4. Two isosceles triangles are possible with area of 120 square unit and length of edges integers. One of these two triangles has sides of lengths 1717 and 16. Determine the length of edges of second one.

[Hint: In ABC if AB=AC and AD is perpendicular to BC then BD=CD.]

Answer: The height of the isosceles triangle which has three sides of length 17,17,16, is =sqrt((17))−sqrt((8.0))=15 The area of this triangle is 120. Let x,y be the height and base of the new triangle respectively. if x=16/2=8 [Half of the base the first triangle] and y=15*2=30 [double of the height of the first triangle] then the area of the new triangle is 120 and the length of the two equal sides will be sqrt(15)+sqrt((8.0))=17 so the length of the sides of the new triangle are 17,17,30.