Draw a line segment of length 7cm . Find a point p on it which divide

1.	Draw a line segment of length 7cm . Find a point p on it which divide it in the ratio 3:5
Answer» We have to draw a line segment of length 7 cm.Then,we have to find a point P on it, which divides it in the ratio 3 : 5.Steps of construction:\tDraw a line segment {tex}AB=7{/tex} cm.\tDraw a ray AX, making an acute {tex}\\angle BAX{/tex} with AB.\tMark {tex}3+5=8{/tex} points, i.e, {tex}A_1, A_2, A_3, A_4 ...A_8{/tex} on AX, such that {tex}AA_1 = A_1A_2 = A_2A_3 = A_3A_4 ... = A_7A_8{/tex}\tJoin A{tex}_8{/tex}B\tFrom A3, draw {tex}A_3P \|\| A_8B{/tex} which intersects AB at point P [ by making an angle at A3 equal to {tex}\\angle AA_8B{/tex}\xa0\tThen, P is the point on AB which divides it in the ratio 3:5. So, {tex}AP : PB = 3:5{/tex}Justification: In {tex}\\triangle ABA_8{/tex}, we have {tex}A_3P \|\| A_8B{/tex}\xa0{tex}\\therefore \\frac{AP}{PB}=\\frac{AA_3}{A_3A_8}{/tex} [ by basic proportionality theorem]By construction, {tex}\\frac{AA_3}{A_3A_8}=\\frac{3}{5}{/tex}\xa0Hence, {tex}\\frac{AP}{PB}=\\frac{3}{5}{/tex}

Draw a line segment of length 7cm . Find a point p on it which divide it in the ratio 3:5