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[marioalb]

A = ½ab = ½cx = 1014 cm²
p = a + b + c = 156 cm
a² + b² = c²

a + b = 156 - c
(a + b)² = (156 - c)²
(a² + b²) + 2ab = 156² - 312c + c²
c² + 4(½ab) = 156² -312c + c²
4*1014 = 156² - 312c
312c = 156² - 4*1014
c = 65 cm
x = 31, 2 cm ( Solved √ )

[murdock]

156 = 12(13) = 3(13) + 4(13) + 5(13); ab = 2028 = 3(13)4(13) = 12(169) → c = 5(13) → x = ab/c = (12/5)13

[AmirgabYT]

Let perpendicular to each other sides ba a and b and hypothenuse be c.
Area of triangle:
ab/2=1014
Perimeter of triangle:
a+b+c=156
Let's make a system:
av/2=1014
a+b+c=156

ab=2028
a+b+c=156

c=√(a²+b²)
a+b+√(a²+b²)=156
a²+b²=(a+b)²-2ab
a+b+√((a+b)²-2ab)=156
Let a+b=t.
t+√(t²-2ab)=156
We also know that ab=2028.
t+√(t²-2•2028)=156
√(t²-4056)=156-t
t²-4056=24336-312t+t²
312t=28392
t=91
a+b=91
91+c=156
c=65
a+b=91
a²+b²=65²

a=91-b
(91-b)²+b²=4225
8281-182b+b²+b²=4225
2b²-182b+4056=0
b²-91b+2028=0
b(1)=39
b(2)=52
a(1)=52
a(2)=39

a=39 b=52 c=65
So we know all sides of triangle.
x is height drawn from a vertex of right angle to the hypotenuse. There exists special formula:
x=ab/c
x=39•52/65=39•4/5=156/5=31, 2

Answer: 31, 2
Pin pls 🙏

[marcgriselhubert]

BC = a and BA = c. We have a.b = 2028 and a + b +sqrt(a^2 + b^2) = 156. Let's also note P = a.b and S = a + b.
We have P = 2028 and S + sqrt(S^2 - 2.P) = 156, or P = 2028 and S + sqrt(S^2 - 4056) = 156.
Let's focus on the second equation which is equivalent to sqrt(S^2 - 4056) = 156 - S, or to S^2 - 4056 = S^2 -312.S + 24336 for S < 156
We then have -312.S + 24336 = -4056, which gives that S = 91.
S = 91 and P = 2028, so a and c are solutions of x^2 - 91.x + 2028 = 0. Delta = 91^2 - 4.2028 = 169 = 13^2, so a = (91 +13)/2 = 52 and c = (91 - 13)/2 = 39
(or c = 39 and a = 52). Then AC = sqrt(a^2 + c^2) = sqrt(1521 + 2704) = sqrt(4225) = 65.
Finally the double of the area of ABC is a.c or x. AC, so 2028 = x.65, and x = 2028/65 = 156/5.

[SrisailamNavuluri]

1014=13×13×6
156=13×12=13(3+4+6)
1/2(3×4)=6
The sides are 3×13, 4×13, 5×13=39, 52, 65
Sum=39+52+65=156

Area =1/2(65×x)=1014
65x=2028, x=2028/65=156/5=31-2 units

[santiagoarosam]

AB=a ; BC=b ; CA=c.
Comprobación previa: (a+b+c)/n=3+4+5=12→ n=13 → a=13*3=39 ; b=13*4=52 ; c=13*5=65 → Perímetro=39+52+65=156 y Área=39*52/2=1014 → La hipótesis de que ABC es un triángulo de lados proporcionales a 3/4/5 es correcta y los valores obtenidos también. → a*b=c*X→ X=2*1014/65=2028/65=156/5 =31, 20 ud².
Gracias y un saludo.

[HassanLakiss]

I worked out from the area using ab=2x1014=39x52 and used the perimeter to find c=65 and checked that Pythagoras worked and hence, used xc=2x1014 to find x=31.2. Thank you

[solimana-soli]

I solved finding the radius of the inscribed circle that is Area/semiperimeter
r = 1014/(156/2) = 13
2 AC = Perimeter - 2r = 156 - 26 = 130 (the reason is the two tangent theorem)
then
AC = 130/2 = 65
x is the height of ABC then Area(ABC) = AC*x*1/2 => x = 2*Area/AC
x = 2*1014/65 = 31, 2
it's being a while that I'm showing this alternative method instead of the pythagorean method, on these kind of question about right triangles, but without success...😟

[marioalb]

Area is approximately x²
x ≈ √A ≈ √1014 = 31, 8 cm aprox.
If you need the exact result, see video

[satrajitghosh]

a + c + √ ( a^2 + c^2) = p
a c = 2 A
Again
a + c - √ ( a^2 + c^2)
= 2 a c /( a + c + √ ( a^2 + c^2))
= 4 A / p
Hereby
√ ( a^2 + c^2) = p /2 - 2 A / p

x = 2 A / √ ( a^2 + c^2)
= 2 A /( p /2 - 2 A / p )
In present problem
p = 156 = 13 * 12
A = 1014 = 13 * 7 8 = p * 13 /2
p /2 - 2 A/ p = 13 * 6 - 13 = 13 * 5

Hereby
x = p * 13 / ( 13 * 5) = p /5 = 156 /5

[yalchingedikgedik]

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

[Birol]

My way of solution ▶
The triangle ΔABC
[AB]= a
[BC]= b
[CA]= c
⇒
P(ΔABC)= 156 length units
a+b+c= 156
a+b= 156-c

A(ΔABC)= 1014 square units
a*b/2= 1014
ab= 2028

(a+b)²= a²+2ab+b²
a²+b²= c²
ab= 2028
2ab= 4056
⇒
(156-c)²= c²+4056
24336-312c+c²= c²+4056
312c= 24.336 -4056
c= 65

A(ΔABC)= c*x/2
1014= 65*x/2
x= 2028/65
x= 31, 2 length units

[misterenter-izrz]

xc=ab=2028, a+b+c=156, c^2=a^2+b^2, c^2=(156-c)^2-4056, -156(2c-156)=4056, 156-2c=26, c=65, x=2028/65=31.2.😢

[peteroleary]

A quick look at the perimeter tells us this is a Pythagorean triple. In fact it's 13X scale of a 3, 4, 5 triangle. Therefore x=31.2. QED

[devondevon]

x=31.2
a * b = 2028 ( 2* 1014)
2ab = 4056
a+ b+ c = 156
a+ b = 156- c

a^2 + b^2 + 2ab = 156^2 + c^2 - 312c
a^2 + b^2 + 4056 = 156^2 + c^2 - 312c
a^2 + b^2 = 156^2 - 4056 + c^2 -312c
= 20280 + c^2 - 312 c
c^2 =20280 + c^2 -312 c ( Pythagorean a^2 + b^2 = c^2)
312 c = 20280
c = 65

Hence, a + b = 91 ( 156- 65)
Since ab =2028
then, a = 2028/b
then 2028/b + b = 91
2028 + b^2 = 91 b
b^2 -91 + 2028b =0
(b-39)(b-52) = 0
b=39 a = 52 and c = 65

Since ab/2 = 1014 = area
and c=65
let c = p + r
then area = xp/2 + xr/2= 1014
x/2 ( p + r) = 1014 (factor out x/2)
x (p+r) = 2028 (recall c = p+r)
x (c) = 2028

x = 2028/65
x=31.2

[sergeyvinns]

Russia. Находим гипотенузу с, которая является основанием треугольника, для нахождения его высоты, через радиус вписанной окружности. r=2A/P=2028/156=13. Выразим гипотенузу с, через стороны a, b и r; с=a+b-2r=a+b-26, а так же через периметр, с=156-(a+b); приравняем правые стороны, и получим, a+b=91; находим с=156-91=65; высоту х находим по формуле площади треугольника, которая рана половине произведения основания на высоту; х=2А/с;;
х=2028/65; х=31, 2.

[DB-lgsq]

شكرا لكم على المجهودات
يمكن استعمال
a=BC, b=AC, c=AB
x=2048/b
ac=2048
a+b+c=156
a^2+c^2=b^2
(a+b+c)^2=156^2

b=65
x=2028/65

[himo]

AB=3k BC=4k CA=5k 3K+4k+5k=156 12k=156 k=13
CA=5k=65
65*x*1/2=1014 65x=2028 x=156/5

[unknownidentity]

Let's find x:
.
..
...
....


Since ABC is a right triangle, we can apply the Pythagorean theorem. So from the known area A and the known perimeter P we obtain:

AB + AC + BC = P
AB + BC = P − AC
(AB + BC)² = (P − AC)²
AB² + 2*AB*BC + BC² = P² − 2*P*AC + AC²

A = (1/2)*AB*BC ∧ AC² = AB² + BC²

AB² + 2*AB*BC + BC² = P² − 2*P*AC + AB² + BC²
2*AB*BC = P² − 2*P*AC
4*(1/2)*AB*BC = P² − 2*P*AC
4*A = P² − 2*P*AC
2*P*AC = P² − 4*A
⇒ AC = (P² − 4*A)/(2*P) = P/2 − 2*A/P = 156/2 − 2*1014/156 = 78 − 13 = 65

Now we are able to calculate the value of x:

A = (1/2)*AC*h(AC) = (1/2)*AC*BD = (1/2)*AC*x ⇒ x = 2*A/AC = 2*1014/65 = 156/5 = 31.2

Best regards from Germany

[LuisdeBritoCamacho]

STEP-BY-STEP RESOLUTION PROPOSAL :

01) AB = a

02) BC = b

03) AC = c

04) a^2 + b^2 = c^2

05) a + b + c = 156 ; a + b = 156 - c

06) a * b = 2 * 1.014 ; a * b = 2.028

07) (a + b)^2 = (156 - c)^2 ; a^2 + b^2 + 2ab = 156^2 - 312c + c^2 ; 2ab = 156^2 - 312c ; 4.056 = 156^2 - 312c ; 312c = 156^2 - 4.056 ; 312c = 24.336 - 4.056 ; 312c = 20.280 ; c= 20.280 / 312 ;

c = 65

08) So : c = 65

09) X * c = (2 * 1.014)

10) X * c = 2.028

11) X * 65 = 2.028

12) X = 2.028 / 65

13) X = 31, 2


Therefore,


OUR ANSWER :

The Length of X equal to 31, 2 Linear Units.