Solutions To Yesterday’s Hard Geometry Problems Today I would like to post solutions to the three hard geometry questions I posted yesterday. If you still want further explanation after reading the below solutions please do not hesitate to ask. Level 5 Geometry Problems The lengths of the sides of a triangle are x, 16 and 31, where x is the shortest side. If the triangle is not isosceles, what is a possible value of x? Solution: By the triangle rule, x lies between 31 – 16 = 15 and 31 + 16 = 47. That is, we have 15 < x < 47. But we are also given that x is the length of the shortest side of the triangle. So x < 16.Therefore we can grid in any number between 15 and 16. For example, we can grid in 15.1. For more information on the triangle rule see the following article: The Triangle Rule In the figure above, if AB = 4, BC = 24, and AD = 26, then CD = Solution: The problem becomes much simpler if we “move” BC to the left and AB to the bottom as shown below. We now have a single right triangle and we can either use the Pythagorean Theorem, or better yet notice that 26 = (13)(2) and 24 = (12)(2). Thus the other leg of the triangle is (5)(2) = 10. So we see that CD must have length 10 – 4 = 6. Remark: If we didn’t notice that this was a multiple of a 5-12-13 triangle, then we would use the Pythagorean Theorem as follows. (x + 4)2 + 242 = 262 (x + 4)2 + 576 = 676 (x + 4)2 = 100 x + 4 = 10 x = 6 The figure above shows a right circular cylinder with diameter 6 and height 9. If point O is the center of the top of the cylinder and B lies on the circumference of the bottom of the cylinder, what is the straight-line distance between O and B? Solution: We draw a right triangle inside the cylinder as follows: Note that the bottom leg of the triangle is equal to the radius of the circle (not the diameter) which is why it is 3 and not 6. We can now use the Pythagorean Theorem to find x. x2 = 32 + 92 = 9 + 81 = 90 Taking the square root gives x approximately equal to 9.4868. So we grid in 9.48 or 9.49. More Geometry Practice Problems For more hard geometry problems like these, each with several fully explained solutions, check out the Get 800 collection of test prep books. If you think your friends would like to try these problems, please share: Speak to you soon! Comments comments