{"id":11414,"date":"2022-07-08T22:51:07","date_gmt":"2022-07-08T17:21:07","guid":{"rendered":"https:\/\/mathemerize.com\/?p=11414"},"modified":"2022-07-08T22:51:11","modified_gmt":"2022-07-08T17:21:11","slug":"the-areas-of-two-similar-triangles-are-in-the-ratio-of-the-square-of-the-corresponding-medians","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/the-areas-of-two-similar-triangles-are-in-the-ratio-of-the-square-of-the-corresponding-medians\/","title":{"rendered":"The areas of two similar triangles are in the ratio of the square of the corresponding medians."},"content":{"rendered":"
Given<\/strong> : \\(\\triangle\\) ABC ~ \\(\\triangle\\) DEF and AP, DQ are their medians.<\/p>\n To Prove<\/strong> : \\(area(\\triangle ABC)\\over area(\\triangle DEF)\\) = \\({AP}^2\\over {DQ}^2\\)<\/p>\n Proof<\/strong> : Since the ratio of the area of two similar triangles is equal to the ratio of the squares of any two corresponding sides.<\/p>\n \\(\\therefore\\)\u00a0 \\(area (\\triangle ABC)\\over area (\\triangle DEF)\\) = \\({AB}^2\\over {DE}^2\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ………..(1)<\/p>\n Now,\u00a0 \\(\\triangle\\) ABC ~ \\(\\triangle\\) DEF<\/p>\n \\(\\implies\\)\u00a0 \\(AB\\over DE\\) = \\(BC\\over EF\\) = \\(2BP\\over 2EQ\\) = \\(BP\\over EQ\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0………….(2)<\/p>\n Thus, \\(AB\\over DE\\) = \\(BP\\over EQ\\) and \\(\\angle\\) B = \\(\\angle\\) E\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(because \\(\\triangle\\) ABC ~ \\(\\triangle\\) DEF)<\/p>\n By SAS similarity,<\/p>\n \\(\\triangle\\) APB ~ \\(\\triangle\\) DQE<\/p>\n So, \\(BP\\over EQ\\) = \\(AP\\over DQ\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0……….(3)<\/p>\n From (2) and (3), we get<\/p>\n \\(AB\\over DE\\) = \\(AP\\over DQ\\)\u00a0 \u00a0 \u00a0\\(\\implies\\)\u00a0 \\({AB}^2\\over {DE}^2\\) = \\({AP}^2\\over {DQ}^2\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ………(4)<\/p>\n Substituting the value of \\({AB}^2\\over {DE}^2\\) from (4) in (1), we get<\/p>\n \\(area (\\triangle ABC)\\over area (\\triangle DEF)\\) = \\({AP}^2\\over {DQ}^2\\)<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":" Solution : Given : \\(\\triangle\\) ABC ~ \\(\\triangle\\) DEF and AP, DQ are their medians. To Prove : \\(area(\\triangle ABC)\\over area(\\triangle DEF)\\) = \\({AP}^2\\over {DQ}^2\\) Proof : Since the ratio of the area of two similar triangles is equal to the ratio of the squares of any two corresponding sides. \\(\\therefore\\)\u00a0 \\(area (\\triangle ABC)\\over area …<\/p>\n