Ex.6.4 Q3 Triangles Solution - NCERT Maths Class 10

Go back to  'Ex.6.4'


In Fig. below, \(ABC\) and \(DBC\) are two triangles on the same base \(BC.\) If \(AD\) intersects \(BC\) at \(O,\) show that \(\begin{align}\frac{{\rm {Area\, of\,}}{\Delta ABC}}{{\rm {Area\,of\,}}{\Delta DBC}} = \frac{{AO}}{{DO}}\end{align}\)


Text Solution



AA criterion


In \(\,\Delta ABC\)

Draw \(\,AM \bot BC\)

In\(\,\,\Delta DBC\)

Draw \(\,DN \bot BC\)

Now in \(\,\,\Delta AOM,\,\,\Delta DON\)

\[\begin{align}{\angle AMO}&={\angle D N O}={90^{\circ}} \\ {\angle A O M}&={\angle D O N(\text { Vertically opposite angles })} \\ {\Rightarrow} \qquad {\Delta A O M} &\sim {\Delta D O N \text { (AA criterion) }} \\ {\Rightarrow \qquad\frac{A M}{D N}}&={\frac{O M}{O N}=\frac{A O}{D O}  \ldots (1)}\end{align}\]


\[\begin{align} Area\,\,of\Delta ABC&=\frac{1}{2}\times base\times height \\ & =\frac{1}{2}\times BC\times AM \\ Area\,\,of\,\Delta DBC& =\frac{1}{2}\times BC\times DN \\\frac{Area\,of\,\Delta ABC}{Area\,of\,\Delta DBC}&=\frac{\frac{1}{2}\times BC\times AM}{\frac{1}{2}\times BC\times DN}\\\frac{Area\,of\,\Delta ABC}{Area\,of\,\Delta DBC}&=\frac{AM}{DN}\\\frac{Area\,of\,\Delta ABC}{Area\,of\,\Delta DBC}&=\frac{AO}{DO} \quad (\text{from}\,\text{(1)}) \end{align}\]

Learn math from the experts and clarify doubts instantly

  • Instant doubt clearing (live one on one)
  • Learn from India’s best math teachers
  • Completely personalized curriculum