{"id":11391,"date":"2022-07-08T01:19:38","date_gmt":"2022-07-07T19:49:38","guid":{"rendered":"https:\/\/mathemerize.com\/?p=11391"},"modified":"2022-07-08T01:20:58","modified_gmt":"2022-07-07T19:50:58","slug":"a-vertical-pole-of-length-6-m-casts-a-shadow-4m-long-on-the-ground-and-at-the-same-time-a-tower-casts-a-shadow-28-m-long-find-the-height-of-tower","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/a-vertical-pole-of-length-6-m-casts-a-shadow-4m-long-on-the-ground-and-at-the-same-time-a-tower-casts-a-shadow-28-m-long-find-the-height-of-tower\/","title":{"rendered":"A vertical pole of length 6 m casts a shadow 4m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of tower."},"content":{"rendered":"
Given<\/strong> : Let AC be a tower casts a shadow BC = 28 m and DF = 6 m be a pole casts a shadow EF = 4m<\/p>\n To Find<\/strong> : Height of the tower<\/p>\n Procedure<\/strong> : Now, In triangle ABC and DEF<\/p>\n \\(\\angle\\) ACB = \\(\\angle\\) DEF\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (each 90)<\/p>\n At the same time the rays of sun have same inclination<\/p>\n \\(\\implies\\)\u00a0 \u00a0\\(\\angle\\) ABC = \\(\\angle\\) DEF<\/p>\n By AA similarity,<\/p>\n \\(\\triangle\\) ABC ~ \\(\\triangle\\) DEF<\/p>\n \\(\\implies\\)\u00a0 \\(AC\\over DF\\) = \\(BC\\over EF\\)<\/p>\n \\(\\implies\\)\u00a0 \\(AC\\over 6\\) = \\(28\\over 4\\)<\/p>\n \\(\\implies\\)\u00a0 AC = 42 m<\/p>\n Hence, the height of tower is 42 m<\/strong>.<\/p>\n","protected":false},"excerpt":{"rendered":" Solution : Given : Let AC be a tower casts a shadow BC = 28 m and DF = 6 m be a pole casts a shadow EF = 4m To Find : Height of the tower Procedure : Now, In triangle ABC and DEF \\(\\angle\\) ACB = \\(\\angle\\) DEF\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (each …<\/p>\n