{"id":11511,"date":"2022-07-12T23:13:37","date_gmt":"2022-07-12T17:43:37","guid":{"rendered":"https:\/\/mathemerize.com\/?p=11511"},"modified":"2022-07-12T23:13:41","modified_gmt":"2022-07-12T17:43:41","slug":"in-an-equilateral-triangle-prove-that-three-times-the-square-of-one-side-is-equal-to-four-times-the-square-of-one-its-altitudes","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/in-an-equilateral-triangle-prove-that-three-times-the-square-of-one-side-is-equal-to-four-times-the-square-of-one-its-altitudes\/","title":{"rendered":"In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one its altitudes."},"content":{"rendered":"
Let ABC be and equilateral triangle and let AD \\(\\perp\\) BC.<\/p>\n
In \\(\\triangle\\) ADB and ADC, we have :<\/p>\n
AB = AC\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (given)<\/p>\n
AD = AD\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(common side of triangle)<\/p>\n
and \\(\\angle\\) ADB = \\(\\angle\\) ADB\u00a0 \u00a0 \u00a0 \u00a0 (each 90)<\/p>\n
By RHS criteria of congruence, we have :<\/p>\n
\\(\\triangle\\)\u00a0 ADB \\(\\cong\\) \\(\\triangle\\) ADC<\/p>\n
So, BD = DC or\u00a0 BD = DC = \\(1\\over 2\\) BC\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0……..(1)<\/p>\n
Since \\(\\triangle\\)\u00a0 ADB is a right triangle, angled at D, by Pythagoras theorem, we have :<\/p>\n
\\({AB}^2\\) = \\({AD}^2\\) + \\({BD}^2\\)<\/p>\n
\\({AB}^2\\) = \\({AD}^2\\) + \\(({1\\over 2}BC)^2\\)\u00a0 \u00a0 \u00a0 \u00a0 (from 1)<\/p>\n
\\({AB}^2\\) = \\({AD}^2\\) + \\({1\\over 4}{BC}^2\\)<\/p>\n
\\({AB}^2\\) = \\({AD}^2\\) + \\({AB}^2\\over 4\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (\\(\\therefore\\)\u00a0 BC = AB)<\/p>\n
\\({3\\over 4}{AB}^2\\) = \\({AD}^2\\)\u00a0 or\u00a0 \\(3{AB}^2\\) = \\(4{AD}^2\\)<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":" Solution : Let ABC be and equilateral triangle and let AD \\(\\perp\\) BC. In \\(\\triangle\\) ADB and ADC, we have : AB = AC\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (given) AD = AD\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(common side of triangle) and \\(\\angle\\) ADB = \\(\\angle\\) ADB\u00a0 \u00a0 \u00a0 \u00a0 (each 90) By RHS criteria of …<\/p>\n