I did it by starting with the observation that because the green triangle was twice the size of the blue triangle and both have the same length of x, the height of the green triangle must be twice the height of the blue triangle, which allowed me to state the length of the yellow triangle as x-z, and the height of the yellow triangle as x-2z. I then solved for z instead of solving for x and plugged that back into x=4/z but ultimately got the same result, 7.
My HS Geometry Teacher used to put problems like this. He would include some circles and label a few of the angles, line segments and some congruent symbols here and there. Our goal was to figure out as much as possible (with justification): angles, sides, chords ...you name it. I really looked forward to these! Great Work! ❤😂🎉
Randomly found your channel and it gives me life. In another life I would've gotten my degree in Math instead of Finance and gone on to be a math professor. Your channel made me nerd out again over math.
My technique to solve this doesn't involve Algebra but rather just Geometric Logic 1. Double the 3cm² triangle to make a 6cm² rectangle and removing a part of the Red Triangle. 2. Double the 4cm² triangle to make a 8cm² rectangle, removing a part of both Red Triangle and the 2cm² triangle. 3. Move top part of the 2cm² triangle to what is remaining of the Red Triangle, you can see it all fits nicely on the square now. 4. Add all of the new values to get the SquArea 8cm²+6cm²+2cm²=16cm² 5. Subtract the original triangle areas from the SquArea to get the area of the Red Triangle 16cm²-(4cm²+3cm²+2cm²)=16cm²-9cm²=[7cm²]
Yeah same. It’s very apparent that the blue triangle only checks out as 4/1, which gives you the are of the square easily, then filling in the rest of the triangles by doing area formulas in reverse to subtract.
@@justinmplayz8809it doesn’t assume. It’s really easy to work out. Only whole numbers are used, which basically means all the sides must be whole numbers as well. If the math checks, it checks The X/Z triangle has only one solution, 4:1. Therefore Y must be .5X
@@lancercool1992 no it’s just very (like extraordinarily) unlikely that numbers this low are going to operate with decimal values and it’s the easiest thing to check. A right triangle with an area of 2 is pretty clearly 2 and 1.
Area of Pink Δ = √{(A+B+C)² - 4.A.B} = √{(4+2+3)² - 4.(4).(2)} = √{9²- 4(8)}= √{81- 32}= √49 = 7. Area (Rect.)= 4+2+3+7=16. Here A = area of Δ with one side being the width of the rectangle and B = area of Δ with one side being the height of the rectangle. Let Δ= area of Pink triangle, A = area of 27, B = area of 15 and C= area of 12. Let u=horizontal side of C, v= vertical side of C, a=width, b=height of rectangle. Then A+B+C+Δ=ab. Also uv=2C=2(12), b(a-u)=2B=2(15), and a(b-v)=2A=2(27). So bu=ab-2B & av=ba-2A. ∴ 0=(ab-2B)(ba-2A) - ab(uv) = (ab-2B)(ab-2A) - ab(2C). So (ab)² -2(ab)(A+B+C) + 4AB=0. ∴ (A+B+C+Δ)² -2(A+B+C+Δ).(A+B+C) + 4AB=0. So Δ² = (A+B+C)² - 4AB. Hence Δ = √{(A+B+C)² - 4.A.B}, since Δ>0. Q.E.D. Let D=A+B+C. Then (D+Δ)² -2(D+Δ).D +4AB = 0. ∴ D²+2DΔ+Δ² -2D² -2ΔD +4AB=0. So Δ² = D² - 4AB = (A+B+C)² - 4AB. ∴ Δ= √{(A+B+C)² - 4AB}, since Δ>0. .
I used the blue triangle. The side of the square is x and the base of the blue triangle y. Thus xy divided by 2 is equal to 2 and xy is 4. Or thus xand y are both 2 or x and y are 4 and 1 respectively. The side is larger than the base thus x is 4 and y is 1. And used the rest of the information to calculate the area of the red triangle. there is no need to use such calculations. but still I appreciate your math skills very much and I am your fan. I like all those geometry challenges and I will subscribe you. ❤❤❤❤
x*z=8 x*y=4 (x-y)*(x-z)=6 x^2-x*z-x*y+y*z=6 x^2+y*z=18 x^4+x*y*x*z=18x^2 x^4-18x^2+32=0 (x^2-16)(x^2-2)=0 The area of the square is more than the sum of the areas of the traingle x^2>2+3+4=9. So the area of the square is x^2=16. And the area of the red triangle is 7.
You mean how do you factor the x^4-18x^2+32? into (x^2-2)(x^2-16)? He's just skipping a bit but when you multiply (x+a)(x+b) you get x^2+(a+b)x+(a*b). You can first do a replacement step of "u = x^2" so you have u^2-18u+32 so it's clearer. Then you can try to find two numbers, a and b, where a+b is -18 and a*b is 32. -16 and -2 satisfy this. -> (u-2)(u-16), then replace back (x^2-2)(x^2-16)
both blue and green triangles have the same side so it's obvious that y is 2z. and in a second you realize looking at areas that z is 1 and x is 4 . The equation is solved in seconds.
I just spend new year’s eve trying to solve this, only to find out I forgot x^4-18x+32 can become (x^2-16)(x^2-2), and it annoys the heck out of me! The good news is I probably won’t forget it anymore
I did it mentally anyway. In the blue triangle, double 2 is 4, and the factors are 1 and 4 or 2 and 2. Since the sides of the triangle are not equal, you can rule out the two and two option leaving the sides of the square equal to 4. You can double the area of the other triangles and divide it by 4. Allowing a much easier way to do it mentally. Finally, 4² is 16, subtract the sum of the triangle and you get 7 cm².
@cliptomaniac2562 z = 4/x can be rewritten as x = 4z. then, since area is 1/2bh, rewrite as 1/2(4z) * z = 2, or 2z * z = 2. rewrite again, 2z squared = 2, divide both sides by 2, z squared = 1. root of 1 is 1 so z = 1. which means x must = 4 because x equals 4z. but once you know how to do this stuff quickly, it pops out at you immediately that z must be 1 as soon as you see the area so you can skip right to the end of all these steps the way he did it is more complicated but also more fun
I think this explanation, was over complicated, all I did was observed that the square's total area, ignoring the triangles, would be Green triangle ×4 Here the green triangle represents one side Therefore 4×4=16 After that you add the areas of all the triangles and minus that answer from 16 This should give you seven ,which is the area of the pink triangle.
Finally, someone else who hates fractions
I prefer decimals because I mess fractions up and end up having the wrong answer.
Well for me, I like fractions because decimals are not 100% accurate. Sometimes, you will end up with x.999..
fractions arent even that bad lol
@@heco.fractions are more accurate, but the way you add and subtract them makes me hate it so much
Lil bit of practise can make it easy for u@@HopefullyJustMe
I did it by starting with the observation that because the green triangle was twice the size of the blue triangle and both have the same length of x, the height of the green triangle must be twice the height of the blue triangle, which allowed me to state the length of the yellow triangle as x-z, and the height of the yellow triangle as x-2z. I then solved for z instead of solving for x and plugged that back into x=4/z but ultimately got the same result, 7.
Nice. You may want to change the word 'square' to 'triangle'
@@kasigerito Oops lol, lemme fix
This was my approach too
Nice, I did it this way too :)
I did like that too
My HS Geometry Teacher used to put problems like this. He would include some circles and label a few of the angles, line segments and some congruent symbols here and there. Our goal was to figure out as much as possible (with justification): angles, sides, chords ...you name it. I really looked forward to these! Great Work! ❤😂🎉
I took a screenshot of this and solved it on my own. I almost did the exact same things you did to find the area!
Got most of the way there I just forgot how to factor :(
nice vdeo, I was wondering where you get questions like these, Just so I could practice them for myself
Your channel is exactly what I need right now
How exciting. 😀
This is exciting❤
Randomly found your channel and it gives me life. In another life I would've gotten my degree in Math instead of Finance and gone on to be a math professor. Your channel made me nerd out again over math.
"I don't really like fractions" So relatable my boy!!
Andy could do quantum physics and I’d still watch.
2:40 why do you try to calculate x at all? You lost sight of the endgame here. Your answer is x2-9 so having two values for x2 is good enough.
My technique to solve this doesn't involve Algebra but rather just Geometric Logic
1. Double the 3cm² triangle to make a 6cm² rectangle and removing a part of the Red Triangle.
2. Double the 4cm² triangle to make a 8cm² rectangle, removing a part of both Red Triangle and the 2cm² triangle.
3. Move top part of the 2cm² triangle to what is remaining of the Red Triangle, you can see it all fits nicely on the square now.
4. Add all of the new values to get the SquArea
8cm²+6cm²+2cm²=16cm²
5. Subtract the original triangle areas from the SquArea to get the area of the Red Triangle
16cm²-(4cm²+3cm²+2cm²)=16cm²-9cm²=[7cm²]
I just realized this assumes that x=2y so it isn't really a sure solution
Yeah same. It’s very apparent that the blue triangle only checks out as 4/1, which gives you the are of the square easily, then filling in the rest of the triangles by doing area formulas in reverse to subtract.
@@justinmplayz8809it doesn’t assume. It’s really easy to work out. Only whole numbers are used, which basically means all the sides must be whole numbers as well. If the math checks, it checks
The X/Z triangle has only one solution, 4:1. Therefore Y must be .5X
"it doesn't assume"
assumes whole numbers @@Nameandaddresswithheld
@@lancercool1992 no it’s just very (like extraordinarily) unlikely that numbers this low are going to operate with decimal values and it’s the easiest thing to check. A right triangle with an area of 2 is pretty clearly 2 and 1.
Absolutely stoked that we solved this the same way
Area of Pink Δ = √{(A+B+C)² - 4.A.B} = √{(4+2+3)² - 4.(4).(2)}
= √{9²- 4(8)}= √{81- 32}= √49 = 7. Area (Rect.)= 4+2+3+7=16.
Here A = area of Δ with one side being the width of the rectangle
and B = area of Δ with one side being the height of the rectangle.
Let Δ= area of Pink triangle, A = area of 27, B = area of 15 and C= area of 12.
Let u=horizontal side of C, v= vertical side of C, a=width, b=height of rectangle.
Then A+B+C+Δ=ab. Also uv=2C=2(12), b(a-u)=2B=2(15), and a(b-v)=2A=2(27).
So bu=ab-2B & av=ba-2A. ∴ 0=(ab-2B)(ba-2A) - ab(uv) = (ab-2B)(ab-2A) - ab(2C).
So (ab)² -2(ab)(A+B+C) + 4AB=0. ∴ (A+B+C+Δ)² -2(A+B+C+Δ).(A+B+C) + 4AB=0.
So Δ² = (A+B+C)² - 4AB. Hence Δ = √{(A+B+C)² - 4.A.B}, since Δ>0. Q.E.D.
Let D=A+B+C. Then (D+Δ)² -2(D+Δ).D +4AB = 0. ∴ D²+2DΔ+Δ² -2D² -2ΔD +4AB=0.
So Δ² = D² - 4AB = (A+B+C)² - 4AB. ∴ Δ= √{(A+B+C)² - 4AB}, since Δ>0.
.
😅 wait but how did u write the first equation
@@Aditya_196searched up the symbols and then copied the symbols, probably
You can find x by just considering which values make the area of the blue triangle, however your method is much more rigorous
"someone even emailed it to me asking me to solve it... *so let's just solve it"*
🗿
Where is this question? It was a squire questinn.I can't find it.
I used the blue triangle. The side of the square is x and the base of the blue triangle y. Thus xy divided by 2 is equal to 2 and xy is
4. Or thus xand y are both 2 or x and y are 4 and 1 respectively. The side is larger than the base thus x is 4 and y is 1. And used the rest of the information to calculate the area of the red triangle. there is no need to use such calculations. but still I appreciate your math skills very much and I am your fan. I like all those geometry challenges and I will subscribe you. ❤❤❤❤
Finding yz is so annoying. My brain is being fried trying to find it.
x*z=8
x*y=4
(x-y)*(x-z)=6
x^2-x*z-x*y+y*z=6
x^2+y*z=18
x^4+x*y*x*z=18x^2
x^4-18x^2+32=0
(x^2-16)(x^2-2)=0
The area of the square is more than the sum of the areas of the traingle x^2>2+3+4=9. So the area of the square is x^2=16. And the area of the red triangle is 7.
why didn't you just do the addition first oops i realized that you had to find the area of the square
haha this isnt even edited why did you comment this 😂
i don't understand how you factor it that way at 2:07 wouldn't there be an extra number in the step before
You mean how do you factor the x^4-18x^2+32? into (x^2-2)(x^2-16)?
He's just skipping a bit but when you multiply (x+a)(x+b) you get x^2+(a+b)x+(a*b). You can first do a replacement step of "u = x^2" so you have u^2-18u+32 so it's clearer.
Then you can try to find two numbers, a and b, where a+b is -18 and a*b is 32. -16 and -2 satisfy this. -> (u-2)(u-16), then replace back (x^2-2)(x^2-16)
@@Tordek oh ok thanks dawg
i overcomplicated this SOOO hard, and ended up forgetting that red was scalene
I think he doesn't like fractions
0:45 "I don't really like fractions"
0:58 procedes to willingly make fractions
well yeah u cant really solve it otherwise...
@@Horse_In_A_Suit my point still stands
both blue and green triangles have the same side so it's obvious that y is 2z. and in a second you realize looking at areas that z is 1 and x is 4 . The equation is solved in seconds.
I just spend new year’s eve trying to solve this, only to find out I forgot x^4-18x+32 can become (x^2-16)(x^2-2), and it annoys the heck out of me! The good news is I probably won’t forget it anymore
Wait you can just do that to fractions?!?
What happens to your voice after a sentenced? Do you have corona?
What would it look like if x was {2}?
God dangit I really gotta retake geometry, I got everything in terms of the y and that got me a ? area of 128/(17+√(33))
What was the 2 meant for? Was it a
“I don’t really like fractions…”
*proceeds to create more fractions immediately after getting rid of the original fractions.
But do you know how to use the 3 sea shells?
7 cm2
square of 2 is square root of 2😮
i just realised, i am stupid
i got 7 cm^2 tooo
Another eazy method
After we get the triangel split the hyp into x and 5-x then apply phygoras thorem
But area one is more eazy
I managed to get it from the thumbnail. I feel smart for once 😅
I was literaly able to figur it out in my head why all of this
without pen: ua-cam.com/video/XMzVJiuoqV4/v-deo.html
I did it mentally anyway. In the blue triangle, double 2 is 4, and the factors are 1 and 4 or 2 and 2. Since the sides of the triangle are not equal, you can rule out the two and two option leaving the sides of the square equal to 4. You can double the area of the other triangles and divide it by 4. Allowing a much easier way to do it mentally. Finally, 4² is 16, subtract the sum of the triangle and you get 7 cm².
What do you mean the factors are 1 and 4 or 2 and 2? Isn’t that an assumption?
@cliptomaniac2562 z = 4/x can be rewritten as x = 4z. then, since area is 1/2bh, rewrite as 1/2(4z) * z = 2, or 2z * z = 2. rewrite again, 2z squared = 2, divide both sides by 2, z squared = 1. root of 1 is 1 so z = 1. which means x must = 4 because x equals 4z. but once you know how to do this stuff quickly, it pops out at you immediately that z must be 1 as soon as you see the area so you can skip right to the end of all these steps
the way he did it is more complicated but also more fun
solved
I also dont like fractions
Alot of work for nothing. You can pretty much conform that s=4 from the blue triangle and subtract 2, 3, and 4 from 4^2. 16-2-3-4=7
i just count the square probably : (2+3+4+?)
3.3 = 9 ( It cant be )
4.4 = 16 ( probably )
5.5 = 25 ( to far )
then 16 - (2+3+4) = 7
I think this explanation, was over complicated, all I did was observed that the square's total area, ignoring the triangles, would be
Green triangle ×4
Here the green triangle represents one side
Therefore 4×4=16
After that you add the areas of all the triangles and minus that answer from 16
This should give you seven ,which is the area of the pink triangle.
What did you use to determine the green triangle is 1/4 the area of the whole square?
Try Vegan Please
I don't really like fractions - word! 🤌