Question

In the given figure, ∠ABC = ∠DCB = 90° AB = 6, DC = 8 then $\frac{\mathrm{A}\left(∆\mathrm{ABC}\right)}{\mathrm{A}\left(∆\mathrm{DCB}\right)}=?$

Answer 1

Given:
∠ABC = ∠DCB = 90°
AB = 6
DC = 8
$$\begin{gathered} \mathrm{Now},\frac{\mathrm{A}\left(∆\mathrm{ABC}\right)}{\mathrm{A}\left(∆\mathrm{DCB}\right)}=\frac{{\displaystyle \frac{1}{2}\times \mathrm{AB}\times \mathrm{BC}}}{{\displaystyle \frac{1}{2}\times \mathrm{DC}\times \mathrm{BC}}} \\ =\frac{{\displaystyle 6}}{{\displaystyle 8}} \\ =\frac{{\displaystyle 3}}{{\displaystyle 4}} \end{gathered}$$