{"id":11405,"date":"2022-07-08T22:24:38","date_gmt":"2022-07-08T16:54:38","guid":{"rendered":"https:\/\/mathemerize.com\/?p=11405"},"modified":"2022-07-08T22:24:43","modified_gmt":"2022-07-08T16:54:43","slug":"diagonals-of-trapezium-abcd-with-ab-dc-intersect-each-other-at-the-point-o-if-ab-2cd-find-the-ratio-of-the-areas-of-triangles-aob-and-cod","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/diagonals-of-trapezium-abcd-with-ab-dc-intersect-each-other-at-the-point-o-if-ab-2cd-find-the-ratio-of-the-areas-of-triangles-aob-and-cod\/","title":{"rendered":"Diagonals of trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2CD, find the ratio of the areas of triangles AOB and COD."},"content":{"rendered":"
In triangles AOB and COD,<\/p>\n
\\(\\angle\\) AOB = \\(\\angle\\) COD\u00a0 \u00a0 \u00a0 \u00a0(vertically opposite angles)<\/p>\n
\\(\\angle\\) OAB = \\(\\angle\\) OCD\u00a0 \u00a0 \u00a0 \u00a0 (corresponding angles)<\/p>\n
\\(\\therefore\\) By AA similarity,<\/p>\n
\\(\\triangle\\) AOB ~ \\(\\triangle\\) COD<\/p>\n
\\(\\implies\\)\u00a0 \\(area(\\triangle AOB)\\over area(\\triangle COD)\\) = \\({AB}^2\\over {DC}^2\\)<\/p>\n
\\(\\implies\\)\u00a0 \\(area(\\triangle AOB)\\over area(\\triangle COD)\\) = = \\({2DC}^2\\over {DC}^2\\) = \\(4\\over 1\\)<\/p>\n
Hence, Area of triangle AOB : Area of triangle COD = 4 : 1<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":" Solution : In triangles AOB and COD, \\(\\angle\\) AOB = \\(\\angle\\) COD\u00a0 \u00a0 \u00a0 \u00a0(vertically opposite angles) \\(\\angle\\) OAB = \\(\\angle\\) OCD\u00a0 \u00a0 \u00a0 \u00a0 (corresponding angles) \\(\\therefore\\) By AA similarity, \\(\\triangle\\) AOB ~ \\(\\triangle\\) COD \\(\\implies\\)\u00a0 \\(area(\\triangle AOB)\\over area(\\triangle COD)\\) = \\({AB}^2\\over {DC}^2\\) \\(\\implies\\)\u00a0 \\(area(\\triangle AOB)\\over area(\\triangle COD)\\) = = \\({2DC}^2\\over {DC}^2\\) = \\(4\\over …<\/p>\n