MATHMATHmathhhh.?

Question

the second side of a triangle is four more than the first side, and the third side is five more than the second. if the product of the first and third sides is added to the product of the second and third sides, the result is 124 more than twice the square of the first side. find the perimeter of the triangle.

Answer 1

Let the first side, second side and third side of the triangle be x, (x + 4) and (x + 9) respectively.

x(x + 9) + (x + 4)(x + 9) = 2(x)^2 + 124
x^2 + 9x + x^2 + 9x + 4x + 36 = 2x^2 + 124
2x^2 + 22x + 36 = 2x^2 + 124
22x = 88
x = 4

Perimeter Of Triangle --> (x) + (x + 4) + (x + 9)
= 3x + 13
= 3(4) + 13
= 25 units

Answer 2

Let the first side be X
Let the second side be X+4
Let the third side be X+4+5 or X+9
Product of the 1st and 3 side X^2+9X
Product of the 2d and 3d side X^2+13X+36
Sum of the two products 2X^2+22X+36
Square of the first side X^2
Sum of 124 + 2*above or 2X^2+124 = 2X^2+22X+36
You can do the rest (solve for a positive X). The perimeter of the triangle is 3X+13.

Answer 3

let first side of trangel = x
then second side od trangel is = 4+x
then third side is = 9+x
product of first and third = x^2+9x...(1)
pruduct of second and third = (4+x)(9+x)=x^2+13x+36...(2)
now as question said
(x^2+9x)+(x^2+13x+36)=124+2x^2
2x^2 + 22x + 36 = 2x^2 + 124
22x=124-36
22x = 88
x = 4

Perimeter Of Triangle
=(x)+(x + 4)+(x + 9)
= 3x + 13
= 3(4) + 13
= 25 units

Answer 4

S = 4+F => F = S-4......(1)
T = 5+S ......(2)
F*T+S*T = 124+2F^2......(3)
Substitute (1) and (2) in (3),
(2S-4)(S+5) = 124+2(S-4)^2
2S^2+6S-20 = 124+2S^2-16S+32
22S = 176
S = 8
T = 13
F = 4
Perimeter = S+T+F = 25 units
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Ideas: Use F for the first side, S for the second side, and T for the third side so that they make more sense in your mind. You can solve for any of the three unknowns first.

Answer 5

let P = x + y + z
x,y,z are sides of the triangle

y=x+4
z=5+y

x(z)+(y)(z)=124+2x^2
z(x+y)=124+2x^2
(5+y)(x+y)=124+2x^2
(5+x+4)(x+x+4)=124+2x^2
(x+9)(2x+4)=124+2x^2
2x^2+18x+4x+36=124+2x^2
22x=124-36=88
x=4
y=x+4=8
z=5+y=5+8=13
P=x+y+z=4+8+13=25 units

Answer 6

LET a,b,c be three sides(1st, 2nd and 3rd resp),
given,
a+4=b
b+5=c
ac+bc- 2a^2=124
to find a+b+c perimeter

b=a+4
c=5+a+4
=a+9
a(a+9)+(a+4)(a+9)-2a^2=124
a^2+9a+a^2+13a+36-2a^2=124
22a+36=124
22a=88
a=4
-> b=8
-->c=13--------> perimeter=25units