{"id":11412,"date":"2022-07-08T22:46:43","date_gmt":"2022-07-08T17:16:43","guid":{"rendered":"https:\/\/mathemerize.com\/?p=11412"},"modified":"2022-07-08T22:46:47","modified_gmt":"2022-07-08T17:16:47","slug":"d-e-f-are-the-mid-points-of-the-sides-bc-ca-and-ab-respectively-of-a-triangle-abc-determine-the-ratio-of-the-areas-of-triangle-def-and-triangle-abc","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/d-e-f-are-the-mid-points-of-the-sides-bc-ca-and-ab-respectively-of-a-triangle-abc-determine-the-ratio-of-the-areas-of-triangle-def-and-triangle-abc\/","title":{"rendered":"D, E, F are the mid-points of the sides BC, CA and AB respectively of a \\(\\triangle\\) ABC. Determine the ratio of the areas of \\(\\triangle\\) DEF and \\(\\triangle\\) ABC."},"content":{"rendered":"
Since D and E are the mid-points of the sides BC and CA respectively of \\(\\triangle\\) ABC.<\/p>\n
\\(\\therefore\\)\u00a0 DE || BA<\/p>\n
\\(\\implies\\)\u00a0 DE || FA\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0……….(1)<\/p>\n
Since D and F are the mid-points of the sides BC and AB respectively of \\(\\triangle\\) ABC. Therefore,<\/p>\n
DF || CA\u00a0 \\(\\implies\\)\u00a0 DF || AE\u00a0 \u00a0 \u00a0 \u00a0 \u00a0………..(2)<\/p>\n
From (1) and (2), we get AFDE is a parallelogram<\/p>\n
Similarly, BDEF is a parallelogram.<\/p>\n
In triangle DEF and ABC,<\/p>\n
\\(\\angle\\) FDE = \\(\\angle\\) A\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (opposite angles of ||gm AFDE)<\/p>\n
\\(\\angle\\) DEF = \\(\\angle\\) B\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (opposite angles of ||gm BDEF)<\/p>\n
\\(\\therefore\\)\u00a0 By AA similarity,<\/p>\n
\\(\\triangle\\) DEF ~ \\(\\triangle\\) ABC<\/p>\n
\\(\\implies\\)\u00a0 \\(area(\\triangle DEF)\\over area(\\triangle ABC)\\) = \\({DE}^2\\over {AB}^2\\) = \\(1\\over 4\\)<\/p>\n
(The areas of two similar triangles are in the ratio of the squares of the corresponding sides)<\/p>\n
Hence, Area of triangle DEF : Area of triangle ABC = 1 : 4<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":" Solution : Since D and E are the mid-points of the sides BC and CA respectively of \\(\\triangle\\) ABC. \\(\\therefore\\)\u00a0 DE || BA \\(\\implies\\)\u00a0 DE || FA\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0……….(1) Since D and F are the mid-points of the sides BC and AB respectively of \\(\\triangle\\) ABC. Therefore, DF || CA\u00a0 \\(\\implies\\)\u00a0 DF …<\/p>\n