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Construct a trinagle with sides 5 cm 6cm and 7 cm then another triangle whos sides 7/5 of thia

Answer» To construct: To construct a triangle of sides 5 cm, 6 cm and 7 cm and then a triangle similar to it whose sides are of\xa0{tex}\\frac{7}{5}{/tex} the corresponding sides of the first triangle.Steps of construction:\tDraw a triangle ABC of sides 5 cm, 6 cm and 7 cm.\tFrom any ray BX, making an acute angle with BC on the side opposite to the vertex A.\tLocate 7 points B1, B2, B3, B4, B5, B6 and B7 on BX such that BB1 = B1 B2 = B2 B3 = B3 B4 = B4 B5 = B5 B6 = B6 B7.\tJoin B5 C and draw a line through the point B7, draw a line parallel to B5 C intersecting BC at the point C\'.\tDraw a line through C\' parallel to the line CA to intersect BA at A\'.\tThen, A\'BC\' is the required triangle.\tJustification :\t{tex}\\because CA||C\'A\'{/tex} [By construction]\t{tex}\\therefore {/tex}\xa0{tex}\\triangle ABC \\sim \\triangle A\'BC\'{/tex}\xa0[AA similarity]\t{tex}\\therefore \\frac{{A\'B}}{{AB}} = \\frac{{A\'C\'}}{{AC}} = \\frac{{BC\'}}{{BC}}{/tex}\xa0[By Basic Proportionality Theorem]\t{tex}\\because {B_7}C\'||{B_5}C{/tex}\xa0[By construction]\t{tex}\\therefore \\triangle B{B_7}C\' \\sim \\triangle B{B_5}C{/tex}\xa0[AA similarity]\tBut {tex}\\frac{{B{B_5}}}{{B{B_7}}} = \\frac{5}{7}{/tex}\xa0[By construction]\tTherefore, {tex}\\frac{{BC}}{{BC\'}} = \\frac{5}{7} \\Rightarrow \\frac{{BC\'}}{{BC}} = \\frac{7}{5}{/tex}\t{tex}\\therefore \\frac{{AB}}{{AB}} = \\frac{{A\'C\'}}{{AC}} = \\frac{{BC\'}}{{BC}} = \\frac{7}{5}{/tex}


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