{"id":11484,"date":"2022-07-12T13:01:12","date_gmt":"2022-07-12T07:31:12","guid":{"rendered":"https:\/\/mathemerize.com\/?p=11484"},"modified":"2022-07-12T13:01:16","modified_gmt":"2022-07-12T07:31:16","slug":"a-guy-wire-attached-to-a-vertical-pole-of-height-18-m-is-24-m-long-and-has-a-stake-attached-to-the-other-end-how-far-from-the-base-of-the-pole-should-the-stake-be-driven-so-that-the-wire-will-be-taut","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/a-guy-wire-attached-to-a-vertical-pole-of-height-18-m-is-24-m-long-and-has-a-stake-attached-to-the-other-end-how-far-from-the-base-of-the-pole-should-the-stake-be-driven-so-that-the-wire-will-be-taut\/","title":{"rendered":"A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut ?"},"content":{"rendered":"
Let AB = 24 m be guy wire attached to a vertical pole BC of height 18 m. To keep the wire taut, let it be fixed to stake at A. Then, ABC is a right angled triangle at C.<\/p>\n
\\(\\therefore\\)\u00a0 \\({AB}^2\\) = \\({AC}^2\\) + \\({BC}^2\\)<\/p>\n
So, \\({24}^2\\) = \\({AC}^2\\) + \\({18}^2\\)<\/p>\n
\\(\\implies\\) \\({AC}^2\\) = 576 – 324<\/p>\n
\\(\\implies\\)\u00a0 \\({AC}^2\\) = 252<\/p>\n
\\(\\implies\\)\u00a0 AC = \\(6\\sqrt{7}\\)<\/p>\n
Hence, the stake may be placed at a distance of \\(6\\sqrt{7}\\) m from the base of pole.<\/p>\n","protected":false},"excerpt":{"rendered":"
Solution : Let AB = 24 m be guy wire attached to a vertical pole BC of height 18 m. To keep the wire taut, let it be fixed to stake at A. Then, ABC is a right angled triangle at C. \\(\\therefore\\)\u00a0 \\({AB}^2\\) = \\({AC}^2\\) + \\({BC}^2\\) So, \\({24}^2\\) = \\({AC}^2\\) + \\({18}^2\\) \\(\\implies\\) …<\/p>\n