AD is 25, by pythagorean theorem, so DB is 25 in an isosceles triangle, then CB is 25+7=32, and thus AB is 40, by pythagorean theorem again, now EB is one half of AB, is 40/2=20, therefore h=15, also applying pythagorean theorem. 😃
I solved it, (probably the same way most others did), by connecting A and D, making that the hypotenuse of the triangle ACD, solving for the length AD, (which is also DB), which gives us BC, from which you can solve for AB, half of AB is BE, then you have all the values to “plug in to” the Pythagorean theorem.
Setting AE=EB=x and DB=y, I constructed two equations, the first by applying the Pythagorean theorem to the triangle ABC (24)^2+(7+y)^2=(2x)^2 I found the second equation by applying the proportions to the two similar triangles ABC and EBD 2x:y=(y+7):x hence 2x^2=y^2+7y whose solutions are x=20 and y=25 Finally, to find h it is enough to apply the Pythagorean theorem again to the triangle EBD, the sides EB and DB being known
ABC is similar to DBE so once you know that ABC is in the proportions of a 3,4,5 triangle (it's 8x3, 8x4, 8x5) you know that DBE must also be in the proportions of a 3,4,5 triangle. You know that the hypotenuse is 5x5 = 25 so that leaves the other sides being 20 and 15. From similarity h must be the shortest side so that's h = 15.
Excellent! Very memorable. I need to do a better job in remembering as you show here that one can actually MANIPULATE figures and shapes to solve them, rather than being just stuck with what is given to try to deductively figure it out. Thank you, PreMath!
This is how math should be taught. Geometry is a lot like poetry, when you're watching someone solve it. If I ever taught an English Class, I'd pull up one of these types of videos, and teach them to apply the same skills in this, to their poems. Nobody thinks like that, though, but that's also why we have a lot of people who can't think. They always want words to have ambiguity, but a good reader will follow the logic of the poem's words on the page, the same way they'd follow this guy's presentation. Like, poetry is exactly like this video. You're being walked step by step to the answer. And each verse breaks the logic down into simpler bits. Good poetry, anyway is like that.
Find AD by Pythagorean theorem , AD = 25. Triangles ADE and BDE are congruent. Hence DB =25. BC =32. AB can be found by Pythagorean theorem, AB= 40. EB=40/2=20 DE can be found by Pythagorean theorem,DE= 15
AD=DB, because triangles ADE and EBD are equal. AD^2=24^2+7^2=576+49=625 AD=DB=25 so CB=32. In triangle ABC AB^2= 24^2+32^2= 576+1024=1600 AB=40 , so DB=40/2=20 . In triangle DBE h^2= 25^2- 20^2= 625 - 400 = 225 h=V225=15
More than one hour, I figured it out, h = 15! 😓 I was connecting line CE = AE = BE = it's a Radius!* (Fail) Then, I connect AD, voila! Much-much Easier! It is Triple Phytagoras!!! Great Maths 🙂 👍
An in-the-head quickie if one recognizes two familiar Pythagorean triples. Spoiler alert. An auxiliary line segment can be imagined linking points A and D. We can immediately recognize a 7-24-25 triangle in the top left, so AD = 25 units. Triangles ADE and BDE are congruent, so DB must also be 25 units. CB = 7 + 25 = 32 units. 24 is three-quarters of 32, so ABC is a scaled 3-4-5 triangle. Therefore, AED is also a scaled 3-4-5 triangle, so: h = 25 (3 / 5) = 15 units. Edited to correct a small typo.
I am terrible in geometry at knowing where to draw additional line. I used pure algebra. Let us call the length of DB = x and the length of EB = y (so AB = 2y). We have two right triangles, ABC and DBE. Because B is present in both of them, they are similar by AAA. The lengths of the sides of these two triangles (SL;LL;H) are 24;7+x;2y and h;y;x. Therefore, comparing the ratio of the long legs and hypotenuses, we get y:7+x=x:2y. When we multiply this, we get 2y^2 = (7+x)*x = x^2 + 7x. By the Pythagorean theorem, 24^2 + (7 + x)^2 = (2y)^2 576 + 49 + 14x + x^2 = 4y^2 If we multiply the first equation by 2 we get 4y^2 = 2x^2 + 14x. Therefore, 625 + 14x + x^2 = 2x^2 + 14x. Substracting x^2 + 14x from both sides, we get 625 = x^2, or x = 25. So now we know that the sides of ABC are 24; 7 + 25 = 32; 2y. Using the Pythagorean formula we get 576 + 1024 = 4y^2 1600 = 4y^2 400 = y^2 20 = y Now we can either use the Pythagorean formula again to get x^2 - y^2 = x^2 25^2 - 20^2 = h^2 625 - 400 = h^2 225 = h^2 15 = h Or we can use the ratios of sides, either 24:h = 7+x:y 24y = (7 + x)h 24 * 20 = 32h 480 = 32h 480/32 = h 15 = h or 24:h = 2y:x 24x = 2y * h 24 * 25 = 2 * 20 * h 600 = 40h 600/40 = h 15 = h
I did the last step without pythagorean theorem: There are 2 ways to calculate the area of the triangle. (24/2*32) = (40/2*h) + (24/2*7) 20*h = 12(32-7) 20*h = 12*25 20*h = 300 h =15
Yeah, nice. I did it another way, by knowing that if two sides of ABD are 24 and 32 that's 8x3 = 24, 8x4 = 32. It has to be in proportion to a 3,4,5 triangle since it is a right-angled triangle and the two shortest sides are in a 3:4 ratio. That means the hypotenuse of ABC *_must_* be 8x5 = 40. ABC is also similar to DBE so once you know that ABC is in the proportions of a 3,4,5 triangle you know that DBE must also be in the proportions of a 3,4,5 triangle. You know that the hypotenuse is 5x5 = 25 so that leaves the other sides being 4x5 = 20 and 3x5 = 15. From similarity of ABC with DBE, the length h must be the shortest side so that's h = 15. It's using 3,4,5 triangles and similarity arguments to get h=15. You only really need the Pythagorean theorem once, unless you happen to know that a 7,24,25 triangle is also a right-angled triangle in which case you would never need the Pythagorean theorem at all, at least not explicitly.
A slightly different approach: integer Pythagorean triangles abound in this one. We could calculate that AD = 25 by adding 7^2 + 24^2 = 625 = 25^2; or we could just remember that 7-24-25 is an integer Pythagorean triangle. This gives us 25 for DB and 25 + 7 = 32 for CB. Hey, maybe we're onto something here.... Similarly, the big triangle ABC is another integer Pythagorean, 24-32-40, so the base AB is 40 and half the base AE (or EB) is 40/2 = 20. And finally, triangle BDE (or ADE) is a 3-4-5 triangle multiplied by 5, so it's 15-20-25; and h = 15. No peeking, no calculators; just my table of integer-sided right triangles. Coraggio. 🤠
let's assume that AE=BE=x △DEB and △ACB triangles are similar We make proportions: EB/BC = CE/AB ⇒ x/BC = (BC - 7)/2x ⇒ 2x² = BC² - 7BC (1) According to the Pythagorean theorem BC² = AB² - AC² = 4x² - 24² (2) Substitute (2) into (1) 2x² = 4x² - 24² -7√(4x² - 24²) 7√(4x² - 24²) = 2x² - 24² We get the biquadratic equation 4x⁴ - 2500x² + 360000 = 0 x₁² = 400; x₂² = 225 x₁ = 20; x₂ = 15 1)Consider the first option: x = 20 AC = 40; BC = √(40² - 24²) = 32; EC = 25 h = √(25² - 20²) = 15 2) Consider the second option: x = 15 AC = 30; BC = √(30² - 24²) = 18; EC = 11 EC < EB the hypotenuse is smaller than the leg. This is impossible. Answer: h = 15
I went trig route. Probably made it harder for myself: CosB=(7+Y)/2X=X/Y 2X^2=Y(7+Y) Mr Pythagoras enters the chat: (7+Y)^2+(24)^2=(2X)^2 =4X^2 =2(2X^2) [Now substitute:] ...=2Y(7+Y) Drop the Algebra hammer: Y^2+14Y+49+576=2Y^2+14Y Combine: Y^2-625=0 Y^2=625 Y=25 Solve for X: 2X^2=7(25)+(25)^2 2X^2=625+175=800 X^2=400 X=20 SinB=24/2X=h/Y h=24Y/2X=24(25)/2(20) h=15
Why is the product of a square root always positive or negative? I've heard you say that hundreds of times but I cannot fathom a reason for that being so.
From problem above: Sqrt[625]=+/-25 (+25)^2=(-25)^2=625 Typically its easier to just think in terms of absolute value, thus positive, but technically it IS [Plus or Minus] Hope that helps.
If I only had recognized the first pythagorean triangle, I would have solved it in 20 seconds. Now it probably took two minutes for my slow brain to calculate the first hypotenuse in my head. Still a bit proud I didn't need pen and paper...
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AD is 25, by pythagorean theorem, so DB is 25 in an isosceles triangle, then CB is 25+7=32, and thus AB is 40, by pythagorean theorem again, now EB is one half of AB, is 40/2=20, therefore h=15, also applying pythagorean theorem. 😃
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I did the same.
BD*AC=CE*AB
I solved it, (probably the same way most others did), by connecting A and D, making that the hypotenuse of the triangle ACD, solving for the length AD, (which is also DB), which gives us BC, from which you can solve for AB, half of AB is BE, then you have all the values to “plug in to” the Pythagorean theorem.
Setting AE=EB=x and DB=y, I constructed two equations, the first by applying the Pythagorean theorem to the triangle ABC
(24)^2+(7+y)^2=(2x)^2
I found the second equation by applying the proportions to the two similar triangles ABC and EBD
2x:y=(y+7):x hence 2x^2=y^2+7y
whose solutions are x=20 and y=25
Finally, to find h it is enough to apply the Pythagorean theorem again to the triangle EBD, the sides EB and DB being known
ABC is similar to DBE so once you know that ABC is in the proportions of a 3,4,5 triangle (it's 8x3, 8x4, 8x5) you know that DBE must also be in the proportions of a 3,4,5 triangle. You know that the hypotenuse is 5x5 = 25 so that leaves the other sides being 20 and 15. From similarity h must be the shortest side so that's h = 15.
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Excellent! Very memorable. I need to do a better job in remembering as you show here that one can actually MANIPULATE figures and shapes to solve them, rather than being just stuck with what is given to try to deductively figure it out. Thank you, PreMath!
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Thanks, it’s intersting. I used AREA to find out h value. Triangle ABD=300, AB=40, 40xhx1/2=300, 20h=300, h=15
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Thanks for the explanation, needs for me to see 100 times until I try by myself but I going to get it, Thanks again
This is how math should be taught. Geometry is a lot like poetry, when you're watching someone solve it. If I ever taught an English Class, I'd pull up one of these types of videos, and teach them to apply the same skills in this, to their poems. Nobody thinks like that, though, but that's also why we have a lot of people who can't think. They always want words to have ambiguity, but a good reader will follow the logic of the poem's words on the page, the same way they'd follow this guy's presentation. Like, poetry is exactly like this video. You're being walked step by step to the answer. And each verse breaks the logic down into simpler bits. Good poetry, anyway is like that.
Thanx your first problem I could work out in my head all the practice is paying off. 🎂
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Find AD by Pythagorean theorem , AD = 25.
Triangles ADE and BDE are congruent. Hence DB =25.
BC =32.
AB can be found by Pythagorean theorem, AB= 40.
EB=40/2=20
DE can be found by Pythagorean theorem,DE= 15
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Super easy. After seeing the Similarity Theorem, the problem became a piece of cake and solved it off the bat before I can finish the video.
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AD=DB, because triangles ADE and EBD are equal. AD^2=24^2+7^2=576+49=625 AD=DB=25
so CB=32. In triangle ABC AB^2= 24^2+32^2= 576+1024=1600 AB=40 , so DB=40/2=20 . In triangle DBE h^2= 25^2- 20^2= 625 - 400 = 225 h=V225=15
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Nice and awesome, many thanks, Sir!
sin(φ) = 3/5 = h/25 → h = DE = 15
Thanks for a great challenge which gave me much pleasure to solve.
More than one hour, I figured it out, h = 15! 😓
I was connecting line CE = AE = BE = it's a Radius!* (Fail)
Then, I connect AD, voila!
Much-much Easier!
It is Triple Phytagoras!!!
Great Maths 🙂 👍
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Yes!! I have solved this problem mentally.
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An in-the-head quickie if one recognizes two familiar Pythagorean triples.
Spoiler alert.
An auxiliary line segment can be imagined linking points A and D.
We can immediately recognize a 7-24-25 triangle in the top left, so AD = 25 units.
Triangles ADE and BDE are congruent, so DB must also be 25 units.
CB = 7 + 25 = 32 units.
24 is three-quarters of 32, so ABC is a scaled 3-4-5 triangle.
Therefore, AED is also a scaled 3-4-5 triangle, so:
h = 25 (3 / 5) = 15 units.
Edited to correct a small typo.
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I am terrible in geometry at knowing where to draw additional line. I used pure algebra.
Let us call the length of DB = x and the length of EB = y (so AB = 2y).
We have two right triangles, ABC and DBE. Because B is present in both of them, they are similar by AAA. The lengths of the sides of these two triangles (SL;LL;H) are 24;7+x;2y and h;y;x. Therefore, comparing the ratio of the long legs and hypotenuses, we get y:7+x=x:2y. When we multiply this, we get 2y^2 = (7+x)*x = x^2 + 7x.
By the Pythagorean theorem, 24^2 + (7 + x)^2 = (2y)^2
576 + 49 + 14x + x^2 = 4y^2
If we multiply the first equation by 2 we get 4y^2 = 2x^2 + 14x.
Therefore, 625 + 14x + x^2 = 2x^2 + 14x.
Substracting x^2 + 14x from both sides, we get 625 = x^2, or x = 25.
So now we know that the sides of ABC are 24; 7 + 25 = 32; 2y.
Using the Pythagorean formula we get 576 + 1024 = 4y^2
1600 = 4y^2
400 = y^2
20 = y
Now we can either use the Pythagorean formula again to get x^2 - y^2 = x^2
25^2 - 20^2 = h^2
625 - 400 = h^2
225 = h^2
15 = h
Or we can use the ratios of sides, either 24:h = 7+x:y
24y = (7 + x)h
24 * 20 = 32h
480 = 32h
480/32 = h
15 = h
or 24:h = 2y:x
24x = 2y * h
24 * 25 = 2 * 20 * h
600 = 40h
600/40 = h
15 = h
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Very nice explanation.
Excellent I got this one.
Great solution!.
Another great video👍
Thanks for sharing😊😊
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I just love this question. Keep up the good work sir.👍🏾
AD=DB 》DB^2=24^2+7^2 》DB=25 》CB=25+7=32 》AB^2=24^2+32^2 》AB=40 》h^2=25^2-(40/2)^2 》h=15
Gracias y un saludo.
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Excellent
BOM DIA PROFESSOR
Agradeço ao SR por ter aprendido geometria
DEUS lhe Abenço
Grato
So nice of you, Alex
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h/20=24/32=3/4=>h=20×3/4=15unit.
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@@PreMath thanks.
Tan BDE = h/20 =24/32 . So h =3/4 x 20 = 15 . Since triangle ABC is similar to triangle BDE !
I did the last step without pythagorean theorem:
There are 2 ways to calculate the area of the triangle.
(24/2*32) = (40/2*h) + (24/2*7)
20*h = 12(32-7)
20*h = 12*25
20*h = 300
h =15
Yeah, nice.
I did it another way, by knowing that if two sides of ABD are 24 and 32 that's 8x3 = 24, 8x4 = 32. It has to be in proportion to a 3,4,5 triangle since it is a right-angled triangle and the two shortest sides are in a 3:4 ratio. That means the hypotenuse of ABC *_must_* be 8x5 = 40.
ABC is also similar to DBE so once you know that ABC is in the proportions of a 3,4,5 triangle you know that DBE must also be in the proportions of a 3,4,5 triangle. You know that the hypotenuse is 5x5 = 25 so that leaves the other sides being 4x5 = 20 and 3x5 = 15. From similarity of ABC with DBE, the length h must be the shortest side so that's h = 15.
It's using 3,4,5 triangles and similarity arguments to get h=15. You only really need the Pythagorean theorem once, unless you happen to know that a 7,24,25 triangle is also a right-angled triangle in which case you would never need the Pythagorean theorem at all, at least not explicitly.
Thanks for video.Good luck sir!!!!!!!!!
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شكرا
يمكن استعمال cos c بصيغ مختلفة
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A slightly different approach: integer Pythagorean triangles abound in this one.
We could calculate that AD = 25 by adding 7^2 + 24^2 = 625 = 25^2; or we could just remember that 7-24-25 is an integer Pythagorean triangle. This gives us 25 for DB and 25 + 7 = 32 for CB.
Hey, maybe we're onto something here....
Similarly, the big triangle ABC is another integer Pythagorean, 24-32-40, so the base AB is 40 and half the base AE (or EB) is 40/2 = 20.
And finally, triangle BDE (or ADE) is a 3-4-5 triangle multiplied by 5, so it's 15-20-25; and h = 15.
No peeking, no calculators; just my table of integer-sided right triangles.
Coraggio. 🤠
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@@PreMath Thanks. I'm in the USA too, and we need all the love and prayers we can get.
Sir I request you to upload some videos on arithmetic.
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Me having an English degree and not having anything to do with Geometry actually listening and enjoying this explanation
let's assume that AE=BE=x
△DEB and △ACB triangles are similar
We make proportions:
EB/BC = CE/AB ⇒ x/BC = (BC - 7)/2x ⇒
2x² = BC² - 7BC (1)
According to the Pythagorean theorem
BC² = AB² - AC² = 4x² - 24² (2)
Substitute (2) into (1)
2x² = 4x² - 24² -7√(4x² - 24²)
7√(4x² - 24²) = 2x² - 24²
We get the biquadratic equation
4x⁴ - 2500x² + 360000 = 0
x₁² = 400; x₂² = 225
x₁ = 20; x₂ = 15
1)Consider the first option: x = 20
AC = 40; BC = √(40² - 24²) = 32; EC = 25
h = √(25² - 20²) = 15
2) Consider the second option: x = 15
AC = 30; BC = √(30² - 24²) = 18; EC = 11
EC < EB the hypotenuse is smaller than the leg. This is impossible.
Answer: h = 15
I love this one just think out the box
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AEDC é um quadrilátero inscritível, correto?
Yay! I solved it.
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Great
You make my calculations fast
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I went trig route. Probably made it harder for myself:
CosB=(7+Y)/2X=X/Y
2X^2=Y(7+Y)
Mr Pythagoras enters the chat:
(7+Y)^2+(24)^2=(2X)^2
=4X^2
=2(2X^2)
[Now substitute:]
...=2Y(7+Y)
Drop the Algebra hammer:
Y^2+14Y+49+576=2Y^2+14Y
Combine:
Y^2-625=0
Y^2=625
Y=25
Solve for X:
2X^2=7(25)+(25)^2
2X^2=625+175=800
X^2=400
X=20
SinB=24/2X=h/Y
h=24Y/2X=24(25)/2(20)
h=15
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Suppose the triangle is Egyptian, then:
AB= (24/3)×5= 40;
BC= (24/3)×4= 32.
Triangle ∆BDE~∆ABC:
BE= AB/2= 40/2= 20;
BD=BC-CD=32-7=25;
DE=h=√(BD^2-BE^2)=
=√(625-400)=√225=15.
Examination.
Let's draw the middle line ∆ABC, the line EF.
BF=32/2=16;
EF=√(BE^2-BF^2)=
=√(20^2-16^2)= 12, i.e.
EF=AC/2=24/2=12.
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@@PreMath Thanks sir 😊
H=15,basta utilizzare la similitudine dei 2 triangoli rettangoli
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I solved this problem with just mental calculations
h=(25^2-20^^2)^1/2=15 unit.ans
how to do in case ED not in the middle of AB?
So easy
Why is the product of a square root always positive or negative? I've heard you say that hundreds of times but I cannot fathom a reason for that being so.
From problem above:
Sqrt[625]=+/-25
(+25)^2=(-25)^2=625
Typically its easier to just think in terms of absolute value, thus positive, but technically it IS [Plus or Minus]
Hope that helps.
Easy! (Again)
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Have we seen this before?😉 Still great example though 👍🏻
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If I only had recognized the first pythagorean triangle, I would have solved it in 20 seconds. Now it probably took two minutes for my slow brain to calculate the first hypotenuse in my head.
Still a bit proud I didn't need pen and paper...
15 if u no ur 3,4,5 sets.
edit:
24/3 = 8
8 x 5 = 40
40/2 = 20
20/4 = 5
5 x 3 = 15
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h=15 unit.ans
15 answer
h=15
15
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Fun problem. 25^2 - 20^2 = 225. √225 = 15.
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asnwer=10cm isit
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يتعذر الاشتراك ولا يوجد جرس-يتعذر التعليق-لا أعلم .
Bruh, I solved it the exact way he did
The famous 3-4-5 triangle....TWICE
利用中垂線性質和畢氏組數。
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Lots of 345 triangles is an easier solution, but not as robust
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Only thing you have to know for this problem is Pythagorean Theorem
h is 15
This solution in VietNam is for grade 6 and you must be solved between 5 to 8 minutes 😅
Only 1 line!)
If solved mathematically, it's ok; but when solved geometrically, it's completely wrong.
Intéressant
Bon.fin
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h=15
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15
h = 15
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h=15
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