Question

ABCD is a trapezium in which AB $\parallel$ CD. The diagonals AC and BD intersect at O. Prove that:
(i) $△AOB\sim △COD$ (ii) If OA = 6 cm, OC = 8cm,

Find:
(a) $\frac{Area\left(△AOB\right)}{Area\left(△COD\right)}$

(b) $\frac{Area\left(△AOD\right)}{Area\left(△COD\right)}$

Answer 1

$ABCD$ is a trapezium in which $AB\parallel CD$.

The diagonals $AC$ and $BD$ intersect at $O$. $OA=6cm$ and $OC=8cm$

In $△AOB$ and $△COD$,

$\angle AOB=\angle COD$ [ Vertically opposite angles ]

$\angle OAB=\angle OCD$ [ Alternate angles ]

$\therefore$ $△AOB\sim △COD$ [ By AA similarity ]

$\left(a\right)$ $\frac{ar\left(△AOB\right)}{ar\left(△COD\right)}=\frac{O{A}^2}{O{C}^2}$ [ By area theorem ]

$\Rightarrow$ $\frac{ar\left(△AOB\right)}{ar\left(△COD\right)}=\frac{(6{)}^2}{(8{)}^2}$

$\Rightarrow$ $\frac{ar\left(△AOB\right)}{ar\left(△COD\right)}=\frac{36}{64}=\frac{9}{16}$

$\left(b\right)$ Draw $DP\perp AC$

$\frac{ar\left(△AOD\right)}{ar\left(△COD\right)}=\frac{\frac{1}{2}\times AO\times DP}{\frac{1}{2}\times CO\times DP}$ [ By area theorem ]

$\Rightarrow$ $\frac{ar\left(△AOD\right)}{ar\left(△COD\right)}=\frac{AO}{CO}$

$\Rightarrow$ $\frac{ar\left(△AOD\right)}{ar\left(△COD\right)}=\frac{6}{8}=\frac{3}{4}$