Question

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

Find:
(a) $\frac{\mathrm{Area}\left(∆\mathrm{AOB}\right)}{\mathrm{Area}\left(∆\mathrm{COD}\right)}$
(b) $\frac{\mathrm{Area}\left(∆\mathrm{AOD}\right)}{\mathrm{Area}\left(∆\mathrm{COD}\right)}$

Answer 1

Given: ABCD is a trapezium in which AB || CD.

The diagonals AC and BD intersect at O.

To prove:

(i)

(ii) If OA = 6 cm, OC = 8 cm

To find:

(a)

(b)

Construction: Draw a line MN passing through O and parallel to AB and CD

(i) Now in ΔAOB and ΔCOD
$$\begin{gathered} \angle \mathrm{OAB}=\angle \mathrm{OCD}\left(\mathrm{Alternate}\mathrm{angles}\right) \\ \angle \mathrm{OBA}=\angle \mathrm{ODC}\left(\mathrm{Alternate}\mathrm{angles}\right) \end{gathered}$$

(ii) (a)We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

(b)We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

$$\begin{gathered} \frac{\mathrm{ar}\left(∆\mathrm{AOD}\right)}{\mathrm{ar}\left(∆\mathrm{COD}\right)}={\left(\frac{\mathrm{OA}}{\mathrm{OC}}\right)}^2 \\ ={\left(\frac{6\mathrm{cm}}{8\mathrm{cm}}\right)}^2 \\ =\frac{9}{16} \end{gathered}$$