Oh, I love this kind of puzzle. ❤❤❤ and your videos make me an expert professional mathematician.

@PreMath

4 ай бұрын

That's awesome! Thanks ❤️

Your solutions are simple and clever! Thank you so much! Alternative approaches: 1/ Drop the height EH to AC. We have EH//=1/2BC so EH= 12; and H is the midpoint of AC. Consider the right triangle AED: sq EH=AH.HD----->144=AH.(25-AH) ---> Sq AH - 25 AH +144 = 0 ----> AH= 32/2----> AC= 32----> AB= 40 ( 3-4-5 triple)----->AE=20-----> X= 15 units (the triangle ADE is also a 3-4-5 triple). 2/The quadrilateral BCDE is cyclic so AE.AB=AD.AC----> sq AB/2=25.(25+DC) -----> sq AB= 50(25+DC) = sq 24 + sq (25+DC) ---->1250 +50DC -= 576+ sq25 + sqDC +50DC sq DC= 49---> DC= 7----> AC= 32 The rest is the same as above.

@PreMath

4 ай бұрын

Thanks ❤️

Excellent sum. How can you imagine such questions!

Let AE=BE=a In triangle ADE AD^2=DE^2+AE^2 x^2+a^^25^2 x^2+a^2=625 (1) Triangle ADE~ABC x/25=24/2a ax=300 (2) (1) and (2) So: x=15 ; x=20 rejected X=15 units. ❤❤❤ Thanks teacher.

@PreMath

4 ай бұрын

Excellent! You are very welcome! Thanks ❤️

@phungpham1725

4 ай бұрын

Very nice!

another great video…my biggest weakness on geometric puzzles is my failure to “add” to the diagrams via extensions etc. you have helped me to take a step back and for ways to “restate” the problem via mods. really appreciate you taking the time snd effort to priduce these.

Thank you!

@PreMath

4 ай бұрын

You are very welcome! Thanks ❤️

Hello! I went a different way: Using properties of similar triangles, I approached the solution by algebraic method. This brought me to the special case of a quartic equation, a bi-quadratic equation. Solving this turned out two solutions for x: X1=15 and x2=20. But plugging in x2=20 causes point C lying between A and D. That means, the resulting shape then will not be a triangle. So x=15 seems to be the only solution making sense. Thanks for the interesting video, best greetings!

Awesome❤

@PreMath

4 ай бұрын

Glad to hear that! Thanks ❤️

Once you have found the base length of 20, to me it is more obvious to once again to apply Pythagorean theorem to find X than using the proportion approach. X ^ 2 = (25 ^ 2) - (20 ^ 2) (X ^ 2) ^ .5 = (225 ^ .5) X = 15

x = 15 Let the hypotenuse = 2y and the remaining leg = n , then the triangles are similar then: 24 /2y = x/25 Hence, 24 * 25 = x* 2y 12 * 25 = xy 300 = xy 300/y =x Using Pythagorean Theorem and the sides 25, 300, and y 25^2 - (300/y)^2 = y^2 625 - 90,000/y^2 = y^2 625y^2 - 90,000 = y^4 (multiply both sides by y^2) let y^2 = n , then 625 n - 90,000 = n^2 n^2 - 625 n + 90,000 =0 (n- 400) ( n- 225) =0 n=400 and n= 225 Hence y^2 = 225 and 400 y = 15 and 20 hence y= 20 and x = 15 as this satisfies the triangle sides a 3-4-5 with sides 40, 32, and 24

@PreMath

4 ай бұрын

Excellent! Thanks ❤️

Construct a perpendicular to AB upward from B. Extend AC until it meets the perpendicular and label the intersection as point F. Note that ΔADB and ΔAFB are similar by angle - angle (common

@PreMath

4 ай бұрын

Thanks ❤️

Draw parallel line from E and use Thales theorem and Pythagoras to get x=15

I used similar approach. But once you have ab/2 you hypotenuse 25^2= 20^2+ x^2 rearrange faktorize if u want to do that in your head. You get 225^0.5 which is 15 225 =3*3*5*5 take the sqr root 15 again Sqr root of ((3^2)*(5^2)) The same as sqr (3^2)*sqr (5^2) Just in case younger audience does not "see" it. Continuous learning is the base of a happy life. Using it is the base of success.

@PreMath

4 ай бұрын

Thanks ❤️

@TheTiberius74

4 ай бұрын

Did not know about the triples. I'm 50 years old forgot a lot since school. Love your videos❤️​@@PreMath

@TheTiberius74

4 ай бұрын

All roads lead to Rome is a saying we have.

I did the same as you till the calculation of AB. Then I used the Pythagorean theorem again to find X. At first sight I was thinking "I could not solve this one", but the key to the solution is to find the length of BD. Once you know this length, you will find the solution using the Pythagorean theorem multiple times.

@PreMath

4 ай бұрын

Thanks ❤️

I did it the hard way with similar triangles, and had to use a substitution of z=x^2 to get a quadratic for z, then sqrt it.

@PreMath

4 ай бұрын

Thanks ❤️

△ACB ~ △AED. AE=√625-x^2. AB=2*√625-x^2. 25/x=(2*√625-x^2)/24. Squaring, solving the biquadrate equation and choosing the appropriate root.

@PreMath

4 ай бұрын

Thanks ❤️

Posto AE=a,le equazioni usate sono a:x=√((2a)^2-24^2):24(triangoli simili)...e a^2=25^2-x^2...x^2=(625-175)/2=225...x=15

@PreMath

4 ай бұрын

Excellent! Thanks ❤️

1st method: draw BD and use the Pythagorean theorem (or Pythagorean triples) three times to find that x = 15. 2nd method: similar triangles ⇒ x / 24 = 25 / (2√(25² − x²)) leading to a quadratic equation and x = 20 or x = 15.

@PreMath

4 ай бұрын

Thanks ❤️

15)

@PreMath

4 ай бұрын

Thanks ❤️

Draw BD. By observation, ∆BDA is an isosceles triangle, so BD = DA = 25. Triangle ∆BCD: a² + b² = c² CD² = 25² - 24² = 625 - 576 = 49 CD = √49 = 7 Triangle ∆BCA: a² + b² = c² AB² = 24² + (25+7)² = 576 + 32² AB² = 576 + 1024 = 1600 AB = √1600 = 40 Triangle ∆AED a² + b² = c² x² = 25² - (40/2)² = 625 - 400 x = √225 = 15

@PreMath

4 ай бұрын

Thanks ❤️

Basic Pythagorean alone like in method 1 worked

A large right-angled triangle in ratio (3,4,5) composed of three small right-angled triangles, two identical in ratio (3,4,5), one in (24,7,25).🎉🎉🎉

@PreMath

4 ай бұрын

Thanks ❤️

Let's use an adapted orthonormal: C(0;0) B(0;24) D(d;0) A(d+25;0) E((d+25)/2;12) VectorAB(d+25;-24) VectorDE(((d+25)/2)-d; 12) or VectorDE(((-d+25)/2;12) VectorAB and VectorDE are orthogonal, so their scalar product is 0, so we have: (d+25).((-d+25)/2) + (-24).(12) = 0 or (625 - d^2)/2 - 288 = 0 or d^2 - 625 = -576 giving that d^2 = 49 and that d = 7. Now we have D(7;0) and E(16;12) and Vector DE(9;12), so DE^2 = 9^2 = 12^2 = 81 + 144 = 225, and finally x = DE = sqrt(225) = 15.

@PreMath

4 ай бұрын

Thanks ❤️

Why is your explanation such a surprise to me?😅 Before you explain it, I tried to solve the problem first. But how do I solve it with the congruence method if one side is unknown? And, i give up😂 and play the video. Then, at 1:48. I realize to used phytagoras. Boom! 🎉 how could this matter be so easy? ❤

@PreMath

4 ай бұрын

Thanks ❤️

When I saw the triangle with right angle and a multiple of 5 the others sides couldn t be different from 20 and 15

@ybodoN

4 ай бұрын

A (3, 4, 5) triangle is a usual suspect. But we also have BCD which is a (7, 24, 25) Pythagorean triple as well...

@PreMath

4 ай бұрын

Thanks ❤️

(25+a)/24=x/b, 25+a=24x/b, b^2+x^2=25^2, b^2=25^2-x^2, 4b^2=(25+a)^2+24^2, 4b^2=(24x/b)^2+24^2, 4b^4=24^2x^2+24^2b^2, 4(25^2-x^2)^2=24^2x^2+24^2(25^2-x^2), 4(25^4-2×25^2 x^2+x^4)=24^2x^2+(24×25)^2-24^2x^2, 4×25^4-8×25^2x^2+4x^4=(24×25)^2, 4x^4-8×25^2x^2+25^2(4×25^2-24^2), ......😅😅😅😅😅😅

@PreMath

4 ай бұрын

Thanks ❤️

I ran out of ink and paper today. All I had too write on was a basketball with a Sharpie too solve this problem . Too say the least, Euclid's took on a whole new dimension. ...bizarre results! 🙂

@PreMath

4 ай бұрын

Thanks a lot ❤️

If the hypotenuse of the big triangle is 2y x^2 + y^2 = 25^2 The small and large triangles are similar 2y/25 = 24/x => y = 25(12)/x x^2 + (25^2)(12^2)/x^2 = 25^2 let x^2 = a a + (25^2)(12^2)/a = 25^2 a^2 - (25^2)a + (25^2)(12^2) = 0 a = ((25^2) +/- ✓(25^4) - 4(25^2)(12^2))/2 = ((25^2) +/- 25✓(25^2) - (24^2))/2 = ((25^2) +/- 25(7))/2 = 25(25 +/- 7)/2 = 25(16) or 25(9) = x^2 x = 5(4) or 5(3) = 20 or 15 x = 15, y = 20

@PreMath

4 ай бұрын

Thanks ❤️

@cyruschang1904

4 ай бұрын

@@PreMath Thank YOU. I had fun doing it 🙂

I have been out. Let's draw a Line between points B and D. This Line BD = AD = 25. The Triangle [ABD] is an Isosceles Triangle!! So, DC^2 = DB^2 - BC^2 DC^2 = 25^2 - 24^2 DC^2 = 625 - 576 DC^2 = 49 DC = 7 AC = AD + DC = 25 + 7 =32 Now! AB^2 = AC^2 + BC^2 AB^2 = 32^2 + 24^2 AB^2 = 1.024 + 576 AB^2 = 1.600 AB = 40 Final Fase: Area of Triangle [ABC] = 24 * 32 / 2 = 384 Area of Triangle [BCD] = 24 * 7 / 2 = 84 Area of Triangle [ADB] = 384 - 84 = 300 40x / 2 = 300 40x = 600 x = 600 / 40 x = 15 Answer: The line x = 15 linear units

Let's find x: . .. ... .... ..... Obviously the triangles ADE and BDE are congruent. Therefore we know that AD=BD. Since the triangle BCD is a right triangle, we can apply the Pythagorean theorem: BD² = BC² + CD² AD² = BC² + CD² 25² = 24² + CD² 625 = 576 + CD² 49 = CD² ⇒ CD = 7 Now we can apply the Pythagorean theorem again for the triangle ABC: AB² = AC² + BC² AB² = (AD + CD)² + BC² AB² = (25 + 7)² + 24² = 32² + 24² = (4*8)² + (3*8)² = (5*8)² ⇒ AB = 5*8 = 40 Finally we apply the Pythagorean theorem a third time to find x: AD² = AE² + DE² AD² = (AB/2)² + x² 25² = (40/2)² + x² = 20² + x² (5*5)² = (4*5)² + x² (3*5)² = x² ⇒ x = 15 Best regards from Germany

@PreMath

4 ай бұрын

Excellent! Thanks ❤️

asnwer=6cm isit

(25)^2=625 (24)^2:=576 (625+576)=1201 (1201°-180°)=√1021 √5^23√^7√1^2 3^√1 23 (x+2x-3)

@PreMath

4 ай бұрын

Thanks ❤️

Most easy sum ...I literally solved it orally

@PreMath

4 ай бұрын

Excellent! Thanks ❤️

15

There is a mistake at 4:22 minutes, you have a^2 + b^2 = c^2, you put a=24 and b=32, how can 24^2 be more than 32^2 as you show in the next equation.

@PreMath

4 ай бұрын

My apologies! Thanks ❤️

Your isosceles triangle's method is too smart, relative to me.😢

@PreMath

4 ай бұрын

Excellent! Thanks ❤️

x = 15. I cheated a bit. I presumed a 3-4-5 right triangle and it ended up working out.

@PreMath

4 ай бұрын

Thanks ❤️

X=15

sqrt(4(25^2-x^2)-24^2)/24=sqrt(25^2-x^2)/x, (1924-4x^2)/24^2=(25^2-x^2)/x^2, 1924x^2-4x^4=24^2×25^2-24^2x^2, x^4-625x^2+12^2×25^2=0, x^2=(625+-175)/2 =400 or 225, x=20 rejected, for 25 is too large, so x=15😅😅😅😅😅😅😅😅😅

@PreMath

4 ай бұрын

Thanks ❤️

I thought AC is 25 then I got X is (12/25)*(1201^0.5)🤣🤣🤣

Too eazy done in 5 min

@PreMath

4 ай бұрын

Thanks ❤️

@tplayz3454

4 ай бұрын

Thank u soooooooo much u made my day love ur content

this was the easiest in ur channel💔

@PreMath

4 ай бұрын

Thanks ❤️

(25)^2=625 (24)^2:=576 (625+576)=1201 (1201°-180°)=√1021 √5^23√^7√1^2 3^√1 23 (x+2x-3)

@PreMath

4 ай бұрын

Thanks ❤️