![\"triangle\"](\"https:\/\/i0.wp.com\/mathemerize.com\/wp-content\/uploads\/2022\/07\/similartriangles6-5-11.jpeg?resize=230%2C202&ssl=1\")
<\/p>\nLet the first plane starts from O and goes upto A towards north.<\/p>\n
Where OA = (\\(1000 \\times {3\\over 2}\\)) km = 1500 km<\/p>\n
Let the second plane starts from O at the same time and goes upto B towards west, where OB = (\\(1200 \\times {3\\over 2}\\)) km = 1800 km<\/p>\n
According to the question, the required distance = BA.<\/p>\n
In right triangle ABC, by Pythagoras theorem, we have :<\/p>\n
\\({AB}^2\\) = \\({OA}^2\\) + \\({OB}^2\\)<\/p>\n
= \\((1500)^2\\) + \\((1800)^2\\)<\/p>\n
= 2250000 + 3240000<\/p>\n
= 5490000<\/p>\n
AB = \\(3\\times 100\\sqrt{61}\\) = \\(300\\sqrt{61}\\).<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"
Solution : Let the first plane starts from O and goes upto A towards north. Where OA = (\\(1000 \\times {3\\over 2}\\)) km = 1500 km Let the second plane starts from O at the same time and goes upto B towards west, where OB = (\\(1200 \\times {3\\over 2}\\)) km = 1800 km According …<\/p>\n