My preferred solution is. Substitute c=126-(a+b) into pythagoras and this gives a+b=73 This then gives c=53 Use (a-b)^2=(a+b)^2-4ab to find a-b=17 Then solve for a and b. This method avoids the factorisation of 1260.
Sides a and b and hypotenuse c where b < a 1. 1260 = ab 2. a + b + c = 126 3. a^2 + b^2 = c^2 Solve system of equations using substitution and quadratic equation. a = 45, b = 28, c = 53. Check: 45 + 28 + 53 = 126. Check: (45 x 28)/2 = 630 The tricky bit is the substitutions.
It is customary to name the sides of a triangle with the small letter opposite the corresponding vertex: a in front of A, b in front of B, c in front of C
Sometimes when ab is a large number and has more factors involved, it is time consuming to find the exact pair which adds up the required value (a + b). In this case, we can use another way (algebraic formulae) to solve it. We have a + b = 73 and ab = 1260. Now (a - b)² = (a + b)² - 4ab = 73² - 4.1260 = 289. This gives (a - b) = 17. Now a = (73 + 17)/2 = 45 and b = (73 - 17)/2 = 28.
I've found a different solution. Known that in a generic triangle the radius of the inscribed circle is: r = A/P where A is the area and P the semiperimeter (126/2 = 63) we can easily find: r = 630/63 = 10 But we know that the same radius in a right triangle is: r = (a + b - c)/2 so we have: 1) a + b + c = 126 (perimeter) 2) a + b - c = 20 (diameter inscribed circle) sum 1) + 2) then a + b = 73 a*b = 1260 (2*area triangle) now we have different way but we can solve this: X² - (sum) X + (product) = 0 X² - 73X + 1260 = 0 getting: a = 28, b = 45 c = 126 - a - b = 126 - 45 - 28 = 53
@@Chocolate12-m9n Yes maybe, but consider that you have only a text and the only data are area and perimeter... so we could draw the triangle in a different way
Since area = rs and s = 63, then 63r = 630 => r = 10 = 5 x 2. Now, if we set up a fibonacci sequence 5,2,7,9 and form the fundamental pythagorean triple with legs 5 x 9 and 2(2 x 7) and hypotenuse 2^2 + 7^2 (where r = 5 x 2), then the sides are 45, 28, and 53.
Катеты 28 45, гипотенуза 53 Умножаем площадь в два раза, и по теореме Виета методом подбора получаем один катет 45, следовательно другой 28 И теоремой пифагора находим гипотенузу и подтверждаем периметром
Herein a + b + √ ( a^2 + b^2) = p a b = 2 A Hereby a + b - √ ( a^2 + b^2) = (a + b) ^2 - (a^2 + b^2) / [ a + b + √ ( a^2 + b^2) ] = 2 a b / / [ a + b + √ ( a^2 + b^2) ] = 4 A / p Hereby a + b = 2 A / p + p /2 √( a^2 + b^2) = - 2 A / p + p/2 and so forth
A property of right triangles is the relationship between the side lengths, the perimeter P and the area A: hypothenuse c = (P^2 - 4A)/2P a,b = 0.5 * {(P-c) +/- SQRT[(P-c)^2 - 8A]} In this particular example, P = 126, A = 630 --> c = 53, a and b = 28 and 45 Note these formulae don't require the quantities being integers.
I saw the thumbnail and wanted to try this problem, I’m in 8th grade so the method I used might not be as good or straight forward but it lead me to the right answer so here’s what I did Eq. 1 a+b+c=126 Eq. 2 (1/2)(ab)=630 Eq. 3 a^2 + b^2 = c^2 In Eq. 2 by multiplying both sides by 2 I got: ab=1260 In Eq. 3 by square rooting both sides I got: C=Sqrt(a^2+b^2) I substitute that C value in to Eq. 1 giving me: a+b+sqrt(a^2+b^2)=126 Then I subtracted a and b from both sides then squared both sides leaving me with: a^2+b^2=(-a-b+126)^2 Foiling the right side left me with: a^2+b^2=a^2+b^2+2ab-252(a+b)+15876 I then substituted the ab value from Eq. 2 which was ab=1260 a^2+b^2=a^2+b^2+2520-252(a+b)+15876 Combining the constant terms left me with: a^2+b^2=a^2+b^2 -252(a+b) + 18396 If a^2+b^2=a^2+b^2 + t (any expression) then t has to equal zero I set -252(a+b)+18396=0 Subtracting 18396 from both sides left me with: -252(a+b)=-18396 Dividing both sides by -252 gave me the expression “(a+b)=73” Substituting 73 for (a+b) in Eq. 1 gave me: 73+c=126 After subtracting 73 on both sides I had found the hypotenuse of the triangle which was 53 Now I set up a system of equations using: Eq. 1 [a+b=73] Eq.2 [ab=1260] In Eq. 2 dividing a on both sides gave me: b=1260/a I substituted b=1260/a into Eq. 1 which gave me: a+1260/a=73 I multiplied a on both sides to give me a^2+1260=73a I then moved the 73a to left so I could use the quadratic formula in terms of a: a^2-73a+1260=0 After plugging in 1, -73, and 1260 into the quadratic formula I got: A = (73 plus or minus sqrt(289))/2 Simplifying the radical gave me (73 plus or minus 17)/2 I then had A=90/2, A=56/2 Simplifying them further I had gotten: A=45, A=28 I substituted A=45 into Eq. 1 and got: 45+b=73 Subtracting 45 on both sides gave me b=23. I checked by plugging in A=45, B=23, and C=53 into the original 3 equations and all of them checked out
It is good that you got the solution, but as you say it isn't the best method. I would substitute c=126-(a+b) into pythagoras. To find a+b. Also, don't square 126, keep it as 126x126 and don't double 126 keep it as 2x126, then cancel as demonstrated in the video, this avoids large numbers and factorisation. Consider (4x10)/2 =40/2=20 this is bad. (4×10)/2=2×10 =20 is much better. If you expand you are going to have to factorise later. So wait and see if you can cancel, which you can in this problem.
Herr professor, I tried to find the three sides of triangles using equations, but, I confess, I couldn't. Then I moved on to the arm: If the area of the triangle is 630 area units, then the area of the rectangle will be double, 1,260 a * b = 1260 Then I decomposed these 1260 finding: 2² * 3² * 5¹ * 7¹ Then it became easy, I tried 4 * 315 - it doesn't work; 9 * 140, it doesn't work; 35 * 36, almost gave; 45 + 28, bingo C = 53 28² + 45² = C² C = 53 !!!!! Bingo
✍prof.(eccellente soluzione ) Esisterebbe a mio avviso tuttavia questa scorciatoia che le propongo. °) scompongo in fattori →⇒1260= (2^2)*(3^2)*(5)*(7) dove; 2-3-5-7 ; sono NP. Essi offrono queste combinazioni⇒ 4*9= 36 ed ⇒ 5*7=35 dove 35*36=1260 tuttavia alla verifica di a^2+b^2 ≄c^2 e quindi 35^2+36^2= 2521 la cui radice vale ⇒√2521 ≃50,2< 53 . quindi si considera l'altra coppia di numeri ⇒ (3^2)*5=45 ed (2^2)*7= 28 che risolvono il problema; infatti 28^2+45^2=784+2025=2809 ⇒√2809= 2809^(1/2)=(53^2)^1/2= 53⇒ OK Coppia di numeri ; 28 e 45 Inoltre poiché s'intravede il teorema del coseno ecco che scriverei che At= c*(b* cos𝛃)*co𝝿/3= =53(45*0,5283)0,5= 53*23,775*0,5≃630 dove 23,775 = h(altezza triangolo rispetto al diametro-ipotenusa (c). Prof.(complimenti) non avevo mai trovato un problema così . Cordialità😌 li, 1 febbraio 2024⏳
My preferred solution is.
Substitute c=126-(a+b) into pythagoras and this gives a+b=73
This then gives c=53
Use (a-b)^2=(a+b)^2-4ab to find a-b=17
Then solve for a and b.
This method avoids the factorisation of 1260.
This method is simple and easy to understand anybody
(a+b) ² - 4 ab = (a-b)²
Sides a and b and hypotenuse c where b < a
1. 1260 = ab
2. a + b + c = 126
3. a^2 + b^2 = c^2
Solve system of equations using substitution and quadratic equation. a = 45, b = 28, c = 53.
Check: 45 + 28 + 53 = 126.
Check: (45 x 28)/2 = 630
The tricky bit is the substitutions.
It is customary to name the sides of a triangle with the small letter opposite the corresponding vertex: a in front of A, b in front of B, c in front of C
Sir, no need to find roots of quadratic equation
(a+b)=73.......(I)
(a-b)= 17.........(I)
by adding a=45 ,b= 28
Sometimes when ab is a large number and has more factors involved, it is time consuming to find the exact pair which adds up the required value (a + b). In this case, we can use another way (algebraic formulae) to solve it. We have a + b = 73 and ab = 1260. Now (a - b)² = (a + b)² - 4ab = 73² - 4.1260 = 289. This gives (a - b) = 17. Now a = (73 + 17)/2 = 45 and b = (73 - 17)/2 = 28.
I've found a different solution. Known that in a generic triangle the radius of the inscribed circle is:
r = A/P where A is the area and P the semiperimeter (126/2 = 63) we can easily find:
r = 630/63 = 10
But we know that the same radius in a right triangle is:
r = (a + b - c)/2
so we have:
1) a + b + c = 126 (perimeter)
2) a + b - c = 20 (diameter inscribed circle)
sum 1) + 2) then
a + b = 73
a*b = 1260 (2*area triangle)
now we have different way but we can solve this:
X² - (sum) X + (product) = 0
X² - 73X + 1260 = 0
getting: a = 28, b = 45 c = 126 - a - b = 126 - 45 - 28 = 53
I think a=45 cause the BC>AB and b=28
@@Chocolate12-m9n Yes maybe, but consider that you have only a text and the only data are area and perimeter... so we could draw the triangle in a different way
@@soli9mana-soli4953 of course
Since area = rs and s = 63, then 63r = 630 => r = 10 = 5 x 2.
Now, if we set up a fibonacci sequence 5,2,7,9 and form the fundamental pythagorean triple with legs
5 x 9 and 2(2 x 7) and hypotenuse 2^2 + 7^2 (where r = 5 x 2), then the sides are 45, 28, and 53.
Катеты 28 45, гипотенуза 53
Умножаем площадь в два раза, и по теореме Виета методом подбора получаем один катет 45, следовательно другой 28
И теоремой пифагора находим гипотенузу и подтверждаем периметром
Alternate method: factor 1260. 4, 5, 7, 9. Combinations (20, 63), (28, 45), (35, 36). Plug into the Pythagorean formula. Only (28, 45) results in an integer solution. 126 - (28 + 45) = 53.
Herein a + b + √ ( a^2 + b^2) = p
a b = 2 A
Hereby
a + b - √ ( a^2 + b^2)
= (a + b) ^2 - (a^2 + b^2)
/ [ a + b + √ ( a^2 + b^2) ]
= 2 a b / / [ a + b + √ ( a^2 + b^2) ]
= 4 A / p
Hereby a + b = 2 A / p + p /2
√( a^2 + b^2) = - 2 A / p + p/2
and so forth
A property of right triangles is the relationship between the side lengths, the perimeter P and the area A:
hypothenuse c = (P^2 - 4A)/2P
a,b = 0.5 * {(P-c) +/- SQRT[(P-c)^2 - 8A]}
In this particular example, P = 126, A = 630 --> c = 53, a and b = 28 and 45
Note these formulae don't require the quantities being integers.
What did the sycamore say to the triangle?
"Gee, I'm a tree."
I saw the thumbnail and wanted to try this problem, I’m in 8th grade so the method I used might not be as good or straight forward but it lead me to the right answer so here’s what I did
Eq. 1 a+b+c=126
Eq. 2 (1/2)(ab)=630
Eq. 3 a^2 + b^2 = c^2
In Eq. 2 by multiplying both sides by 2 I got:
ab=1260
In Eq. 3 by square rooting both sides I got:
C=Sqrt(a^2+b^2)
I substitute that C value in to Eq. 1 giving me:
a+b+sqrt(a^2+b^2)=126
Then I subtracted a and b from both sides then squared both sides leaving me with:
a^2+b^2=(-a-b+126)^2
Foiling the right side left me with:
a^2+b^2=a^2+b^2+2ab-252(a+b)+15876
I then substituted the ab value from Eq. 2 which was ab=1260
a^2+b^2=a^2+b^2+2520-252(a+b)+15876
Combining the constant terms left me with:
a^2+b^2=a^2+b^2 -252(a+b) + 18396
If a^2+b^2=a^2+b^2 + t (any expression) then t has to equal zero
I set -252(a+b)+18396=0
Subtracting 18396 from both sides left me with:
-252(a+b)=-18396
Dividing both sides by -252 gave me the expression “(a+b)=73”
Substituting 73 for (a+b) in Eq. 1 gave me:
73+c=126
After subtracting 73 on both sides I had found the hypotenuse of the triangle which was 53
Now I set up a system of equations using:
Eq. 1 [a+b=73]
Eq.2 [ab=1260]
In Eq. 2 dividing a on both sides gave me:
b=1260/a
I substituted b=1260/a into Eq. 1 which gave me:
a+1260/a=73
I multiplied a on both sides to give me a^2+1260=73a
I then moved the 73a to left so I could use the quadratic formula in terms of a:
a^2-73a+1260=0
After plugging in 1, -73, and 1260 into the quadratic formula I got:
A = (73 plus or minus sqrt(289))/2
Simplifying the radical gave me (73 plus or minus 17)/2
I then had A=90/2, A=56/2
Simplifying them further I had gotten:
A=45, A=28
I substituted A=45 into Eq. 1 and got:
45+b=73
Subtracting 45 on both sides gave me b=23.
I checked by plugging in A=45, B=23, and C=53 into the original 3 equations and all of them checked out
It is good that you got the solution, but as you say it isn't the best method. I would substitute c=126-(a+b) into pythagoras. To find a+b.
Also, don't square 126, keep it as 126x126 and don't double 126 keep it as 2x126, then cancel as demonstrated in the video, this avoids large numbers and factorisation.
Consider (4x10)/2 =40/2=20 this is bad. (4×10)/2=2×10 =20 is much better.
If you expand you are going to have to factorise later. So wait and see if you can cancel, which you can in this problem.
45, 28, 53
Using eq. (i) and (ii) we get c=53,, then find a+b and a-b finally giving a=45 and b=28
Got it! (Almost gave up...)
Using three equation , the length of sides will s1=45 s2= 28 & s3=53 .
Nice solution . In Greece we use Vieta's theory .(RIP)
Thanks.
Bro, in the question u saw that BC is greater than AB so the answer is {a,b,c}={45,23,53} and another answer also has probabilities but not.
Alternate :
b.a/2 = 630
a + b + c = 126
c sq = a sq + b sq
Works well, but can not answer which on the shortest leg. :)
Use area = =square root is s(s-a)(s-b)(s-c(
Yes, Heron's Formula.
There is no need to reinvent the wheel.
Como lo haces en terreno??
I love it!
45+28+53=126)(45×26=1260÷2=630
a=45, b=28, c=53!
a=45 b=28 c=53
length of AB is c. not by b
Easy!!!
However once pythogorean triplet this sum can in no time be solved why such a long process.
45,28,53
solo que es muy largo!!!
30,42,54
c=sqrt(30^2+42^2)≈51.6≠54 🤣
=> Heron :
A=sqrt(63×33×21×9)≈626.8≠630 😂👎
Herr professor, I tried to find the three sides of triangles using equations, but, I confess, I couldn't.
Then I moved on to the arm:
If the area of the triangle is 630 area units, then the area of the rectangle will be double, 1,260
a * b = 1260
Then I decomposed these 1260 finding: 2² * 3² * 5¹ * 7¹
Then it became easy,
I tried 4 * 315 - it doesn't work;
9 * 140, it doesn't work;
35 * 36, almost gave;
45 + 28, bingo C = 53
28² + 45² = C²
C = 53 !!!!! Bingo
A me risulta a=42 b =30 c=54
This answer doesn't satisfy pythagoras theorem. It is obviously incorrect since a^2+b^2 ends with a 4 and c^2 ends with a 6.
process ok
Long method.
Calcul tres lent🤤
Second degre
Du yoù naw math weri long ut ansawar
✍prof.(eccellente soluzione )
Esisterebbe a mio avviso tuttavia questa scorciatoia che le propongo.
°) scompongo in fattori →⇒1260= (2^2)*(3^2)*(5)*(7) dove; 2-3-5-7 ; sono NP.
Essi offrono queste combinazioni⇒ 4*9= 36 ed ⇒ 5*7=35 dove 35*36=1260 tuttavia alla verifica di
a^2+b^2 ≄c^2 e quindi 35^2+36^2= 2521 la cui radice vale ⇒√2521 ≃50,2< 53 .
quindi si considera l'altra coppia di numeri ⇒ (3^2)*5=45 ed (2^2)*7= 28 che risolvono il problema; infatti
28^2+45^2=784+2025=2809 ⇒√2809= 2809^(1/2)=(53^2)^1/2= 53⇒ OK
Coppia di numeri ; 28 e 45
Inoltre poiché s'intravede il teorema del coseno ecco che scriverei che At= c*(b* cos𝛃)*co𝝿/3=
=53(45*0,5283)0,5= 53*23,775*0,5≃630 dove 23,775 = h(altezza triangolo rispetto al diametro-ipotenusa (c).
Prof.(complimenti) non avevo mai trovato un problema così .
Cordialità😌
li, 1 febbraio 2024⏳
28,45,53
Base 28 height 45