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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Комментарии : 44

@harikatragadda 11 месяцев назад

Intersecting Chords theorem at F. √6*FB = (2+√2)*(2-√2) FB = 2/√6 AB = √6 + 2/√6 = 8/√6

@ghep74 11 месяцев назад

I solved it this way too! : )

@ybodoN 11 месяцев назад

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.

@quigonkenny 6 месяцев назад

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

@JLvatron 11 месяцев назад

Brilliant! Thanks!!

@tombufford136 7 месяцев назад

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

@abdulkadirbuyuksoy2076 11 месяцев назад

Perfect

@tayyabamustafanewchannel9564 11 месяцев назад

MashaAllah. Excellent work 👍. Very informative video

@PreMath 11 месяцев назад

Thanks for liking ❤️ 🌹

@santiagoarosam430 11 месяцев назад

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.

@williamwingo4740 11 месяцев назад

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! 🤠

@marioalb9726 11 месяцев назад

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 √ )

@hcgreier6037 11 месяцев назад

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.

@raya.pawley3563 11 месяцев назад

Thank you

@phungpham1725 11 месяцев назад

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

@AmirgabYT2185 8 месяцев назад

AB=(4√6)/3≈3,27 cm

@MathsMadeSimple101 11 месяцев назад

Technically the screen is a rectangular box, so we can’t think outside the box…

@jphilsol6459 11 месяцев назад

With Triangle AOF we can get angle Â with tangent = sqrt2/2, result 35.26° then we have right triangle ABC with hypothenuse AC = 4 and cos Â = AB/AC cos Â = 0.816 So, AB = 4 cos Â, 4*0.816 = 3.264 cm (with hight précision 3.266)

@johnbrennan3372 11 месяцев назад

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

@arnavkange1487 11 месяцев назад

you have a good sense of humor

@marioalb9726 11 месяцев назад

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 √ )

@bhalusingh123 11 месяцев назад

Easy one

@Ibrahimfamilyvlog2097l 11 месяцев назад

Good job sar❤👍

@JSSTyger 11 месяцев назад

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

@georgebliss964 11 месяцев назад

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.

@marioalb9726 11 месяцев назад

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 √ )

@marioalb9726 11 месяцев назад

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 √ )

@ashleydelarue8209 11 месяцев назад

i edge to your videos

@misterenter-iz7rz 11 месяцев назад

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. 😊😊😊

@PreMath 11 месяцев назад

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍

@adarshlignite9254 11 месяцев назад

how you calculated the length of the perpendicular pls explain

@misterenter-iz7rz 11 месяцев назад

@@adarshlignite9254 consider similar triangles.

@Dimitar_Stoyanov_359 11 месяцев назад

[ 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

@alexniklas8777 11 месяцев назад

AB= [r^2+r^2-2×r^2×Cos(

@wackojacko3962 11 месяцев назад

🙂

@arnavkange1487 11 месяцев назад

you can use the formula √2*side of square to calculate the diagonal of square ........

@elmer6123 11 месяцев назад

Similar triangles: AB/4 = 2/√6 AB = 8/√6 = 3.266...cm

@ybodoN 11 месяцев назад

Generalized: AB = 4/3 √(3K) where K is the area of the square.

@giuseppemalaguti435 11 месяцев назад

AB=√6+√6/3=(4/3)√6

@prossvay8744 11 месяцев назад

AB=4√6/3

@PreMath 11 месяцев назад

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍

@unknownidentity2846 11 месяцев назад

Let's find the solution: . .. ... .... ..... 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