Two starting equations: a + b + c = 120 (The triangle's perimeter) a² + b² + c² = 5202 A = (ab)/2 By the Pythagorean Theorem, a² + b² = c². Substitute c² for a² + b² in the second equation. c² + c² = 5202 2c² = 5202 c² = 2601 c = 51 a + b + 51 = 120 a + b = 69 (a + b)² = a² + 2ab + b² 69² = 2601 + 4A 4A + 2601 = 4761 4A = 2160 A = 540 So, the area of the triangle is 540 square units.
120=3×40 5202=9×578 a^2+b^2+c^2=9×578 where a^2+b^2=c^2 a^2+b^2=9×289=9(64+225). =(3×8)^2+(3×15)^2 a=24,b=45,c=3×17=51 a+b+c=120 Area ÷1/2(24×45)=12×45=540 sq units
Ok, this is a right triangle, so a^2 + b^2 = c^2. Sub that into a^2 + b^2 + c^2 = 5202 and we get 2*c^2 = 5202, which leads to c = 51. Then a + b + c = 120 gives us a + b = 69. So b = 69 - a. Sub that into a^2 + b^2 = 2601 to get a^2 + (69 - a)^2 = 2601 a^2 + 4761 - 138*a + a^2 = 2601 2*a^2 - 138*a + 2160 = 0 a^2 - 69*a + 1080 = 0 The roots are 45 and 24, and these are the sought after values of a and b. The area is (45*24)/2 = 540. Q.E.D.
Let's find the area: . .. ... .... ..... Since we have a right triangle, we can apply the Pythagorean theorem: c² = a² + b² c² + c² = a² + b² + c² 2*c² = 5202 c² = 2601 ⇒ c = 51 a² + b² = c² = 51² a + b + c = 120 a + b + 51 = 120 a + b = 69 Now we are able to calculate the area of the triangle: A = ab/2 = 2ab/4 = (a² + 2ab + b² − a² − b²)/4 = [(a + b)² − c²]/4 A = (69² − 51²)/4 = (69 + 51)(69 − 51)/4 = 120*18/4 = 30*18 = 540 Best regards from Germany
PreMath finds the area without finding a and b, but, those who did find that a, b, and c are the Pythagorean triple 24, 45, 51. PreMath really should check that a and b are positive real numbers, since a product of 2 negative numbers would produce a positive number and look correct. It is not necessary that the sides of the triangle be a Pythagorean triple or even be rational numbers. The sum of the squares of the 3 sides need not be 2 times a square number, so c will be irrational in that case. If a circle is constructed through the 3 triangle vertices, the hypotenuse will be the circle's diameter. The minimum value of a + b will be just slightly more than c. The maximum will be for the isosceles right triangle, a = b. In that case a = b = (c√2)/2. As a increases from just slightly more than 0 to (c√2)/2, the perimeter increases continuously from just slightly greater than 2c to (1 + √2)c. Any integer value between those two numbers is valid. A large enough value of c will ensure a valid integer value for the perimeter. So, both the perimeter and the sum of squares for the sides can be integers, but a, b, and c be irrational.
*c^2 = a^2 + b^2, so a^2 + b^2 + c^2 = 2.(a^2 + b^2) = 5202, so a^2 + b^2 = c^2 = 2601. *Then c= sqrt(2601) = 51, and a + b = 120 - c = 69. *2601 = (a^2 + b^2) = (a + b)^2 - 2.a;b = 69^2 - 2.a.b, then 2.a.b = 69^2 - 2601 = 4761 - 2601 = 2160. *Finally, the area of ABC is (1/2).a.b = 2160/4 = 540. *We can also calculate a, b and c, even not asked: a.b = 1080 and a + b = 69, so a and b are solutions of x^2 - 69.x + 1080 = 0. Delta = (-69)^2 -4.1.1080 = 441 = 21^2 a = (69 - 21)/2 = 24, b = (69 +21)/2 = 45 (if a < b), c = 51. So (24, 45, 51) is a Pythagorean triple.
STEP-BY-STEP RESOLUTION PROPOSAL : 01) As : a^2 + b^2 = c^2 ; we must conclude that: 02) a^2 + b^2 + c^2 = c^2 + c^2 03) So: 2c^2 = 5.202 04) c^2 = 5.202 / 2 ; c^2 = 1.601 ; c = sqrt(2.601) ; c = 51 lin un 05) a + b + c = 120 ; a + b = 120 - 51 ; a + b = 69 06) a^2 + b^2 = 5.202 - 1.601 ; a^2 + b^2 = 2.601 07) Solving these two Equations we have : a = 24 lin un and b = 45 lin un 08) Checking Solutions: 09) 24 + 45 + 51 = 120 ; 120 = 120 10) 24^2 + 45^2 + 51^2 = 5.202 ; 576 + 2.025 + 2.601 = 5.202 ; 5.202 = 5.202 11) Area = (a * b) / 2 ; Area = 1.080 / 2 ; Area = 540 sq un Thus, OUR BEST ANSWER IS : Tiangle Area is equal to 540 Square Units.
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Two starting equations:
a + b + c = 120 (The triangle's perimeter)
a² + b² + c² = 5202
A = (ab)/2
By the Pythagorean Theorem, a² + b² = c². Substitute c² for a² + b² in the second equation.
c² + c² = 5202
2c² = 5202
c² = 2601
c = 51
a + b + 51 = 120
a + b = 69
(a + b)² = a² + 2ab + b²
69² = 2601 + 4A
4A + 2601 = 4761
4A = 2160
A = 540
So, the area of the triangle is 540 square units.
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Nice! Eye opening use of substitution.
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120=3×40
5202=9×578
a^2+b^2+c^2=9×578 where a^2+b^2=c^2
a^2+b^2=9×289=9(64+225).
=(3×8)^2+(3×15)^2
a=24,b=45,c=3×17=51
a+b+c=120
Area ÷1/2(24×45)=12×45=540 sq units
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@@PreMath I like your methods of solving.I gave this method to make numbers smaller.Thank you.
@@SrisailamNavuluri
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Your method is so clear and easy to follow Professor !
I just love your demonstrations Man ….!
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Very helpful sir🙋🏻♂️
Just love the way of your presentation🙏🏼👍🏼🙋🏻♂️
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c^2=2601=>c=51=>a+b=120-51=69.now a^2+b^2+2ab=4761=>2ab=2
160=>ab/2=540 sq.unit is ar. of triangle.
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We have :
a² + b² = c²
a² + b² - c² = 0 .....(A)
a² + b² + c² = 5202 ...... (B)
By adding equations (A) and (B) , we find the following :
2 (a²+b²) = 5202
a² + b² = 2601 ..... (E)
We substitute the value of (E) into equation (B) :
2601 + c² = 5202
c² = 2601
c = √2 601
c = 51
____________________________
a + b + c = 120
(a + b)² = (120 - c)²
a² + b² + 2ab = (120 - c)²
S = ab/2 = ((120- c )² - ( a² + b² ))/4
S = 540 square units
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Ok, this is a right triangle, so a^2 + b^2 = c^2. Sub that into a^2 + b^2 + c^2 = 5202 and we get 2*c^2 = 5202, which leads to c = 51. Then a + b + c = 120 gives us a + b = 69. So b = 69 - a. Sub that into a^2 + b^2 = 2601 to get
a^2 + (69 - a)^2 = 2601
a^2 + 4761 - 138*a + a^2 = 2601
2*a^2 - 138*a + 2160 = 0
a^2 - 69*a + 1080 = 0
The roots are 45 and 24, and these are the sought after values of a and b. The area is (45*24)/2 = 540. Q.E.D.
Let's find the area:
.
..
...
....
.....
Since we have a right triangle, we can apply the Pythagorean theorem:
c² = a² + b²
c² + c² = a² + b² + c²
2*c² = 5202
c² = 2601
⇒ c = 51
a² + b² = c² = 51²
a + b + c = 120
a + b + 51 = 120
a + b = 69
Now we are able to calculate the area of the triangle:
A = ab/2 = 2ab/4 = (a² + 2ab + b² − a² − b²)/4 = [(a + b)² − c²]/4
A = (69² − 51²)/4 = (69 + 51)(69 − 51)/4 = 120*18/4 = 30*18 = 540
Best regards from Germany
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PreMath finds the area without finding a and b, but, those who did find that a, b, and c are the Pythagorean triple 24, 45, 51. PreMath really should check that a and b are positive real numbers, since a product of 2 negative numbers would produce a positive number and look correct.
It is not necessary that the sides of the triangle be a Pythagorean triple or even be rational numbers. The sum of the squares of the 3 sides need not be 2 times a square number, so c will be irrational in that case. If a circle is constructed through the 3 triangle vertices, the hypotenuse will be the circle's diameter. The minimum value of a + b will be just slightly more than c. The maximum will be for the isosceles right triangle, a = b. In that case a = b = (c√2)/2. As a increases from just slightly more than 0 to (c√2)/2, the perimeter increases continuously from just slightly greater than 2c to (1 + √2)c. Any integer value between those two numbers is valid. A large enough value of c will ensure a valid integer value for the perimeter. So, both the perimeter and the sum of squares for the sides can be integers, but a, b, and c be irrational.
a+b+c=120
a^2+b^2+c^2=5202
(a^2+b^2)=c^2
2c^2=5202
c^2=5202÷2
=2601
=51^2
c=51
a^2+b^2=2601
a+b=120-51
a+b=69
(??4)^2+(??5)^2
=2601
24^2+45^2=2601
a=24,b=45,c=51
(axb)÷2
(24×45)÷2
=540
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Just used the same approach.👍
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Good exercise.
I knew of the 3-4-5 and the 5-12-13, but it appears that there's a pile of others.
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*c^2 = a^2 + b^2, so a^2 + b^2 + c^2 = 2.(a^2 + b^2) = 5202, so a^2 + b^2 = c^2 = 2601.
*Then c= sqrt(2601) = 51, and a + b = 120 - c = 69.
*2601 = (a^2 + b^2) = (a + b)^2 - 2.a;b = 69^2 - 2.a.b,
then 2.a.b = 69^2 - 2601 = 4761 - 2601 = 2160.
*Finally, the area of ABC is (1/2).a.b = 2160/4 = 540.
*We can also calculate a, b and c, even not asked:
a.b = 1080 and a + b = 69, so a and b are solutions of
x^2 - 69.x + 1080 = 0. Delta = (-69)^2 -4.1.1080 = 441 = 21^2
a = (69 - 21)/2 = 24, b = (69 +21)/2 = 45 (if a < b), c = 51.
So (24, 45, 51) is a Pythagorean triple.
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My way of solution ▶
a+b+c= 120
a²+b²+c²= 5202
a²+b²= c²
⇒
2c²= 5202
c²= 2601
c= √2601
c= 51 length units
a+b= 120-51
a+b= 69
(a+b)²= 69²
69²= a²+b²+2ab
a²+b²= c²
⇒
69²= c²+2ab
2ab= 69²-c²
2ab= 69²-51²
2ab= (69-51)*(69+51)
2ab= 18*120
ab/2= 18*120/4
ab/2= 18*30
ab/2= 540 square units
a+b+c=120 (1)
a^2+b^2+c^2=5202 (2)
a^2+b^2=c^2 (3)
(2) 2c^2=5202
So c=51
(1) a+b=120-51=69
a^2+b^2=2601
b=69-a
a^2+(69-a)^2=2601
a=24 and a=45
b=45 and b=24
So a=24 ;b=45 ; c=51
a=45 ; b=24 ; C=51
So area triangle=1/2(24)(45)=540square units.
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1/2(24)(45)=540 sq.u ;))
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a + b + c = 120
a² + b² + c² = 5.202
Area = ?
Solution
A = ½ b . h
A = ½ ab ... ¹
Pythagorean Theorem:
a² + b² = c² ... ²
a² + b² + c² = 5.202
c² + c² = 5.202
2c² = 5.202
c² = 2.601
*c = 51*
a + b + c = 120
a + b + 51 = 120
a + b = 69 ... ³
(a + b)² = (69)²
a² + 2ab + b² = 4.761
2ab + (51)² = 4.761
2ab + 2.601 = 4.761
2ab = 4.761 - 2.601
2ab = 2.160
ab = 1.080
A = ½ ab
A = ½ 1.080
*A = 540 Square Units*
*=================*
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Some Pythagoras and then some substitution. Not too difficult. Usually you have some geometry or trig as well but still fun.
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STEP-BY-STEP RESOLUTION PROPOSAL :
01) As : a^2 + b^2 = c^2 ; we must conclude that:
02) a^2 + b^2 + c^2 = c^2 + c^2
03) So: 2c^2 = 5.202
04) c^2 = 5.202 / 2 ; c^2 = 1.601 ; c = sqrt(2.601) ; c = 51 lin un
05) a + b + c = 120 ; a + b = 120 - 51 ; a + b = 69
06) a^2 + b^2 = 5.202 - 1.601 ; a^2 + b^2 = 2.601
07) Solving these two Equations we have : a = 24 lin un and b = 45 lin un
08) Checking Solutions:
09) 24 + 45 + 51 = 120 ; 120 = 120
10) 24^2 + 45^2 + 51^2 = 5.202 ; 576 + 2.025 + 2.601 = 5.202 ; 5.202 = 5.202
11) Area = (a * b) / 2 ; Area = 1.080 / 2 ; Area = 540 sq un
Thus,
OUR BEST ANSWER IS :
Tiangle Area is equal to 540 Square Units.
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1/ c= 51-> a+b=69
2/sq(a+b)= sqa+sqb+2ab
-> sq69=sq51+2ab
-> 2ab=sq69-sq51=120x18
ab=60x18
Area= ab/2=9x60=540 sq unitd😅😅😅
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Very very easy sum ...just we have to substitute the values in the equationsand use pt😂
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(69^2-51^2)/4
kesinlikle mükemmel
apslyly perfect
Teşekkürler canım❤️🌹
Too easy 🙂
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2c^2=5202...c^2=2601...c=51....
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Piece of cake! Got it in less than a minute
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540. Sq. Units
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540
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DEI says no skills required to solve! 🙂
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