{"id":11467,"date":"2022-07-12T12:40:10","date_gmt":"2022-07-12T07:10:10","guid":{"rendered":"https:\/\/mathemerize.com\/?p=11467"},"modified":"2022-07-12T12:40:14","modified_gmt":"2022-07-12T07:10:14","slug":"prove-that-the-altitude-of-an-equilateral-triangle-of-side-2a-is-sqrt3-a","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/prove-that-the-altitude-of-an-equilateral-triangle-of-side-2a-is-sqrt3-a\/","title":{"rendered":"Prove that the altitude of an equilateral triangle of side 2a is \\(\\sqrt{3} a\\)."},"content":{"rendered":"
Given<\/strong> : \\(\\triangle\\) ABC, in which each side is of length 2a.<\/p>\n To Find<\/strong> : AD (altitude)<\/p>\n In \\(\\triangle\\) ADB and \\(\\triangle\\) ADC,<\/p>\n AD = AD\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (common)<\/p>\n \\(\\angle\\) 1 = \\(\\angle\\) 2\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(90 each)<\/p>\n AB = AC\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(given)<\/p>\n By RHS,<\/p>\n \\(\\triangle\\) ADB \\(\\cong\\) \\(\\triangle\\)\u00a0 ADC<\/p>\n So,\u00a0 BD = DC\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(By C.P.C.T)<\/p>\n \\(\\implies\\) BD = DC = a<\/p>\n Now, in \\(\\triangle\\) ADB,<\/p>\n \\({AD}^2\\) + \\({BD}^2\\) = \\({AB}^2\\)<\/p>\n \\(\\implies\\)\u00a0 \\({AD}^2\\) + \\(a^2\\) = \\((2a)^2\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (By Pythagoras Theorem)<\/p>\n \\({AD}^2\\) = \\(3a^2\\)\u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a0AD = \\(\\sqrt{3} a\\)<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":" Solution : Given : \\(\\triangle\\) ABC, in which each side is of length 2a. To Find : AD (altitude) In \\(\\triangle\\) ADB and \\(\\triangle\\) ADC, AD = AD\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (common) \\(\\angle\\) 1 = \\(\\angle\\) 2\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(90 each) AB = AC\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 …<\/p>\n