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Question Number 67969 by behi83417@gmail.com last updated on 02/Sep/19

Two triangles △_1 and △_2 are given,such that length of sides of triangle 1,are equail to length of medians of triangle 2. 1.find the ratio of areas of triangles. 2.given that small side of △_1 , be equail to:(√2) and one angle be:90^• . find at least one angle of △_2 . 3.solve part#2,if replace: △_2 with: △_1 . 4.solve part#2,if great side of:△_1 ,be equail to :(√2).

$$\boldsymbol{\mathrm{Two}}\:\boldsymbol{\mathrm{triangles}}\:\bigtriangleup_{\mathrm{1}} \:\boldsymbol{\mathrm{and}}\:\bigtriangleup_{\mathrm{2}} \:\boldsymbol{\mathrm{are}}\:\boldsymbol{\mathrm{given}},\boldsymbol{\mathrm{such}}\: \\ $$$$\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{length}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{sides}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{triangle}}\:\mathrm{1},\boldsymbol{\mathrm{are}}\: \\ $$$$\boldsymbol{\mathrm{equail}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{length}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{medians}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{triangle}}\:\mathrm{2}. \\ $$$$\mathrm{1}.\boldsymbol{\mathrm{find}}\:\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{ratio}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{areas}}\:\boldsymbol{\mathrm{of}}\:\:\boldsymbol{\mathrm{triangles}}. \\ $$$$\mathrm{2}.\boldsymbol{\mathrm{given}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{small}}\:\boldsymbol{\mathrm{side}}\:\boldsymbol{\mathrm{of}}\:\bigtriangleup_{\mathrm{1}} ,\:\boldsymbol{\mathrm{be}}\:\boldsymbol{\mathrm{equail}}\:\boldsymbol{\mathrm{to}}:\sqrt{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{one}}\:\boldsymbol{\mathrm{angle}}\:\boldsymbol{\mathrm{be}}:\mathrm{90}^{\bullet} . \\ $$$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{at}}\:\boldsymbol{\mathrm{least}}\:\boldsymbol{\mathrm{one}}\:\boldsymbol{\mathrm{angle}}\:\boldsymbol{\mathrm{of}}\:\bigtriangleup_{\mathrm{2}} . \\ $$$$\mathrm{3}.\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{part}}#\mathrm{2},\boldsymbol{\mathrm{if}}\:\boldsymbol{\mathrm{replace}}:\:\bigtriangleup_{\mathrm{2}} \boldsymbol{\mathrm{with}}:\:\bigtriangleup_{\mathrm{1}} . \\ $$$$\mathrm{4}.\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{part}}#\mathrm{2},\boldsymbol{\mathrm{if}}\:\boldsymbol{\mathrm{great}}\:\boldsymbol{\mathrm{side}}\:\boldsymbol{\mathrm{of}}:\bigtriangleup_{\mathrm{1}} ,\boldsymbol{\mathrm{be}}\:\boldsymbol{\mathrm{equail}}\: \\ $$$$\boldsymbol{\mathrm{to}}\::\sqrt{\mathrm{2}}. \\ $$

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