In ∆abc with m∠c = 90° the sides satisfy the ratio bc: ac: ab = 4: 3: 5. if the side with middle length is 12 cm, find: 1) the perimeter of ∆abc; 2) the area of ∆abc; 3) the height to the hypotenuse. walk me through the question. 20 points.

There are (as with most math problems) at least a couple of different ways you can approach this. I prefer to think in terms of scale factors.

1) The perimeter of the 4:3:5 triangle is 4+3+5 = 12. The middle length of the 4:3:5 triangle is 4. The middle length of ∆ABC is 12, so is a factor of 12/4 = 3 longer. The perimeter of ∆ABC is a factor of 3 longer than 12, so is 3*12 = 36.

The perimeter of ∆ABC is 36 cm.

2) The area of a right triangle is half the product of the leg lenghts. For the 4:3:5 triangle, that is
  area = (1/2)*4*3 = 6 units²
∆ABC has leg lengths that are a factor of 3 longer, so its area will be the square of that factor times the area of the 4:3:5 triangle.
  area of ∆ABC = 3²×6 = 54 . . . . cm²

The area of ∆ABC is 54 cm².


We have already worked out the scale factor to be 3, so we know the side lengths of ∆ABC are (4×3):(3×3):(5×3) = 12:9:15. Then the area of ∆ABC is
  area ∆ABC = (1/2)(12 cm)(9 cm) = 54 cm².

Yes you do, so £1973.09 - £1473.50 = £499.59 so tim overpaid £499.59 .hope this ; )