Can you find the area of the triangle? | #math #maths

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Can you find the area of the triangle? | #math #maths

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[michaelkouzmin]

I dare recall the algebra that you usually apply in similar cases:
1. We can gather a system of 3 equations:
(1) a+b+51=120;
(2) a^2+b^2=51^2;
(3) Area s= a*b/2;
2. a+b=69
let us square both sides:
(4) (a^2+b^2)+2ab=69^2;
3. Let us put (2) & (3) into (4)
51^2+2*2*s=69^2;
4s=69^2-51^2;
sq. units.

[ChuzzleFriends]

Label legs as a (height) and b (base).
a + b + 51 = 120
a + b = 69
a² + b² = 51² = 2601
The area of the triangle = 1/2 * a * b. Label the area as A.
(a + b)² = a² + 2ab + b²
69² = 51² + 4A
4761 = 2601 + 4A
2160 = 4A
540 = A
The area of the right triangle is 540 square units.

[vacuumcarexpo]

・The area of a triangle is the radius of its inscribed circle multiplied by its semi-perimeter.
・The radius of a right triangle's inscribed circle is half the value subtracting its hypotenuse from the sum of the two sides other than its hypotenuse.

Using the two facts above, we can solve this in our heads.

AB+BC=120-51=69
The radius of its inscribed circle is (69-51)/2=9.
The area of this triangle is 120×9/2=540.

[hervechampagne]

51= 3*17
There is a Pyhagorean triplet (8, 15, 17).
Thus if we make three times each digit of thid Pythagorean triplet we have (24, 45, 51).
24+45+51 = 120. And 24*45/2 = 540
QED

[jimlocke]

Grouping and factoring works nicely when both the variable and coefficients are integers. Here the coefficients are integers, but we don't know about the variable. If the variable is not an integer, we need to use the "brute force" quadratic formula. If the equation can be solved by grouping and factoring integers, which was the case here, then the sides of the triangle are integers and must be a Pythagorean triple, or multiple thereof. So, instead of creating the quadratic equation and trying grouping and factoring, a strategy is to check first for a Pythagorean triple, as follows. The hypotenuse is given as 51, which factors to (3)(17). The Pythagorean triple with 17 is 8, 15, 17, which becomes 24, 45, and 51 when multiplied by 3. Summing these sides, we get the desired perimeter of 120. So we have a solution. The area is (1/2)(24)(45) = 540 sq. units, as PreMath also found.

[AmirgabYT]

a²+b²=51²
a+b+51=120

a²+b²=2601
a+b=69

a=69-b
(69-b)²+b²-2601=0
2b²-138b+2160=0
b²-69b+1080=0
b(1)=24
b(2)=45
a(1)=45
a(2)=24
S=(45•24)/2=540 sq. u.

[prossvay]

Let AB=a; BC=b
a+b+51=120
a+b=120-51=69
a^2+b^2=51^2
a^2+b^2=2601
a=69-b
(69-b)^2+b^2=2601
4761-138b+b^2+b^2=2601
2b^2-138b+2160=0
b^2-69b+1080=9
b=45; b=24
a=69-45=24
So area of the triangle=1/2(45)(24)=540 square units.❤❤❤ Thanks

[solimana-soli]

Perimeter = a+b+c = 120 => semiperimeter = 120/2=60
Being c=51 we can find the diameter of the inscribed circle that is:
2r = a + b - c = a + b + c - 2*c = 120 - 2*51 = 120 - 102 = 18 then
r = 9
knowing that in every triangle the radius of the inscribed circle is:
r = Area/semiperimeter
Area = r * semiperimeter = 9*60 = 540

[jarikosonen]

4:31 x^2-69x+1080=0
Factoring x only:
x*(69-x)=1080,
Divide by 2:
1/2*x*(69-x)=540.
So its done.

[yucangwang]

No need to solve a quadrtic equation, x^2+y^2＝51^2 (1)
x+y＝120-51＝59 (2)

just square (2)and then minus (1), you will find x*y/2. This is easier .

[johnjones]

Several people solved it the same way I did. So rather than repeat what they said, I decided to generalize the problem and came up with:

A = (P(P-2c))/4

[DraCat]

120-51=69
69 ²=51^2+2AB
S=AB/2
S=(69²-51²)/4
S=540

[rvbln]

with c=51 and a+b+c=120 we get a+b=69 (1)
by pythagoras we get
a²+b²=51² |+2ab
a²+b²+2ab=51²+2ab
(a+b)²=51²+2ab
with (1) we get
(69)²=51²+2ab |-51²
69²-51²=2ab |:4
(69²-51²)/4 = ab/2
now on the right side is the formula for calculating the area of the triangle

so 540 = ab/2

[yalchingedikgedik]

Thanks Sir
Thanks PreMath
That’s enjoyable and useful.
Good luck with glades

[devondevon]

540
a^2 + b^2 = 51^2 or 2601 (Pythagorean)
a + b = 69 ( 120-51)
a^2 + b^2 + 2ab = 69^2 = 4761
2601 + 2ab = 4761
2ab = 2160
ab/2 = 2160/4
= 540 Answer

[marcgriselhubert]

Two unknown lengths: a = BC and c = AB.
Two equations. First: a + c + 51 = perimeter = 120, or a + c = 69. Second: The Pythagorean theorem: a^2 + c^2 = 51^2, or a^2 + c^2 = 2601.
From the first equation we get that (a + c)^2 = 69^2 = 4761, so a^2 + c^2 + 2.a.c = 4761 and then 2.a.c = 4761 - 2601 = 2160, and a.c = 1080.
Now we have to find a and c knowing their sum 69 and thein product 1080. There are the solutions of the equation x^2 - 69.x + 1080 = 0
delta = (-69)^2 -4.1080 = 4761 - 4320 = 441, and sqrt(delta) = 21. So a = BC = (69 + 21) / 2 = 45, and c = AB = (69 - 21) / 2 = 24 (assuming that a > b)

[engralsaffar]

I did it this way,

Using pythagoras,
h^2+b^2=51^2=2601 - - (1)

Calculating the perimeter,

h+b+51=120
h+b=120-51=69
(h+b)^2=69^2
h^2+b^2+2*h*b=4761 - - (2)

Subtracting (1) from (2)
2*h*b=2160

A=0.5*h*b=2160/4=540 square units

[wackojacko]

@ 4:29 in other words, the Coefficient of the 2nd degree term x² (and in this case is 1) we are looking for two numbers whose product of 2nd term Coefficient times the constant 1080 (IS EQUAL TO!) the Sum of 69. I like your top down approach for finding feasible factors and keeping in mind absolute values . 🙂

[davideissler]

What is interesting about this problem is that you actually don’t need to find x. The area is 1/2*x*(69-x)=1/2*(69x-x^2). From the Pythagorean theorem you get: 2*x^2-2*69*x+69^2=51^2. Moving the 51^2 to the left and the x terms to the right you get 69^2-51^2=2*(69*x -x^2). Note the right hand side is 4 times the area of the triangle so dividing each side by 4 gives: Area = (69^2-51^2)/4 =

[jeffkowk]

Let’s make it easier:
AB=X, BC=Y, X+Y=120-51=69,
X*X+Y*Y=51*51,
(X+Y)*(X+Y)=69*69

X*Y/2=540