You are very welcome Kenneth! You are absolutely correct Geometry has too many theorems. Honestly, I get stumped once in a while. We are all lifelong learners! I'm glad you liked it! Take care my friend and all the best😃
That‘s a good one. Once you realize, that weckte dealing with multiples of the 3-4-5 Right triangle, which are all alike (and the 25 of the Hypothenuse of the ADF Triangle is the hint), it is clear that the first part of the length x is 20 and and second part 15 times 3/4 (or 11.25). (As explained in the video.) I did it in my head.
@@PreMathI don’t see that. It seems to me switching values of m & n still satisfy GMT. The only way I can see m=16/n=9 is right and m=9/n=16 is wrong is if we know area of yellow triangle is larger than the green one or that AF>DF or AE>DE. But we haven’t been told any of that. Are we assuming the diagram is close enough to scale so we can assume that from looking at the diagram? You didn’t make it clear from the start it’s not to scale like you usually do, I noticed.
So nice of you Umesh dear! You are awesome 👍 I'm glad you liked it! Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃
Dear sir: 12^2 = m(25-m) , this is a theorm. So we have two options for n. So we have AF. So with cos(alpha) [which is basically similar triangles] we have value of x.
ABC vertices right triangle with cathetus x (from left counterclockwise) Similarly AFD right triangle vertices with hypotenuse 25 E = intersection of cathetus 12 with hypotenuse 25 Triangle area AFD = 1/2 * 25 * 12 = 1/2 * AD * DF… 25 * 12 = AD * DF From the Pythagorean theorem ADsquared + DFsquared = 252 (1) or also (AD + DF) squared - 2AD * DF = 625 that is AD + DF = √ (600 + 625) = 35 (2) (1) can also be written in the form (AD-DF) squared + 2AD * DF = 625 hence AD-DF = 5 (3) making a system between (2) and (3) we obtain AD = 20 and DF = 15 applying Euclid's theorem to the AFC triangle, we have 252 = x * AD from which x = 625/20 = 125/4 again by Euclid's theorem applied to the AFC triangle, DFsquared = AD * (x-AD) = 20 * x - ADsquared = 225 + 400 = 20x, x = 625/20 = 125/4 which confirms the result already found
Nice question and clear explanation. However, it’s just as easy applying Pythagoras and using basic trigonometry ratios to find the other lengths, since the triangles are similar. Thank you.
AD = AE + ED. Let t = tan(FAE), then AE = 12/t, and (since angle DFE = angle FAE), ED = 12t, therefore 12/t + 12t = 25, 12t^2 - 25t + 12 = 0 = (4t-3)(3t-4) and t = 3/4 or 4/3. Assuming the drawing is approx to scale, the solution must be t = 3/4 (for the angle to be < pi/4). Therefore, AE = 16, ED = 9. FAE and FED are 3-4-5 triangles, AF = 4*5 = 20, FD = 3*5 = 15. Triangle CDF is similar, and t = FC/FD, giving FC = 15t = 15*3/4 = 45/4 = 11 1/4. Therefore x = AF + FC = 20 + 11 1/4 = 31 1/4.
Since triangle ADC also has angles α, 90, β, then it is similar to triangle AEF, so once you've found AF = 20, you could just use ratios to solve for AC directly: AC/AD = AF/AE → AC/25 = 20/16 → AC = x = 125/4 = 31.25
Note triangles ADC, AEF and FED are congruent via the angle angle angle theorem. The since 12/(25-ED) = ED/12, ED = 16 or 9. Since only ED = 9 makes sense we have all three triangles as 3-4-5 triangles. Since we know AD = 25, we have 4/5 = 25/x, so x=125/4
Why does only ED=9 make sense? No one said the diagram was drawn to scale. The only angles specified are the right angles. Therefore, there are 2 solutions and the author only gets 5/10 for a grade since he missed 50% of the answer.
Or like this: Due to the Geometric mean theorem in the triangle AFD h^2 = pq, so that 12^2 = 144 = p*q, with p+q = 25. That only works for p = 16 and q = 9. The intercept theorem gives 12/16 = CD/25, which results to CD = 18.75. Finally, the Pythagorean theorem leads to x^2 = 25^2 + 18.75^2, so that x = +sqrt(976.6525), which results in x = 31.25
once you did your geometric mean theorem, you could have also then switched to using trigonometric functions taking the inverse tangent of the AEF triangle to get the angle alpha and then using the cosine equation to solve for the X of triangle ACD.
FD =15 and AF 20 (3-4-5 triangle since AD=25) AE=16 from 20^2 - 12^2= AE^2 = 400-144 = sqrt 256 =16 for CD 16/12 =25/CD 16CD= 25 x 12 (cross multiply) 16CD= 300 CD= 18.75 Since we looking AC or x then x^2 = 25^2 + 18.75^2 x = sqrt 976.5625 x=31.25 11:18
Because h x h = m x n, therefore: (h/m) = (n/h) or (h/n) = (m/h). In both cases you get via cross-multiplying 144 = 9 x 16 or 144 = 16 x 9. It doesn't matter 🙂
Hmmm… there are actually 2 solutions, kind of glossed over by PreMath solving the simultaneous equations № 1.1: 𝒎𝒏 = ( 𝒉² → 12² → 144 ) … and № 1.2: ( 𝒎 + 𝒏 = 25 ) → 𝒏 = 25 - 𝒎 Yields № 2.1: 𝒎(25 - 𝒏) = 144 … becoming № 2.2: 𝒎² - 25𝒎 + 144 = 0 … quadratically solved is № 2.3: 𝒎 = [9, 16]; № 2.4: 𝒏 = [16, 9]; respectively; Which is to say, either value. Well, following (𝒎 = 9, 𝒏 = 16) as PreMath did yields PreMath's results. But following (𝒎 = 16, 𝒏 = 9) does not. № 3.1: AF / 𝒏 = AC / (𝒏 + 𝒎) … cross × № 3.2: AC = AF • (𝒏 + 𝒎) / 𝒏 … where 𝒙 = AC Need AF, but that's easy with 𝒉 = 12, and 𝒏 = 9 № 4.1: AF² = 12² ⊕ 9² № 4.2: AF² = 144 + 81 № 4.3: AF² = 225 № 4.4: AF = 15 Now that we got that, plug it in to (№ 3.2) № 3.3: AC = 15 × 25 ÷ 9 № 3.3: AC = 375 ÷ 9 № 3.3: AC = 41.66666666…; And AC = 𝒙, thus … another different answer. The 'other' answer. See what I mean? ⋅-⋅-⋅ Just saying, ⋅-⋅-⋅ ⋅-=≡ GoatGuy ✓ ≡=-⋅
I just found AE to be 16, then used inverse tangent to find angle CAD (came out to roughly 37 degrees). From there, I used the equation cos(37 degrees)=25/x to solve for x.
The second solution (value)? I solved in 5 min for AC = 31.25 but regarded also m = 16 and n = 9 (not m = 9 and n = 16). Then I obtained AF = 15, FD = 20, FC = 26.666666. Is this second solution real?
Are there not two valid solutions? x = 125 / 4 and x = 125 /3 ? One is the mirror of the other. Given the hypotenuse and the height from the hypotenuse to the right angle, the product of the other two sides is determined, but not which side is which. In other words, the portion of x fixed by the given dimensions could be either 15 or 20 etc.
I thought I understood basic geometry. Your demonstration problems are reminders that I should review all plane geometry theorems. Thank You!
You are very welcome Kenneth! You are absolutely correct Geometry has too many theorems. Honestly, I get stumped once in a while. We are all lifelong learners!
I'm glad you liked it! Take care my friend and all the best😃
@@PreMath plz make a video on how you apply similarity in triangles
That‘s a good one. Once you realize, that weckte dealing with multiples of the 3-4-5 Right triangle, which are all alike (and the 25 of the Hypothenuse of the ADF Triangle is the hint), it is clear that the first part of the length x is 20 and and second part 15 times 3/4 (or 11.25). (As explained in the video.) I did it in my head.
Philip you are amazing!
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1).Is it possible that m=16 and n=9 ?
2). Since we knew the ratios of the triangle are 3 : 4 : 5, we can directly find AC = (5/4) * 25 = 31.25.
Dear Ilyas, according to Geometric Mean Theorem, n=16 and m=9.
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Thanks for the reply🙏
@@PreMathI don’t see that. It seems to me switching values of m & n still satisfy GMT. The only way I can see m=16/n=9 is right and m=9/n=16 is wrong is if we know area of yellow triangle is larger than the green one or that AF>DF or AE>DE. But we haven’t been told any of that.
Are we assuming the diagram is close enough to scale so we can assume that from looking at the diagram? You didn’t make it clear from the start it’s not to scale like you usually do, I noticed.
What a super example, that's taught me something new .. the geometric theorem and the sum product thing, excellent thanks again 👍🏻
A good way to solve this problem is using metric relations. Good video :)
Nice example, many thanks. A slightly different approach via Euclid’s height theorem:
h² = pq →144 = pq = p(25 - p) → p1 = 9; p2 = 16 → 3-4-5-triangles → (3/4) = k/25
→ k = (75/4); ADC triangle: (25/4)(3 : 4 : 5) → x = 125/4=31,25 🙂
Hello, i think, there are two solution, you choose one solution m=9 an n=16, when you put m=16 and n=9, x= 125/3 = 41.66666
Thankyou very much sir for such a helpfull vedio it helps me to understand the basic concepts.
Again thank you very much.😃😃
So nice of you Umesh dear! You are awesome 👍 I'm glad you liked it! Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃
Dear sir: 12^2 = m(25-m) , this is a theorm. So we have two options for n. So we have AF. So with cos(alpha) [which is basically similar triangles] we have value of x.
ABC vertices right triangle with cathetus x (from left counterclockwise)
Similarly AFD right triangle vertices with hypotenuse 25
E = intersection of cathetus 12 with hypotenuse 25
Triangle area AFD = 1/2 * 25 * 12 = 1/2 * AD * DF… 25 * 12 = AD * DF
From the Pythagorean theorem ADsquared + DFsquared = 252 (1)
or also (AD + DF) squared - 2AD * DF = 625
that is AD + DF = √ (600 + 625) = 35 (2)
(1) can also be written in the form (AD-DF) squared + 2AD * DF = 625
hence AD-DF = 5 (3)
making a system between (2) and (3) we obtain AD = 20 and DF = 15
applying Euclid's theorem to the AFC triangle, we have 252 = x * AD from which x = 625/20 = 125/4
again by Euclid's theorem applied to the AFC triangle, DFsquared =
AD * (x-AD) = 20 * x - ADsquared = 225 + 400 = 20x,
x = 625/20 = 125/4
which confirms the result already found
Nice question and clear explanation. However, it’s just as easy applying Pythagoras and using basic trigonometry ratios to find the other lengths, since the triangles are similar. Thank you.
AD = AE + ED. Let t = tan(FAE), then AE = 12/t, and (since angle DFE = angle FAE), ED = 12t, therefore 12/t + 12t = 25, 12t^2 - 25t + 12 = 0 = (4t-3)(3t-4) and t = 3/4 or 4/3. Assuming the drawing is approx to scale, the solution must be t = 3/4 (for the angle to be < pi/4). Therefore, AE = 16, ED = 9. FAE and FED are 3-4-5 triangles, AF = 4*5 = 20, FD = 3*5 = 15. Triangle CDF is similar, and t = FC/FD, giving FC = 15t = 15*3/4 = 45/4 = 11 1/4. Therefore x = AF + FC = 20 + 11 1/4 = 31 1/4.
Since triangle ADC also has angles α, 90, β, then it is similar to triangle AEF, so once you've found AF = 20, you could just use ratios to solve for AC directly:
AC/AD = AF/AE → AC/25 = 20/16 → AC = x = 125/4 = 31.25
Note triangles ADC, AEF and FED are congruent via the angle angle angle theorem. The since 12/(25-ED) = ED/12, ED = 16 or 9. Since only ED = 9 makes sense we have all three triangles as 3-4-5 triangles. Since we know AD = 25, we have 4/5 = 25/x, so x=125/4
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Why does only ED=9 make sense? No one said the diagram was drawn to scale. The only angles specified are the right angles. Therefore, there are 2 solutions and the author only gets 5/10 for a grade since he missed 50% of the answer.
@@brianfriend238 The drawing would like quite different and the answer would be x = 125/3
Draw the two solutions to scale. You will see they both work. The reason you get two solutions is because you are solving a quadratic equation.
@@brianfriend238 yes, you swap the lengths of AE and ED. AD goes from the long leg to the short leg of a 3-4-5 trangle with x as its hypotenuse.
In this channel, all answers are present clearly. It nay not need to be the simplest one. Therefore we have the room of thinking for better approach
Thanks Fong for your kind and encouraging words.
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Thank for PreMath
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Or like this: Due to the Geometric mean theorem in the triangle AFD h^2 = pq, so that 12^2 = 144 = p*q, with p+q = 25. That only works for p = 16 and q = 9. The intercept theorem gives 12/16 = CD/25, which results to CD = 18.75. Finally, the Pythagorean theorem leads to x^2 = 25^2 + 18.75^2, so that x = +sqrt(976.6525), which results in x = 31.25
once you did your geometric mean theorem, you could have also then switched to using trigonometric functions taking the inverse tangent of the AEF triangle to get the angle alpha and then using the cosine equation to solve for the X of triangle ACD.
Can't AE be 9 and ED be 16? Why must it be 16 and then 9?
Make a video on how you apply similarity in triangles
Sure Ajay.
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How can you say that , ED = 9; it may be 16
FD =15 and AF 20 (3-4-5 triangle since AD=25)
AE=16 from 20^2 - 12^2= AE^2 = 400-144 = sqrt 256 =16
for CD 16/12 =25/CD
16CD= 25 x 12 (cross multiply)
16CD= 300
CD= 18.75
Since we looking AC or x then x^2 = 25^2 + 18.75^2
x = sqrt 976.5625
x=31.25 11:18
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The 3,4,5 triangle has to be proved first as shown in the video.
Lost you at the mn=144 and m+n=144. How did you come up with the value of m and n so quickly?
ADC is also a 3-4-5 right triangle so 25/4*5 = 31.25
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N can also equal 9 and m equal 16. Therefore there are two solutions.
I agree. If we were told that angle alpha< beta there would be a unique solution but we are not told that.
Why not m=16 and n=9 instead of m=9 and n=16? If beta>alpha no problem,but is it?
Because h x h = m x n, therefore: (h/m) = (n/h) or (h/n) = (m/h). In both cases you get via cross-multiplying 144 = 9 x 16 or 144 = 16 x 9. It doesn't matter 🙂
Vvnice true thanks Dr.JPP
this one had me trying to solve using Pythagoras, unsuccessful, understand your step-by-step solution, great job
Thanks my friend.
Cheers😀
Highly recommended. Easy to understand. Thank you so much for saving my time.
Thank you sir
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Lets think about similar (but much harder) conditions: AB=10 and AF=BC=? ...are you able to solve it? Thank you for (maybe) nice explanation
Step 4 :
🔺️AEF is similar to 🔺️ADC.
So AC/AF = AD/AE
x/20 = 25/16 , so x= 20*25/16
x= 31.25
Alternate method
Excellent George
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Very simple logic
Thank you
watched and liked the video
It's more confusing sir.
We know AD = 25
You find. AE = 16
Then. AF = 20
Simply find AC = AD / AE * AF
AC = 25 / 16 * 20 = 31.25
I have calculated CD by using the theorem of intersecting lines...
Hmmm… there are actually 2 solutions, kind of glossed over by PreMath
solving the simultaneous equations
№ 1.1: 𝒎𝒏 = ( 𝒉² → 12² → 144 ) … and
№ 1.2: ( 𝒎 + 𝒏 = 25 ) → 𝒏 = 25 - 𝒎
Yields
№ 2.1: 𝒎(25 - 𝒏) = 144 … becoming
№ 2.2: 𝒎² - 25𝒎 + 144 = 0 … quadratically solved is
№ 2.3: 𝒎 = [9, 16];
№ 2.4: 𝒏 = [16, 9]; respectively;
Which is to say, either value. Well, following (𝒎 = 9, 𝒏 = 16) as PreMath did yields PreMath's results. But following (𝒎 = 16, 𝒏 = 9) does not.
№ 3.1: AF / 𝒏 = AC / (𝒏 + 𝒎) … cross ×
№ 3.2: AC = AF • (𝒏 + 𝒎) / 𝒏 … where 𝒙 = AC
Need AF, but that's easy with 𝒉 = 12, and 𝒏 = 9
№ 4.1: AF² = 12² ⊕ 9²
№ 4.2: AF² = 144 + 81
№ 4.3: AF² = 225
№ 4.4: AF = 15
Now that we got that, plug it in to (№ 3.2)
№ 3.3: AC = 15 × 25 ÷ 9
№ 3.3: AC = 375 ÷ 9
№ 3.3: AC = 41.66666666…;
And AC = 𝒙, thus … another different answer. The 'other' answer.
See what I mean?
⋅-⋅-⋅ Just saying, ⋅-⋅-⋅
⋅-=≡ GoatGuy ✓ ≡=-⋅
x=31,25
I just found AE to be 16, then used inverse tangent to find angle CAD (came out to roughly 37 degrees). From there, I used the equation cos(37 degrees)=25/x to solve for x.
Nice tip!
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Thanks for formula m.n = h^2.
The second solution (value)?
I solved in 5 min for AC = 31.25 but regarded also m = 16 and n = 9 (not m = 9 and n = 16). Then I obtained AF = 15, FD = 20, FC = 26.666666.
Is this second solution real?
The purple part may not be necessary.
You are correct!
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Are there not two valid solutions? x = 125 / 4 and x = 125 /3 ? One is the mirror of the other. Given the hypotenuse and the height from the hypotenuse to the right angle, the product of the other two sides is determined, but not which side is which. In other words, the portion of x fixed by the given dimensions could be either 15 or 20 etc.
cos(A)=4/5. x=AD/cos(A)=31.25
Easy - the value of side x is that without it, there would be no triangle ABC.
This is only the first part of the solution because there's a second solution, where AE = 9 and ED = 16 => AC ~ 41.67 !!!
EF = 12
AD = 25
1) AE + ED = 25
AE = 25 / ED
ED = 25 / AE
2) AD² = AF² + FD² (pythagore for AFD)
3) FD² = 144 + ED² (pythagore for DEF)
4) AF² = AE² + 144 (pythagore for AEF)
2)+3)+4) :
AD² + FD² + AF² = AF² + FD² + 144 + ED² + AE² + 144
AD² = 144 + ED² + AE² + 144
625 = 288 + ED² + AE²
=> 5) ED² + AE² = 337
(ED + AE)² = ED² + 2*ED*AE + AE²
AD² = ED² + AE² + 2*ED*AE
625 = 337 + 2ED*AE
ED*AE = 144
(AD - AE)*AE = 144
-AE² + 25AE = 144
AE² - 25AE + 144 = 0
( delta = 625 - 4*144 = 49 )
AE = 25 +- sqrt(49) / 2
AE = ( 25 + 7 ) / 2 = 16
OR
AE = ( 25 - 7 ) / 2 = 9
2nd solution => AE = 9 => AC ~ 41.67
Geogebra.org : www.geogebra.org/m/kd9gxtz6
(you can move the crusor "a" to see both solutions :-) )
Note. N cannot be 9 as |ae|=must be>12
Why AE > EF?
I solved it by other way according to Pythagorean theorem 😉
Why so complicated?
After you got AE = 16 , you can find CD by similar triangles. CD/25 = 12/16
CD =18.75
Then use Pythagorean theorem to find AC
yes
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....FD=15. 25/12=X/15. X=25*15/12=31,25
31.25
wonderful!
3.75
20:16 = x:25
X= 31.25 units .
Simple 3: 4: 5 method.
Great job Mumtaz!
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I think 31/25
there is an other solution ! 125/3
m=16;n=9........?
144 = 12x12 = 3x4x3x4 = 3x3x4x4 = 9x16 (simply factoring) 🙂
Too long.. complicated...just after finding m &n... use similar triangle theream and ratikn AF/FC = (AE/ED)...
As EF & CD are parallel lines
Ez
Your Explainations are Too Slow!
You can Cut duration to half: 6' Minutes Top!
Save Our Times & Datas. 😇
11.25