I used the tangents to the radius X to form the Pythagorean Relationship (x+126)^2 + (x+264)^2 = 390^2 . Your solution eliminates the need for most of the large multiples and the quadratic formula. I worked the problem using a calculator and got the same answer but I appreciate the use of multiple variables to reduce the amount of computation of large numbers to the last step. I hope to acquire this ability through practice.
Tò semplify computations you can divide for G.C.D. between 126 and 264 that Is 6, then find radius in a smaller similar triangle. At the end moltiply the area for 36
This is nice, but you took an extremely roundabout way, I was expecting to see the ever-present Pythagorean theorem: (264 + 126)² = (264 + x)² + (126 + x)² x = -462 (no) x = 72 (yes) [(264 + 72) • (126 + 72)] / 2 = 33264
Wow! What an elegant solution. The author just show and prove very rare formula of the area of the right triangle, while we know only the length of segments of hypotenuse formed by point of tangency of inscribed circle. It is much simple way to find the area of the triangle, rather to find out first the radius of the circle with such ugly digits then find the area of the triangle
Area of right triangle is equal to the multiplication of both segments of hypotenuse A = 264 x 126 = 33264 cm² ( Solved √ ) Demonstration was done several times in previous videos.
Thank you very much for your very interesting math problems and your excellent, pedagogically meticulous, sometimes surprising solutions which makes it possible even in my age of 89 to learn new ways of solving such problems.
Others beat me to it! Yes, just apply the Pythagorean theorem and the theorem "tangents to a circle from a point have equal length", as well as determining that BE = BF = r, solve for r, compute the lengths of the 2 sides and use the formula for area of a right triangle.
(x+y)^2 = (x+r)^2+(y+r)^2 so x^2+2xy+y^2 = x^2+2xr+r^2+y^2+2yr+r^2 then we can drop x^2 and y^2 and then divide by 2 and we obtain xy= r^2+ r(x+y) eq 1. The triangle area is (x+r)(y+r)/2=( xy+(x+y)r+ r^2)/2. Using eq 1 we notice that the triangle area is xy!
(126+264)^2=(126+r)^2+(264+r)^2, 390^2=2r^2+780r+126^2+264^2, r^2+390r-33264=0, r=72 or 462 rejected, therefore the answer is (1/2)(72+126)(72+264)=33264.😊
Much better to explicitly find the area as 0.5 x (M+r) x (N+r). From the diagram, we have, (M+r)^2 + (N+r)^2 = (M+N)^2 => r^2 + r(M+N) - MN = 0 => r^2 + r(M+N) + MN - MN - MN = 0 => r^2 + r(M+N) + MN = 2MN => (M+r) (N+r) = 2MN dividing by 2 on both sides we get the LHS as in the form of aforementioned area, hence => Area of triangle = 0.5 (M+r) (N+r) = MN
Let the long section of the hypothenuse be x and the short section be y. The area of such a triangle is simply x·y = 264·126 = 33264. As the general calculation shows 2·(x·r)/2 + 2·(y·r)/2 + r² = (x+r)·(y+r)/2 x·r + y·r + r² = (x·y + x·r + y·r + r²)/2 2·x·r + 2·y·r + 2r² = x·y + x·r + y·r + r² x·r + y·r + r² = x·y one can see that the left side is the area of the triangle, as shown in the 2nd line. So there is a general relationship between the area of the triangle and those hypotenuse sections that the triangle's incircle contact point creates at the hypothenuse.
33,264 A different method This is 33,56,65 triplet scaled up by 6 Draw a straight line from F to the circle's center AF=AD tangent circle theorem Draw a straight line from D to the circle's center CD=BF tangent circle theorem EB=FB radius of the circle Using Pythagorean c= 390 (126 +164 a= 126+ r b= 264+r 390^2 = (126 + r)^2 + (264 + r)^2 0 =2r^2 + 780r -66,528 0= r^2 + 390 r - 33,264 divide both sides by 2 0= (r + 462) (r -72) r= 72 Hence, a = 126 + 72 = 198 Hence, b = 264 +72 =336 Hence Area of triangle = 336 * 198 * 1/2 = 33,264
I'm going for the tangents and Pythagoras methods. (126+r)^2 + (264+r)^2 = 390^2. That should give a value for r. r^2 + 252r + 15876 + r^2 + 528r + 69696 = 152100 2r^2 + 780r + 85572 = 152100 Simplify a bit by halving everything: r^2 + 390r + 42786 = 76050. Zero the RHS: r^2 + 390r - 33264 = 0 It may be possible to factorise that, but I'll go for the quadratic formula as I have a calculator in my hand.(-390+or-sqrt(390^2 - 4* -33264)/2 = 72 r=72. This makes the two short sides 198 and 336 (198 * 336)/2 = 33264 square units. That's interesting as that number cropped up earlier. I have a feeling there was a simpler way for this one. Ho-hum. EDIT: I just skimmed over your video. I will take a longer look later, but that looks like it's worth my time studying as it can greatly shorten the calculations.. Thank you.
Excellent presentation sir
Many many thanks ❤️
You are awesome. Keep it up 👍
I used the tangents to the radius X to form the Pythagorean Relationship (x+126)^2 + (x+264)^2 = 390^2 . Your solution eliminates the need for most of the large multiples and the quadratic formula. I worked the problem using a calculator and got the same answer but I appreciate the use of multiple variables to reduce the amount of computation of large numbers to the last step. I hope to acquire this ability through practice.
Tò semplify computations you can divide for G.C.D. between 126 and 264 that Is 6, then find radius in a smaller similar triangle. At the end moltiply the area for 36
-Tangents from the same point are equal in length:
390^2 = (r+126)^2 + (r+ 264)^2
152100 = r^2 + 252r+15876 + r^2 + 528r + 69696
152100 = 2r^2 + 780r + 85572
2r^2 + 780r-66528 = 0
R = 72
Side 1 = 126 + 72 = 198
Side 2 = 264+ 72=336
1/2 *336*198 = 33,264
You presented a new way of finding area! So neat. No words to explain, I just loved it!
Thanks a lot 😊 ❤️
Amazing!
❤❤❤❤❤
Thanks dear ❤️
Stay blessed 🙏
This is nice, but you took an extremely roundabout way, I was expecting to see the ever-present Pythagorean theorem:
(264 + 126)² = (264 + x)² + (126 + x)²
x = -462 (no)
x = 72 (yes)
[(264 + 72) • (126 + 72)] / 2 = 33264
Thanks for your feedback! Cheers! 😀
You are awesome. Keep it up 👍
Excellent working 👍
Wow! What an elegant solution. The author just show and prove very rare formula of the area of the right triangle, while we know only the length of segments of hypotenuse formed by point of tangency of inscribed circle. It is much simple way to find the area of the triangle, rather to find out first the radius of the circle with such ugly digits then find the area of the triangle
Area of right triangle is equal to the multiplication of both segments of hypotenuse
A = 264 x 126 = 33264 cm²
( Solved √ )
Demonstration was done several times in previous videos.
Thank you very much for your very interesting math problems and your excellent, pedagogically meticulous, sometimes surprising solutions which makes it possible even in my age of 89 to learn new ways of solving such problems.
Thank you
Others beat me to it! Yes, just apply the Pythagorean theorem and the theorem "tangents to a circle from a point have equal length", as well as determining that BE = BF = r, solve for r, compute the lengths of the 2 sides and use the formula for area of a right triangle.
(x+y)^2 = (x+r)^2+(y+r)^2 so x^2+2xy+y^2 = x^2+2xr+r^2+y^2+2yr+r^2 then we can drop x^2 and y^2 and then divide by 2 and we obtain xy= r^2+ r(x+y) eq 1. The triangle area is (x+r)(y+r)/2=( xy+(x+y)r+ r^2)/2. Using eq 1 we notice that the triangle area is xy!
(126+264)^2=(126+r)^2+(264+r)^2, 390^2=2r^2+780r+126^2+264^2, r^2+390r-33264=0, r=72 or 462 rejected, therefore the answer is (1/2)(72+126)(72+264)=33264.😊
Excellent!
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍
That is how I did it. The video was way too complicated for me.
r = -462 rejected.👍
Generally for any length of hypotenuse a+b, the area of the right-angled triangle is simply ab.😘
What a lovely formula! 😍
An enjoyable and instructive problem.
Spoiler alert.
Labelling a generalized figure:
𝑎 = AD and AF;
𝑏 = CD and CE;
𝑟 = DO, OE, OF, BF and BE.
Additional line segments:
AO, OC, DO, OE, OF.
𝑟 (𝑟 + 𝑎 + 𝑏) = (𝑟 + 𝑎) (𝑟 + 𝑏) / 2;
2𝑟 (𝑟 + 𝑎 + 𝑏) = (𝑟 + 𝑎) (𝑟 + 𝑏);
2𝑟 (𝑟 + 𝑎 + 𝑏) = 𝑟 (𝑟 + 𝑎 + 𝑏) + 𝑎𝑏;
2𝑟 (𝑟 + 𝑎 + 𝑏) − 𝑟 (𝑟 + 𝑎 + 𝑏) = 𝑎𝑏;
𝑟 (𝑟 + 𝑎 + 𝑏) = 𝑎𝑏.
As 𝑟 (𝑟 + 𝑎 + 𝑏) is the area of the triangle, 𝑎𝑏 must be the same area.
In our example, 𝑎 = 264 and 𝑏 = 126.
So, the area of the triangle = 𝑎𝑏 = 264 ∙ 126 = 33,264 square units.
For completeness:
𝑟 = 72 (worked out with the quadratic formula).
Area = 𝑟 (𝑟 + 𝑎 + 𝑏)
= 72 (72 + 264 + 126)
= 72 (462)
= 33,264.
Area = (72 + 264) (72 + 126) / 2
= (336) (198) / 2
= (336) (99)
= 336 ∙ 100 − 336
= 33,600 − 336
= 33,264.
This is theorem! Very smart!)
asnwer=145 cm isit
هل يمكننا أن نعتبرها قاعدة حيث أن مساحة المثلث القائم الزاوية تساوي حاصل ضرب جزأين الوتر نتيجة رسم دائرة تلامس الأضلاع الثلاثة للمثلث ؟؟؟
صحيح جدا! ❤️
Much better to explicitly find the area as 0.5 x (M+r) x (N+r).
From the diagram, we have,
(M+r)^2 + (N+r)^2 = (M+N)^2
=> r^2 + r(M+N) - MN = 0
=> r^2 + r(M+N) + MN - MN - MN = 0
=> r^2 + r(M+N) + MN = 2MN
=> (M+r) (N+r) = 2MN
dividing by 2 on both sides we get the LHS as in the form of aforementioned area, hence
=> Area of triangle = 0.5 (M+r) (N+r) = MN
Use Pythagorean theorem much easier to find out area
Let the long section of the hypothenuse be x and the short section be y.
The area of such a triangle is simply x·y = 264·126 = 33264.
As the general calculation shows
2·(x·r)/2 + 2·(y·r)/2 + r² = (x+r)·(y+r)/2
x·r + y·r + r² = (x·y + x·r + y·r + r²)/2
2·x·r + 2·y·r + 2r² = x·y + x·r + y·r + r²
x·r + y·r + r² = x·y
one can see that the left side is the area of the triangle, as shown in the 2nd line.
So there is a general relationship between the area of the triangle and those hypotenuse sections that the triangle's incircle contact point creates at the hypothenuse.
Per la similitudine dei triangoli rettangoli risulta (264+r)^2+(126+r)^2=(264+126)^2...r=72...A=bh/2=(264+72)(126+72)/2=336*99=33600-336=33264
Smart solution!
I went with Pythagore, and got a nice quadratic equation to find r.
Smartly avoided impossible quadratic equation. :)
Handy 7482 is Right. Exactly my Solution.
😉💯👍
33264 square unit
Think x(1/6) s=(1/2)(44+r)(21+r)=(44+21+r)r=(65+r)r r^2+65r-924=0 s=924 S=36s=33264
J'étais parti faire du Pythagore, mais avec des grands nombres (390²), c'était avec la calculatrice obligatoire ^^
I used tangents but I’m too lazy to do the actual math;-;
I did it the same way, but failed to see that you don‘t need to calculate the radius (which is 72, btw…!)😂
It is class 10th question
33,264
Excellent!
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍
The triangle isn't actually blue.
Very large figures.😢
Thanks for your feedback! Cheers! 😀
You are awesome. Keep it up 👍
It is a 33, 56, 65 Pythagorean triplet right triangle scaled up by 6. Therefore,
you could use 44 and 21 instead by dividing 264 and 126 by 6.
33,264
A different method
This is 33,56,65 triplet scaled up by 6
Draw a straight line from F to the circle's center
AF=AD tangent circle theorem
Draw a straight line from D to the circle's center
CD=BF tangent circle theorem
EB=FB radius of the circle
Using Pythagorean c= 390 (126 +164
a= 126+ r
b= 264+r
390^2 = (126 + r)^2 + (264 + r)^2
0 =2r^2 + 780r -66,528
0= r^2 + 390 r - 33,264 divide both sides by 2
0= (r + 462) (r -72)
r= 72
Hence, a = 126 + 72 = 198
Hence, b = 264 +72 =336
Hence Area of triangle = 336 * 198 * 1/2 = 33,264
I'm going for the tangents and Pythagoras methods. (126+r)^2 + (264+r)^2 = 390^2. That should give a value for r.
r^2 + 252r + 15876 + r^2 + 528r + 69696 = 152100
2r^2 + 780r + 85572 = 152100
Simplify a bit by halving everything: r^2 + 390r + 42786 = 76050.
Zero the RHS: r^2 + 390r - 33264 = 0
It may be possible to factorise that, but I'll go for the quadratic formula as I have a calculator in my hand.(-390+or-sqrt(390^2 - 4* -33264)/2 = 72
r=72.
This makes the two short sides 198 and 336
(198 * 336)/2 = 33264 square units.
That's interesting as that number cropped up earlier. I have a feeling there was a simpler way for this one. Ho-hum.
EDIT: I just skimmed over your video. I will take a longer look later, but that looks like it's worth my time studying as it can greatly shorten the calculations.. Thank you.
33,264