PQ= 2x+3, QR= 3x+8, PR=6x+9

Answer 1

One thing about the triangle lengths is the value of x can't make any side zero or negative

It has to be a positive length, so x>-3/2 x>-8/3 x>-9/6=-3/2 from setting each side >0 x>-3/2 covers all 3.

Use Heron's formula for the area of a triangle

A=square root of (p)(p-a)(p-b)(p-c) where p=1/2 the perimeter, and the perimeter = a+b+c

Perimeter = 11x+20 p=11x/2+10

p-a= 11x/2+10-(2x+3)=11x/2-2x+10-3=7x/2+7 =p-PQ

p-b=11x/2+10-(3x+8)=5x/2+2 =p-QR

p-c=11x/2+10-(6x+9)=-x/2+1 =p-PR

now multiply p by each of those 3 factors

(11x/2+10)/2(7x/2+7)(5x/2+2)(-x/2+1)

When you multiply that out, then take the square root to get the Area

I'm just guessing you want the perimeter and area. If you want the angles and sides of the triangle it gets harder. This seems like a sadistic problem. Maybe someone else has some other ideas on this one.

One other idea is let x=0 and see how it looks. Then the sides are 3, 8 and 9. Perimeter=20, p=10

The area then is square root of 10(7)(2)(1) or square root of 140 =10(1.4)^1/2 = slightly less than 12

Law of sines is a/sinA = b/sinB = c/sinC ratio of sides to their opposite angles are all equal

3/sinA=8/sinB=9/sinC It's nearly an isosceles triangle so the last two are close. 3 is less so it's an angle less than 60 degrees for A.

Okay, I give up. Do they really give problems like this? It's interesting though. You could spend hours on it, and maybe come up with something, depending on what you're looking for.

Another thing you could do is find the x value that makes the triangle isosceles, two sides and 2 angles equal.

3x+8=6x+9

3x=-1

x=-1/3 This triangle has 2 sides each = 7 The third side is -2/3+3=7/3 An isosceles triangle two sides=7 base of 3, height of a little over 6 or square root of 40 49=9+40 hypotenuse squared equals sum of squares of the other two sides. Height and base are perpendicular.

As x gets very large or approaches infinity, the triangle sides approach a ratio of 2:3:6

Another question might be for what value of x is the triangle a right triangle, such that one side squared equals the sum of the other two sides squared. (2x+3)^1/2 + (3x+8)² = (6x+9)²