A triangle has sides 10, 17, and 21. A square is inscribed in the triangle. One side of the square lies on the longest side of the triangle. The other two vertices of the square touch the two shorter sides of the triangle. What is the length of the side of the square? Choices:
A) 148 / 29  B) 168 / 29  C) 168 / 26  D) 158 / 29  E) 198 / 29

Rationale: by heron ' s formula , the area , a , of a triangle with sides a , b , c is given by a = square root [ s ( s − a ) ( s − b ) ( s − c ) ] , where s = ½ ( a + b + c ) is the semi - perimeter of the triangle . then s = ½ ( 10 + 17 + 21 ) = 24 , and a = 84 . a square inscribed in a 10 by 17 by 21 triangle , with perpendicular dropped onto the side of length 21 . now drop a perpendicular of length h onto the side of length 21 . we also have a = ½ × base × perpendicular height . hence a = 21 h / 2 = 84 , from which h = 8 . notice that the triangle above the square is similar to the whole triangle . ( this follows because its base , the top of the square , is parallel to the base of the whole triangle . ) let the square have side of length d . considering the ratio of altitude to base in each triangle , we have 8 / 21 = ( 8 − d ) / d = 8 / d − 1 . therefore the length of the side of the square is 168 / 29 . answer : b
Answer: B