Lost and Found?
As soon as Agent Pathfinder climbed down the ladder (Agent Pathfinder finds a path!!!), her captors sprang out of nowhere and captured her again. She writhed, protested and pleaded, but to no avail. She was taken to a new room as the previous one wasn't "secure" enough. So, they trapped her in a suspended square cage in a triangular room of dimensions 10 metres, 17 metres and 21 metres respectively. The cage was suspended in such a way that two vertices of the cage touched the two shorter walls while one side of the cage touched the room's longest wall. The guards sitting outside the door tried to kill time by calculating the length of the side of the cage. Help them. Also provide a detailed explanation along with your answer.
Views: 937
Solution:
5.73 metres (approx). According to 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[= (1/2)*(a + b + c)] is the semi-perimeter of the triangle. Thus, s = (1/2)*(10 + 17 + 21) = 24, and A = 84. Let's draw a perpendicular of length h to the side of length 21 m from the vertex opposite it. We know that, area of a triangle, A = (1/2) * base * height Hence, A = (21 * h)/2 = 84 Thus, h = 8. Now, the top of the square, is parallel to the base of the whole triangle. Hence, the triangle above the square is similar to the whole triangle. Let the side of the square be d metres long. By similarity, we can consider the ratio of altitude to base in each triangle Thus, we get 8/21 = (8 − d)/d = 8/d − 1. Therefore, the length of the side of the square is 168/29 = 5.73 m (approx).