in triangle ABC right angled at C altitude CH and median CM trisect the right angle. if the area of triangle CHM is k then the area of triangle ABC is

Given: CM is median and CH is the altitude of the triangle ABC, CH and CM trisect the right angle, area of triangle CHM is equal to k.

To find: Area of triangle ABC

$$\begin{gathered} \mathrm{As} \mathrm{CH} \mathrm{and} \mathrm{CM} \mathrm{trisect} \mathrm{the} \mathrm{right} \mathrm{angle} \mathrm{ACB} \\ \therefore \angle \mathrm{ACH}=\angle \mathrm{HCM}=\angle \mathrm{MCB}=30° \\ \mathrm{Considering} △\mathrm{CHA} \mathrm{and} △\mathrm{CHM} \\ \angle \mathrm{CHA}=\angle \mathrm{CHM}=90° \\ \angle \mathrm{ACH}=\angle \mathrm{MCH}=30° \\ \therefore △\mathrm{CHA}~△\mathrm{CHM} \\ \Rightarrow {\left(\frac{\mathrm{CH}}{\mathrm{CH}}\right)}^2={\left(\frac{\mathrm{HA}}{\mathrm{HM}}\right)}^2={\left(\frac{\mathrm{CA}}{\mathrm{CM}}\right)}^2=\frac{\mathrm{Area}\left(△\mathrm{CHA}\right)}{\mathrm{Area}\left(△\mathrm{CHM}\right)} \\ \Rightarrow \frac{\mathrm{Area}\left(△\mathrm{CHA}\right)}{\mathrm{Area}\left(△\mathrm{CHM}\right)}=1 \\ \Rightarrow \mathrm{Area}\left(△\mathrm{CHA}\right)=\mathrm{Area}\left(△\mathrm{CHM}\right)=\mathrm{k} \\ \therefore \mathrm{Area}\left(\mathrm{CMA}\right)=2\mathrm{k} \\ \mathrm{As} \mathrm{we} \mathrm{know} \mathrm{that} \mathrm{median} \mathrm{of} \mathrm{triangle} \mathrm{divide} \mathrm{the} \mathrm{triangle} \mathrm{into} \mathrm{equal} \mathrm{areas}. \\ \mathrm{Area}\left(\mathrm{CMA}\right)=\mathrm{Area}\left(\mathrm{CMB}\right)=2\mathrm{k} \\ \Rightarrow \mathrm{Area}\left(\mathrm{ABC}\right)=2\mathrm{k}+2\mathrm{k}=4\mathrm{k} \end{gathered}$$