Awesme 🎉

@PreMath

11 ай бұрын

So nice of you, dear Thank you! Cheers! 😀

There is a general formula for the radius of a circle inscribed in a triangle: _r = √((s − a)(s − b)(s − c) / s)_ where _s_ is the semiperimeter and _a, b_ and _c_ are the sides of the triangle. In the special case of a right triangle, this formula reduces to _½ (a + b − c)_ where _c_ is the hypotenuse. Another formula for the special case of a right triangle is _r = ab/(a + b + c)_ where _c_ is the hypotenuse.

Nice to discover I can still do this question without paper, pen, or calculator.

Sweet, I figured it out thanks to all of the previous posts done by PreMath. Getting very confident with this, thanks! I like doing a couple or three daily to keep mentally fit...

Thanks for video.Good luck sir!!!!!!!!!!!

using pythg theorem, we can find CB = 51 Then we can find that CD = CE = 34 AD = 51 and EB = 17 drop a line from D perpendicular to AB , let the point of intersection be G DG and AG can be calculated using similar triangles. Area of triangle ADG can then be found Trapezium DEBG 's area can be found. add the areas to find answer.

@soli9mana-soli4953

11 ай бұрын

Once known AG as you suggested, doing AB - AG we get the height of DEC triangle to calculate its area and then subtract it from the area of ABC

Very nice.thanks

Fabulous work!🥂👍

@PreMath

11 ай бұрын

Thanks for the visit So nice of you, my dear friend

👍 great

Wow!!I didn't expect to find the same result with all these calculations but in the end i was correct!

Excellent! Thanks!

@PreMath

11 ай бұрын

Excellent! Thank you! Cheers! 😀

I got this one by (mostly) using your method, but with some variation near the end.

Thanks

@PreMath

11 ай бұрын

You are very welcome! So nice of you, dear Thank you! Cheers! 😀

Excellent !🙏🙏🙏

@PreMath

11 ай бұрын

Thank you 🙌

Good refresher.

@PreMath

11 ай бұрын

Excellent! Thank you! Cheers! 😀

Yay! I solved the problem.

i learn more theorems in this videos thanks

@PreMath

11 ай бұрын

Great! You are awesome. Keep it up 👍

Nice mixture of calculations, great example

@PreMath

11 ай бұрын

Super! You are awesome. Keep it up 👍

You are awesome, thx!

@PreMath

11 ай бұрын

You're welcome!

The harmony of congruence is music too my ears! 🙂

@PreMath

11 ай бұрын

Excellent! Thank you! Cheers! 😀

Wow awesome

@PreMath

11 ай бұрын

Many many thanks

Save a step by letting x be length CD or equal (by two tangent theorem) length CE. To avoid the use of the trigonometric sin() function, drop a perpendicular from point D to CB and call the intersection point G. Using CE as the base of ΔCDE, length DG is the height, call it h. ΔCDG and ΔCAB are similar. Therefore, CD/CA=DG/AB, so 34/85 =h/68 and h=27.2. So area ΔCDE = (1/2)(27.2)(34) = 462.4 As PreMath did, wrap up by subtracting 462.4 from 1734 to get the yellow area, 1271.6.

Why not look for LCD of hypotenuse and side, 17, to recognize this as a 3•4•5 right triangle? Ooh, I commented too soon. Thanks for all your thorough videos.

I used BE as x as well, but I found the ratio between AD and CD, and CE and BE to find out the area instead of using trigonometry.

@PreMath

11 ай бұрын

Many approaches are possible to find the solution to this problem! Thanks for sharing! Cheers!

@naaz5395

11 ай бұрын

How... Can you explain

@privateinformation4401

11 ай бұрын

​@@naaz5395 after finding BE, as shown in the video. I found CE, CD, and AD. then, I drew the line BD. Now, we have three triangles, and we can calculate the ratio between their area because ABD and CBD are both lying on AC, therefore they have the same height, and we can do the same for BDE and CDE because they are both lying on BC.

😊🙏

Height of triangle h² = 85² - 68² h = 51 cm Radius of inscribed circle: (68-r)+(51-r)=85 2r = 68+51-85 r= 17 cm Side of isosceles triangle: s = 51 - 17 = 34 cm Area of isosceles triangle: A = ½ s² sin α A = ½ 34² . 68 / 85 A = 462,40 cm² Area of bigger right triangle: A = ½ b. h A = ½ 68 . 51 A = 1734 cm² Yellow Shaded Area : A = A₁ - A₂ A = 1734 - 462,4 A = 1271,6 cm² ( Solved √ )

By pythagorean theorem, the height of the triangle is 51, then the area of the 0:03 triangle is 34x51, it is also equal to r(85+68+51)/2=102r, where r is the radius, so r=34x51/102=17 hence the area of the rhombus is 17x34, yhe length of two diagonals of the rhombus are 17xroot 5 and 17x34/17xroot 5=34/root 5, then the height of the white triangle is sqrt(4x17^2-17^2/5)=17 sqrt(19/5), therefore the answer is 68x51/2-34/root 5x17sqrt(19/5)/2=68x51/2-34x17xroot(19)/5=1230 approximately 😅.

A fork within the solution radius of a circle inscribed in a triangle (r=x in terms of the Teacher) r = 2*Aabc/(AB+BC+CB) = 2*1734/(86+68+51) =17

@ybodoN

11 ай бұрын

To be exact, this formula applies to circles inscribed in a *right* triangle.

Risulta r=17...A=(289/2)(5+3/rad10)... Ho rifatto i calcoli A=289*22/5

Yellow area = 1;276 and blue = 462.4 This is a 3-4-5 triangle scaled up by 17; hence the 3 side is 51 (17*3) Finding the radius of the circle Let r= the radius Line DA = 68-r = FA tangent circle theorem Line DC = 51-r tangent circle theorem Equation 1 Hence the hypotenuse or AC = (68- r + 51-r) = 119 -2r Since Line DA= 68-r an line AC=85, then line DC also = 85 - (68-r) = 17-r; hence line CE = 17- r tangent line theorem Since line CB = CE + EB (and EB=r) , then line CB= 17 + r +r = 17 + 2r Using these values: 119- 2r, 17+ 2r and 68, and Pythagorean theorem will give the value for r (119- 2r)^2 = (17+ 2r)^2 + 68^2 14,161 + 4r^2 - 476r = 289 +68r+4r^2 + 4,624 14,161-289-4624=476r + 68 r (not the 4r^2 canceled out) 9,248 = 544r 9,248/544= r r= 17 The radius of the circle = 17 since DC = 51- r then DC= 51-17 = 34 note that DC= EC (tangent circle theorem) Angle C = 53.13 degrees since this is a 3-4-5 triangle scaled up by 17, and the angles of 3-4-5 (right triangle) are 90, 53.13, and 36.87 Since angles D and E are congruent, then D or E = (180-53.13)/2 =126.87/2 = 63.435 Using 53.13 degrees, 34, and 63.435 degrees and the law of sine DE= 30.4 Using 30.4, 34, and 34 and Heron's formula yields 462.4 for the area of blue (a triangle) Since the area of the Large triangle is 68 * 51* 1/2 or 34* 51 = 1,734 the area of yellow is 1,734 - 462.4 = 1,276

BC = 51. Let BF = BE = X. 68 - X + 51 - X = 85. X = 17. EC = 34, Angel C = 53.13 Area CDE = 34*34*sin 53.13*0.5 = 462.4 Yellow Area = Area ABC - 450 = 1734 - 462.4 = 1271.6

2 good

R=(AB+BC-AC)÷2=(68+51-85):2=44:2=22. 😀😉

6358/5

My solution: - Found BC using Pythagoras, which is 51 - ABC's area, which is 1,734 - Circle's radius, using Area = Radius x Semiperimeter. Radius equals 17 - CE by subtracting 17 from 51, which equals CD and 34 - With two similar triangles using the circle's center and a line from center to C, found DE and CDE's height, in order to find CDE's area, which is 2312/5 - Finally, to find yellow area, subtracted 2312/5 (CDE's area) from 1,734 (ABC's area), which equals 6358/5 or 1,271,6

DC = DE = x and FB = EB = r 68-r = 85-x and so x = r+17 85² = 68²+(2r+17)² 85² = 68²+4r²+17²+68r 7225 = 4624+4r²+289+68r 2312 = 4r²+68r 4r²+68r-2312 = 0 r = [-68+sqrt(68²-4(4)(-2312))]/8 r = [-68+204]/8 = 17 and x = 34 So the triangle has side lengths of 68, 51, and 85 Angle C = sin-1(68/85) = sin-1(4/5) = 53.13° DE = 2xsin(53.13/2) = 30.41 Blue area = (30.41)(xcos(53.13/2)/2 = 462.4 Yellow area = (1/2)(68)(51)-462.4 = 1271.6

1156 square units

Below 6.2k, Learn how to find the area of the blue. However, since the area of the triangle is 34 * 51= 1734 if the area of the blue is 462.4 then the area of the yellow is 1735 -462.4 = 1271.6

X=r =17

17*4=68 y 17*5=85 → BC=17*3=51 → 3*4/2=6=(4-r)r+r²+(3-r)r→r²-7r+6=0→r=1 → AB=3+1→BC=1+2→CA=2+3 →CD=CE=2→EB=BF=r=1→AF=AD=3 → Si "G" es la proyección vertical de "D" sobre AB→DG/3=3/5→DG=9/5 → Si "O" es el centro del círculo→La altura "h" del triángulo DOE es h=DG-r=(9/5)-1=4/5 → (Área amarilla)/17² =ADOF+OEBF+DOE=(3*1)+(1*1)+[1*(4/5)*(1/2)]=3+1+(4/10)=44/10 → Área amarilla =(44/10)*17²=1271.60 Ud² Interesante problema. Gracias y saludos.

I afraid you calculated sine of wrong angle.

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