 = 3,266 cm ( Solved √ )
Side length of square = root 2.
Then diagonal length of square = 2 (Pythagoras.)
This is the radius OE & also AO.
In triangle FAO, tan FAO = FO /AO.
Tan FAO = (root 2) / 2.
Tan FAO = 0.7071.
FAO = 35.264 degrees.
Joining point O to B makes iscoseles triangle AOB.
Dropping perpendicular from point O to AB at point P bisects it.
Then in triangle AOP, cos 35.264 = AP /OA.
0.8165 = AP / 2.
AP = 1.633.
AB = 2 x AP = 3.266.
With Triangle AOF we can get angle  with tangent = sqrt2/2, result 35.26° then we have right triangle ABC with hypothenuse AC = 4 and
cos  = AB/AC cos  = 0.816
So, AB = 4 cos Â, 4*0.816 = 3.264 cm (with hight précision 3.266)
Similar triangles:
AB/4 = 2/√6
AB = 8/√6 = 3.266...cm
🙂
AB= [r^2+r^2-2×r^2×Cos(
Side of square,
and radius of circle:
A = S² = ½R² = 2 cm²
S = √2 cm
R = 2 cm
tan α = S/ R = √2 / 2
α = 35,2644°
AB = 2R cos α
AB = 3,266 cm ( Solved √ )
Side of square,
and radius of circle:
A = S² = ½R² = 2 cm²
S = √2 cm
R = 2 cm
tan α = S / R = √2 / 2
α = 35,2644°
β = 180° - 2α
β = 109,4712°
Cosine rule:
AB² = 2R² (1-cosβ)
AB = 3,266 cm ( Solved √ )
[ 3:30 ] For △AOF
AO / OF = 2 / √2 = √2 = cotg(α)
⇒ OF = AO / √2
Similarly, for △ABC (~ △AOF)
AB / BC = √2 = cotg(α)
⇒ BC = AB / √2
Now if we apply the Pythagorean theorem for △ABC:
4² = AB² + (AB / √2)² =
= AB² + AB² / 2 = 3·AB² / 2
⇒ (3 / 2)·AB² = 16
⇒ AB² = 16·2 / 3
⇒ AB = 4·√2 / √3 = (4 / 3)√6
Generalized: AB = 4/3 √(3K) where K is the area of the square.
Note that the radius is sqrt(2(sqrt(2))^2)=2, then consider the perpendicular from the center to the chord, its length is sqrt(2)/sqrt(6) × 2=(2/3)sqrt(3) therefore the answer is2× sqrt(4-4/9×3) =2×sqrt(4-4/3)=2×sqrt(8/3)=4×sqrt(2/3)=(4/3)sqrt(6) =3.266 approximately. 😊😊😊
Excellent!
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍
how you calculated the length of the perpendicular
pls explain
@@adarshlignite9254 consider similar triangles.
AB=√6+√6/3=(4/3)√6
AB=4√6/3
Excellent!
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍
Let's find the solution:
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A(ODEF) = 2cm²
⇒ OD = DE = EF = FO = (√2)cm
⇒ R = OE = √2*(√2)cm = 2cm
Let's assume that O is the origin of the coordinate system and that AC is located on the x-axis. Then we can conclude from the intercept theorem:
y(B)/(x(B) + R) = OF/OA = OF/R = (√2)cm/(2cm) = 1/√2
⇒ y(B) = (x(B) + R)/√2
⇒ y²(B) = (x(B) + R)²/2
Since the point B is located on the circle, we know:
x²(B) + y²(B) = R²
y²(B) = R² − x²(B)
(x(B) + R)²/2 = R² − x²(B)
(x(B) + R)² = 2R² − 2x²(B)
x²(B) + 2Rx(B) + R² = 2R² − 2x²(B)
3x²(B) + 2Rx(B) − R² = 0
x(B) = (−2R ± √((2R)² − 4*3*(−R²)))/(2*3)
x(B) = (−2R ± √(4R² + 12R²))/6
x(B) = (−2R ± √(16R²))/6
x(B) = (−2R ± 4R)/6
The expression (−2R−4R)/6 leads to x(B)=−R=x(A). Therefore the correct value is:
x(B) = (−2R + 4R)/6 = 2R/6 = R/3
Now we use the intercept theorem again together with the Pythagorean theorem to calculate the length of the chord AB:
AF² = OA² + OF² = R² + OF² = 4cm² + 2cm² = 6cm²
⇒ AF = (√6)cm
AB/AF = (x(B) + R)/R = (R/3 + R)/R = (4R/3)/R = 4/3
⇒ AB = (4/3)AF = (4√6/3)cm ≈ 3.266cm
Best regards from Germany
Perfect