The diagonals of a Trapezium ABCD intersect each other at O. If AB = 2CD then find the ratio of ar(AOB)/ar(COD).
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Given :-
- Diagonals of a trapezium ABCD with AB || CD, intersect each other at point 'O' .
- AB = 2CD .
To Find :-
- Area ∆AOB : Area ∆COD .
Solution :-
from image we have,
→ ABCD is a trapezium .
→ AB ∣∣ DC .
now, in ∆AOB and ∆COD , we have,
→ ∠ABO = ∠CDO (Alternate angles.)
→ ∠BAO = ∠DCO (Alternate angles.)
→ ∠AOB = ∠COD (Vertically opposite angles.)
so,
→ ∆AOB ~ ∆COD (By AAA Similarity.)
now, we know that,
- If two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides.
then,
→ Ar.∆AOB : Ar.∆COD = AB² : CD²
→ ∆AOB / ∆COD = (AB/CD)²
given that, AB = 2CD
→ Ar.∆AOB / Ar.∆COD = (2CD/CD)²
→ Ar.∆AOB / Ar.∆COD = (2/1)²
→ Ar.∆AOB / Ar.∆COD = 4 / 1
→ Ar.∆AOB : Ar.∆COD = 4 : 1 (Ans.)
Learn more :-
In ABC, AD is angle bisector,
angle BAC = 111 and AB+BD=AC find the value of angle ACB=?
https://brainly.in/question/16655884
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