the trapezium intersects the quadrilateral abcd at ab || cd and at point p. if ab = 10 cm, cd = 4 cm, ac = 7 cm and bd = 10.5 cm, find pa, pb, pc and pd.
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Answer:
21 cm
Step-by-step explanation:
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
.
Solve any question of Triangles with:-
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