Question

ABCD is a trapezium in which AB||CD. the diagonals AC & BD intersect each other at O. prove:

1.triangle AOB ~ triangleCOD

2. if OA=6cm , OC=8cm, find

(a)ar(triangle AOB) / ar(triangle COD)

(b)ar(triangle AOD) / ar(triangle COD)


Please really urgent I want the answer of second part....

Answer 1

ABCD is a trapezium.

In which diagonal AC and BD intersect each other at O.

OB = OD

AO = OC

(1)To prove :- Triangle AOB and COD are similar.

angle AOB = DOC. [vertically opposite angle].
Side AB = CD.

Angle OAB = OCD. [ Each equal to 90°]

BY similarity criteria ASA

Triangle AOB similar to Triangle COD.

(2) OA= 6cm
and OC = 8cm
Let the base be X cm

Therefore,

Area of Triangle AOB/ Area of triangle COD = 1/2×X×6/1/2×Xm × 8.

Ar(AOB)/Ar(COD)= 6/8= 3/4

Answer 2

Here, OA = OC and OB = OD

In triangle AOB and triangle COD

Angle AOB = angle COD ( Vertically opposite angle)

Angle OAB = Angle OCD ( alternate angles)

Therefore triangle AOB similar to triangle COD ( by AA)

A. Ar( triangle AOB)/ ar( triangle COD)

= OA square / OC square = 6 square / 8 square = 36/64 = 9/16

B. Ar( triangle AOD) /Ar( COD) = 1/2 × AO × DP / 1/2 × CO × DP = AO/CO = 6/8 = 3/4