In triangle LMN, LM = 8cm, MN = 6cm and LMN = 90°X and Y are the midp

1.	In triangle LMN, LM = 8cm, MN = 6cm and LMN = 90°X and Y are the midpoints of MN and LN respectively.Find YXN and YN.
Answer» $\textbf{Converse of Basic proportionality theorem:}$ $\text{If a line divides two sides of a <klux>TRIANGLE</klux> in the same ratio,}$ $\text{then the line is parallel to the third side}$ $\textbf{Given:}$ $\text{In $\triangle$LMN, LM=8cm, MN=6 cm $\<klux>ANGLE</klux>{LMN}=90^{\circ}$}$ $\textbf{To <klux>FIND</klux>:}$ $\angle{YXN}\;\text{and}\;YN$ $\text{In right angled $\triangle$LMN, by pythagoras theorem}$ $LN^2=LM^2+MN^2$ $LN^2=8^2+6^2$ $LN^2=64+36$ $LN^2=100$ $LN=10\;\text{cm}$ $\text{Since X and Y are midpoints of MN and LN,}$ $MX=XN=3\;\text{cm}\;\text{and}\;NY=LY=5\;\text{cm}$ $\implies\dfrac{NX}{MX}=1\;\text{and}\;\dfrac{NY}{LY}=1$ $\implies\bf\dfrac{NX}{MX}=\dfrac{NY}{LY}$ $\text{By converse of basic proportionality theorem, }$ $\text{XY is parallel to ML}$ $\implies\bf\angle{YXN}=90^{\circ}$ $\text{($\because$ corresponding angles are equal})$ $\therefore\textbf{YN=5 cm and $\bf\angle{YXN}=90^{\circ}$}$ Find more: In triangle ABC angle BAC is 90 degree seg BL and seg CM are medians of triangle ABC. Then prove that 4( BL²+CM²)= 5 BC² brainly.in/question/8096479 The perimeter of an isosceles right-angled triangle is 6(√2+1) cm . Find its area. Please solve it with explanation. brainly.in/question/2120508