= 24. || In the given figure, AB DC, area of AAOD = 5 sq cm and area of ABCD = 8 sq cm. Calculate D C A B. (A) Find the area of AOCD: (a) 2 (b) 3 (c) 4 (d) 5 (B) Find the ratio of BO: OD: (a) 1:2 (b) 5:3 (c) 3:4 (d) 6:7 (C) Find the area of AOAB:
Answers
Answered by
0
Answer:
It is given that ABCD is a trapezium and AB∣∣DC.
In △AOB and △COD
∠ABO=∠CDO (Alternate angles)
∠BAO=∠DCO (Alternate angles)
∠AOB=∠COD (Vertically opposite angles)
Therefore, △ABC∼△COD
We know that the arc of similar triangles are proportional to squares of their corresponding altitude, therefore with Ar(△AOB)=84 cm
2
and AB=2CD, we have,
Ar(△COD)
Ar(△AOB)
=
CD
2
AB
2
⇒
Ar(△COD)
84
=
CD
2
4CD
2
⇒
Ar(△COD)
84
=4
⇒Ar(△COD)=
4
84
⇒Ar(△COD)=21
Hence, area of △COD is 21 cm
2
.
Similar questions