Can you find Perimeter and Area of the triangle? | (Right Triangle) | #math #maths | #geometry

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Can you find Perimeter and Area of the triangle? | (Right Triangle) | #math #maths | #geometry

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[uwelinzbauer]

Thanks for the interesting video.
I spent several time to examine pythagorean triples, making up a table systematically, until I noticed a scheme that you can take the 53, subtract 1, divide by 2 to find n=26, add 1 to find m=27, then generate the missing values by calculating 2mn and m²+n².
But your solution is obviously faster and better, just one has to get the right idea 💡
Thanks for sharing 👍

[kidas]

This task is undefined, it has infinite solutions. The solution you accepted 2809*1 can also be taken as 1404.5*2, or 702.25*4 and so on.
Your solution can be considered correct, but it is not the only solution.

[prossvay]

Let a=53 ; BC=b ; AC=C
So c^2=53^2+b^2
c^2-b^2=2809
(c-b)(c+b)=2809(1)
So c-b=1
c+b=2809
So 2c=2810
c=1405
b=2809-1405=1404
Perimeter=1405+1405+53=2862 units
Area=1/2((35)(1404)=37206 square units.❤❤❤

[MathAddiction]

Here perpendicular = 53 units (an odd integer)
so base = 53^2/2 - 0.5 = 1404 and (use n^2 / 2 - 0.5)
hypotenuse = 53^2 / 2 + 0.5 = 1405 (use n^2 / 2 + 0.5)
Perimeter = p + b + h = 2862 units
Area = 1/2 p x b = 37206 sq units.

[shadoweducation]

Inconceivable !
We can prove 3-4-5 right angle triangle in this way. Thanks a lot.

[himo]

BC=x AC=y
53²+x²=y²　　　　　　　y²-x²=2809 (y+x)(y-x)=2809
y+x=2809 y-x=1 2x=2808 x=1404 y=1405

Perimeter = 53+1404+1405 = 2862 Area = 53*1404*1/2 = 37206

[laxmikantbondre]

This is very easy. If the smallest side of a Right Angle Traingle is given and is an odd number and other sides are integers, there is a standard way to find them.
1. Square the smallest side.
2. Subtract 1 from that number.
3. Divide it by 2. It wil be the other side.
4. Add one to that. It will be the hypotuneus.

So for above
53 sqr = 2809
So other side will be 1404
And Hypotuneous will be 1405


All Pythagorus Triplet in sequence follow this
3 4 5
5 12 13
7 24 25
9 40 41
And so on

[monroeclewis]

Using the difference of 2 squares formula, the only way for 53^2 to equal ((c—b)(c+b) is for ((c—b) = 1, and (c+b)^2 =53^2 or 2809. Therefore, c—b=1, c=b+1. Substitute back in: ((b+1)^2 =b^2 + 2809. Solving that, b=1404. C=b+1=1405. The rest is simple.

[ManojkantSamal]

^=read as to the power
h^2 - b^2=(53)^2
(h+p)(h-p)=2809
(h+p)(h-p)=2809×1
So,
h+p=2809....eqn1
Eqn2
Adding both the equations, we shall get the following
h+p+h-p=2809+1=2810
2h=2810
h=1405
P=2810-1405=1404
Perimeter = 1405+1404+53=2862
Area =1/2(1404×53)
=702×53
=37206
Numerous answers are there

[florianbuerzle]

Use number theory 😉 As the side lengths are known to be positive integers, this implies a Pythagorean triple. So we can use the parametrization:
b = 53 = m² - n²
a = 2mn
c = m² + n²
Using difference of squares:
(m - n)(m + n) = 53.
As 53 is a prime, the only factorization in positive integers is 53 ∙ 1, so we can solve the system:
m - n = 1
m + n = 53
Adding the equations:
2m = 54
and finally
m = 27 and (by substitution into one of the equations above) n = 26.
So we get:
a = 1404 and
c = 1405.
Finally, Area = 37206 and Perimeter = 2862.

[alster]

P= a+b+c
P= a+c+53
c²= a²+b²
c²=a²+(53)²
c²-a²= 2809
(c+a)(c-a)= (2809)(1)

c+a= 2809
c-a= 1

2c= 2810

c= 1405
a= 1404

P= 1405+1403+53
P= 2861 units

A= (702)(53)

A= 37206 units²

[unknownidentity]

Let's do some math:
.
..
...
....


The triangle ABC is a right triangle. Therefore we can apply the Pythagorean theorem:

AB² + BC² = AC²
53² + BC² = AC²
53² = AC² − BC²
53² = (AC + BC)(AC − BC)

The unknown side lengths AC and BC should be positive integers. Since 53 is a prime number, we obtain only one solution:

AC + BC = 53² = 2809
AC − BC = 1

⇒ AC = (2809 + 1)/2 = 1405
∧ BC = (2809 − 1)/2 = 1404

Now we are able to calculate the perimeter P and the area A of this triangle:

P = AB + BC + AC = 53 + 1404 + 1405 = 2862
A = (1/2)*AB*BC = (1/2)*53*1404 = 37206

Best regards from Germany

[murdock]

φ = 30° → sin⁡(3φ) = 1; ∆ ABC → sin⁡(BCA) = 1; BC = 53 = k → AB = (k^2 + 1)/2 = 1405 → AC = (k^2 - 1)/2 = 1404

[misterenter-izrz]

53^2+b^2=c^2, 53=(c-b)(c+b), c=b +1, 2b+1=53^2, b=(53^2-1)/2=1404, c=1405, so 53^3+1404^2=1405^2, the area is 1/2×53×1404=37206, the perimeter is 53+1404+1405=2862.🎉

[yalchingedikgedik]

Very nice and useful
Thanks Sir for understanding .
You are very very good
with my respects
❤❤❤❤❤

[henrykoplien]

Very nice. Did not thought about primes, giving the way to go.

[cvb-bmdg]

Very interesting problem! Since the triangle represents a Pythagorean triple, it may be possible to use the Pythagorean triple formula:
When c is the hypotenuse of a right triangle, c=p^2+q^2, b=p^2-q^2 and a=2pq (p, q ∈ ℤ, p>q). Since 53 is not even, 53=p^2-q^2. In this case p=27, q=26.

[alscents]

The value of c should have been substituted in c - a = 1 to get a = 1404 rather than in the second equation.

[rcnayak_]

There is another short-cut method. In a right angle triangle, if one of the sides of right angle is an odd number say, a, the diagonal and the other side are respectively (a²+1)/2 and (a²-1)/2. Because [(a²+1)/2]² - [(a²-1)/2]² = a². In this case, the sides are 1405 and 1404. The rest follows.

[josehidalgo]

Hello, how are you I need your help. I have a problem. Triangle ABC, AM is a median, BH divide AC in proportion 1 to 2 (AH = 1, HC = 2). P is the intersection to AM with BH, what is the area APH?, thank you.