Can you find Area of the triangle? | (Algebra) | #math #maths | #geometry

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Can you find Area of the triangle? | (Algebra) | #math #maths | #geometry

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[ChuzzleFriends]

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.

[SrisailamNavuluri]

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

[jamestalbott]

Nice! Eye opening use of substitution.

[abeonthehill]

Your method is so clear and easy to follow Professor !
I just love your demonstrations Man ….!

[BBMathTutorials]

Very helpful sir🙋🏻‍♂️
Just love the way of your presentation🙏🏼👍🏼🙋🏻‍♂️

[yalchingedikgedik]

That’s very nice and enjoyable
Thanks Sir
With my respects
❤❤❤❤❤

[amitavadasgupta]

a^2+b^2+2ab=4761=>2ab=2
160=>ab/2=540 sq.unit is ar. of triangle.

[kennethkan]

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

[maisseraboudjema]

We have :
a² + b² = c²
a² + b² - c² = 0
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

[jimlocke]

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.

[phungpham]

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😅😅😅

[Birol]

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

[prossvay]

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 units.
❤❤❤

[unknownidentity]

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

[arnavkange]

Very very easy sum ...just we have to substitute the values in the equationsand use pt😂

[calvinmasters]

Good exercise.
I knew of the 3-4-5 and the 5-12-13, but it appears that there's a pile of others.

[stanbest]

Some Pythagoras and then some substitution. Not too difficult. Usually you have some geometry or trig as well but still fun.

[marcgriselhubert]

*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.

[sergioaiex]

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*

[alster]

Piece of cake! Got it in less than a minute