Draw a triangle LMN with <M=90* , MN=6cm, NL=8cm. Draw another triangle 4/7th its size.

​Please give me a rough sketch for the first triangle. How to draw a right triangle when hypotenuse and base is only given????

Dear student
$$\begin{gathered} \mathrm{Given}: \angle \mathrm{M}=90°,\mathrm{MN}=6\mathrm{cm} \mathrm{and} \mathrm{NL}=8\mathrm{cm} \\ \mathrm{So}, \mathrm{By} \mathrm{pythagoras} \mathrm{theorem} \\ {\mathrm{MN}}^2+{\mathrm{ML}}^2={\mathrm{LN}}^2 \\ \Rightarrow 6^2+{\mathrm{ML}}^2=8^2 \\ \Rightarrow \mathrm{ML}=\sqrt{64-36} \\ \Rightarrow \mathrm{ML}=\sqrt{28}=2\sqrt{7}\mathrm{cm} \end{gathered}$$

Draw a line MX and draw three equal radius arc on line MX that intersect at M1,M2,... and M7.As
   MM1 =M1M2 = M2M3=...=M6M7    (Because our scale factor is $\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$7$}\right.)$
2)darw any acute angle XMN and draw line MN as 12 cm. and ∠M = 90 and length of LM and LN area $2\sqrt{7}$ and 8
4) Join M7 to N and than draw a line from M2 as parallel to M3N that intersect our line MN at N'.
5) Draw line From N' as parallel to LN that intersect line LM at L' 
6) Join line N'L' and form a triangle L'MN'  that is our required triangle.
Regards