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@PreMath

2 ай бұрын

Welcome aboard🌹 You are very welcome! Thanks dear ❤️

Since the rectangle is 21 × 28, its diagonal is 35, making two 7× (3, 4, 5) triangles. Thus, the radius of the inscribed circles is 7 and the line segment AB is 7√5 units.

@PreMath

2 ай бұрын

Excellent! Thanks ❤️

@SkinnerRobot

2 ай бұрын

That's how I solved it.

@marcgriselhubert3915

2 ай бұрын

@@SkinnerRobot That's the simplest.

As FC = DE = 21 = 3(7) and EF = CD = 28 = 4(7), ∆FCD and ∆DEF are 7:1 ratio 3-4-5 right triangles, and DF = 5(7) = 35. By observation, HC = CG = ME = EN = r. By two tangent theorem, GF = FP = ND = DQ = 21-r, and QF = FM = PD = DH = 28-r. FD = FP + PD 35 = (21-r) + (28-r) 35 = 49 - 2r 2r = 14 r = 7 Draw AT and TB, where AT is parallel with CD and TB is parallel with DE. Triangle ∆ATB: TB² + AT² = AB² (21-2(7))² + (28-2(7))² = AB² AB² = 7² + 14² = 49 + 196 AB = √245 = 7√5 Bit of a nitpick, but TB in the video was determined by observation as equal to r, when it should have been determined as being 21-2r. Both are forrect in this one case, but in the instance of a somilar problem where the ratio of r to the height of the rectangle is not 3, but close, someone who has seen this explanation might also "eyeball" that inner triangle height to be r as well, rather than determining it from the distances from the top and bottom of the rectangle.

@PreMath

2 ай бұрын

Excellent! Thanks ❤️

Not difficult, but need a bit computation, 21-r+28-r=49-2r=35, then r=7, then AB=sqrt(7^2+14^2)=7sqrt(5).😊

@PreMath

2 ай бұрын

Excellent! Thanks ❤️

Alternative method of computing r: Construct AF and AP. ΔAGF and ΔAPF are congruent by side-side-side (AG=AP = r, GF=PF by 2 tangents theorem, AF common). Therefore,

@PreMath

2 ай бұрын

Excellent! Thanks ❤️

I - Find the Length of the Diagonal DF = D. a) 21^2 + 28^2 = D^2 ; 441 + 784 = D^2 ; D^2 = 1.225 ; D = sqrt(1.225) ; D = 35 II - Find the Area of Rectangle [CDEF] and the Area of the equal Triangles [CDF] and [DEF] a) Area of Rectangle [CDEF] = A = 21 * 28 ; A = 588 b) Area of Triangle [CDF] ; A = 588 / 2 ; A = 294 III - Find the Length of Radius (R) a) 21R/2 + 28R/2 + 35R/2 = 294 ; 84R/2 = 294 ; 84R = 294 * 2 ; 84R = 588 ; R = 588/84 ; R = 7 IV - Finding the Distance AB. NOTE: If we draw a Vertical Line dividing the Rectangle [CDEF] in two Equal Rectangles with Sides 21 and 14, one can see that this Line is Tangent to both Circles. And if we draw two Horizontal Lines passing the Centre of each Circles; point A and B, one can see that Line passing in Centre B is Tangent to the top of the other Circle, and the Line passing in Centre A is Tangent to the bottom of the other Circle. So: a) AB^2 = 7^2 + 14^2 ; AB^2 = 49 + 196 ; AB^2 = 245 ; AB = sqrt(245) ; AB = 7*sqrt(5) lin un V - Answer The Distance AB is equal to 7sqrt(5) Linear Units or approx. equal to 15,6525 Linear Units

@PreMath

2 ай бұрын

Excellent! Thanks ❤️

√[21^2+28^2]=√1225=35 21-r+28-r=35 2r=14 r=7 21-7-7=7 28-7-7=14 AB=√[7^2+14^2]=√245=7√5

@PreMath

2 ай бұрын

Excellent! Thanks ❤️

1/Let R be the radius of inscribed circle Because the FC=3x7, and CD=4x7 so FCD and FED are 3-4-5 triple so, to make the calculating simpler just consider them as 3-4-5 ones of which the radius r=R/7--->r= 1 ----> AT=2 and BT=1-----> sq AB= 4+1=5---> AB= sqrt5 2/ Answer: The distance AB in the given problem= 7sqrt5

@PreMath

2 ай бұрын

Excellent! Thanks ❤️

Radius of the yellow in circles of the two identical right triangle halves of the outer rectangle = 7 units, which can be calculated by the formula for the inradius of a triangle: 2*area/perimeter Then AB is the hypotenuse of a right triangle whose sides are 21 - 2*7 = 7 and 28 - 2*7 = 14. Hence AB length = 7√5 by Pythagoras theorem

@PreMath

2 ай бұрын

Excellent! Thanks ❤️

@rabotaakk-nw9nm

2 ай бұрын

r=2A/P ??? 🥵 r=(a+b-c)/2 !!! 😊

At 6:51, when you extend GA to point T on vertical line through B, it turns out that BT is a radius, but this will not be true for all rectangles. T could be above or below bottom of circle. We really should go through the process of calculating BT in the same way we calculate AT. If S is point directly below T on line CD, then TS = AH = r, therefore BT = MS - MB - TS = 21-2r = 21-14 = 7. In this case, BT is indeed equal to r, but this will not always be the case.

AB=√14^2+7^2=7√5 units=15.65 units. Thanks Sir.❤❤❤

@PreMath

2 ай бұрын

Excellent! You are very welcome! Thanks ❤️

@ybodoN

2 ай бұрын

Without the brackets √14^2+7^2 = 63 🧐

Awesome! 🙂

@PreMath

2 ай бұрын

Thank you! Cheers!❤️

... Happy Friday, My strategy ... I AB I = SQRT[ (28 - 2r)^2 + (21 - 2r)^2 ] ... so, we want to find the value of " r " ... Angle(GFP) = ARCTAN(4/3) deg. , so Angle(GFA) = 1/2 * ARCTAN(4/3) deg. & Triangle(FGA) is a right triangle, where I GA I = r and I FG I = 21 - r ... TAN(Angle(GFA) = I GA I / I FG I ... TAN(1/2 * ARCTAN(4/3)) = r / (21 - r) = 1/2 ... after a few basic algebraic steps .... r = 7 u. ... finally I AB I = SQRT[ (28 - 14)^2 + (21 - 14)^2 ] = SQRT(245) = 7*SQRT(5) u. ... thanking you as always for posting great geometric exercises together with clear and instructive strategy video's .... best regards, Jan-W

@PreMath

2 ай бұрын

Excellent! Keep rocking 🌹 Thanks ❤️

My Way to find radius: CDF is a triangle, inner circle: DF = 35, 2s = 21+28+35 --> s = 42 A = r•s r = A ÷ s r = (1/2)•21•28 ÷ 42 = 7 The Rest is the Same^ Good Problem, never find this problem at my old school^.👍 29/03/14_7am_North Sumatra

Find the length of the diagonal DF. (21, 28, d) (7 * 3, 7 * 4, 7 * 5) d = 7 * 5 = 35 Draw radii AG, AH, AP, BM, BN, & BQ. By the Circle Theorem, ∠AGC, ∠AHC, ∠APF, ∠BME, ∠BNE, & ∠BQD are right angles. This forms squares AGCH & BMEN with side length r. Then, FG = 21 - r & DH = 28 - r. By the Two-Tangent Theorem, FP = 21 - r & DP = 28 - r. The aforementioned segments form the diagonal DF. (21 - r) + (28 - r) = 35 49 - 2r = 35 2r + 35 = 49 2r = 14 r = 7 The radius of the identical circles is 7 u. Extend segment GA to a point J on segment DE and segment MB to a point K on segment CD. They should be perpendicular to the sides of rectangle CDEF, to form two smaller rectangles that overlap each other. Label the intersection of segments GJ & KM as L. ∠ALB is a right angle by the Perpendicular Transversal Theorem. Since AG = JL = 7, AL = 14. And since BM = KL = 7, BL = 7. You could use the side lengths of rectangle CDEF. Apply the Pythagorean Theorem on △ALB. a² + b² = c² 14² + 7² = (AB)² 196 + 49 = (AB)² (AB)² = 245 AB = √245 = √(5 * 7 * 7) = 7√5 So, the length AB = 7√5 units (exact), or about 15.65 units (approximation).

@PreMath

2 ай бұрын

Excellent! Thanks ❤️

(21-r)r+(28-r)r+r²=21*28/2→ r=7 → AB=√(28-2r)²+(21-2r)² =7√5 = 15,6524. Gracias por el vídeo. Un saludo cordial.

@PreMath

2 ай бұрын

Excellent! You are very welcome! Thanks ❤️

Thank you!

@PreMath

2 ай бұрын

You are very welcome! Thanks ❤️

Thanks

Good one. I used little PA(Origin) right triangles once I found FD and r so it became 2*sqrt((3.5 sq)+ (7 sq)). Many thanks!

@PreMath

2 ай бұрын

You are very welcome! Thanks ❤️

What about the heron formula? Could not be easier to find the radius?

AB=7√5≈15,68

@PreMath

2 ай бұрын

Excellent! Thanks ❤️

How do you know the circles have the same area? I really enjoy your channel! Thank you.

@MarieAnne.

Ай бұрын

The diagonal of any rectangle divides the rectangle into 2 congruent right triangles, so any circles inscribed by the 3 sides of these 2 triangles must also be congruent.

@erich1843

Ай бұрын

Thanks!@@MarieAnne.

أول ما ترى شكل بهذا شيء مكون من مثلثات دوائر جواب يكون واضح ستشتغل على مثلثات وشعاع حيث سؤال مبني على شعاع دائرتين....

Can also use heron’s theorem to find r.

@ybodoN

2 ай бұрын

Of course, the general formula always works. But since we are dealing with right triangles, we can use (a + b − c) / 2 or ab / (a + b + c).

@PreMath

2 ай бұрын

Many approaches are possible to find the solution to this problem! Thanks ❤️

If it is an identical circles that GT can find r 21/3=7.

@PreMath

2 ай бұрын

Thanks ❤️

@ilyashick3178

2 ай бұрын

Sure, and follow up !

How do you know that AT is parallel to CD?

@PreMath

2 ай бұрын

We are dealing with lines with right angles! Thanks ❤️

Almost

@PreMath

2 ай бұрын

Thanks ❤️

(28)^2=784 (21)^2=441 {784+441}=1225 3(30°)=90° 3(30°)=90° 3(30°)=90° 3(30°)=90° {90°+90°+90°+90°}=360° {1225-360°}=√865 √10^√8 5^√13 √5^√2 √2^√4 5^√13^1 √1^√1 √1^√2^2 5^√1^√1√ 2^1^2 √1^√1^2 5 25 (x+2x-5)

@PreMath

2 ай бұрын

Thanks ❤️

(28)^2=784 (21)^2=441 {784+441}=1225 3(30°)=90° 3(30°)=90° 3(30°)=90° 3(30°)=90° {90°+90°+90°+90°}=360° {1225-360°}=√865 √10^√8 5^√13 √5^√2 √2^√4 5^√13^1 √1^√1 √1^√2^2 5^√1^√1√ 2^1^2 √1^√1^2 5 25 (x+2x-5)

@PreMath

2 ай бұрын

Thanks ❤️