SOLUTION: A triangular sheet of paper is 28 square inches, one side of the triangle is 2 inches more than 3 times the length of the altitudeto that side. Find the length of that side and th

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: A triangular sheet of paper is 28 square inches, one side of the triangle is 2 inches more than 3 times the length of the altitudeto that side. Find the length of that side and th      Log On


   



Question 41909: A triangular sheet of paper is 28 square inches, one side of the triangle is 2 inches more than 3 times the length of the altitudeto that side. Find the length of that side and the altitude to the side. please i donot know what formula to use of how to solve. this is not form the texbook.
Found 2 solutions by AnlytcPhil, venugopalramana:
Answer by AnlytcPhil(1807) About Me  (Show Source):
You can put this solution on YOUR website!
A triangular sheet of paper is 28 square inches, one side 
of the triangle is 2 inches more than 3 times the length 
of the altitudeto that side.  Find the length of that 
side and the altitude to the side. 

       C
      /|'
     / |    '
    /  |y       '
   /   |            '
 A ŻŻŻŻDŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻB
  |--------x-----------|

Let AB = x and CD = y


>>...A triangular sheet of paper is 28 square inches...<<

     Area = 1/2 × a side × altitude to that side

       28 = 1/2 × AB × CD 

       28 = (1/2)(x)(y)
       28 = xy/2
      
Multiply both sides by 2 to clear the fraction:

       56 = xy
 
>>...one side of the triangle is 2 inches more than 3 times
the length of the altitude to that side...<<

       AB is 2 inches more than 3 times CD

        x = 3y + 2

So you have the system of equations

56 = xy
 x = 3y + 2

Can you solve that? If not post again

Answer (x,y) = (14,4) and (-12, -14/3)

Discard the answer with negative numbers, 

so solution is

length of side = 14 inches, length of altitude = 4 inches

Edwin
AnlytcPhil@aol.com


Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
A triangular sheet of paper is 28 square inches,
AREA=A=28=0.5*B*H WHERE B IS BASE AND H IS ALTITUDE.
one side of the triangle is 2 inches more than 3 times the length of the altitudeto that side.
3 TIMES ALTITUDE=3H
2 MORE =3H+2=B
AREA =A=28=0.5*(3H+2)H=1.5H^2+H..MULTIPLYING WITH 2 ..WE GET
3H^2+2H=56
3H^2+2H-56=0
3H^2+14H-12H-56=0
H(3H+14)-4(3H+14)=0
(H-4)(3H+14)=0
H-4=0...THAT IS H=4...3H+14=0..LEADS TO NEGATIVE VALUES..SO NOT CONSIDERED.
B=3H+2=3*4+2=14
Find the length of that side and the altitude to the side. please i donoknow what formula to use of how to solve. this is not form the texbook.