if AD = 12 and AC= 4y-36, find the value of y. then find AC and DC

Question

its very confusing and i dont understand

Answer 1

This can be a trigonometry question, that sum of the two sides must be greater than the third side and difference of the two sides must be less than the third side in order to be a triangle

 

AD = 12 and AC= 4y-36, 

Let DC = x,

 

then AC-AD < DC < AC+AD, or 4y - 48 < x < 4y - 24 ------{1}

however, x > 0,  so the left side of {1} 4y-48 >0, or y>12  ------{2}

 

and,   AD + DC > AC, or 12 + x > 4y-36

  simplify,  x - 4y > -48 ------ {3}

and   AC + DC > AD, or 4y- 36 + x > 12

  simplify,  x + 4y > 48 ------ {4}

 

Subtract {3} from {2},  -4y > -96, or y < 24  ------{5}

From {2} and {5}, 12 < y < 24 ------{6}

 

for any value y in {6} and x value x in {1} be the side of the triangle.