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UPSC 2015 CDS II Math Paper
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© examsnet.com
Question : 83
Total: 100
In the above figure, ABCD is a parallelogram. P is a point on BC such that PB : PC = 1 : 2 . DP and AB when both* produced meet at Q. If area of triangle BPQ is 20 square unit, the area of triangle DCP is
[2015 CDS-II]
20 square unit
30 square unit
40 square unit
None of the above
Validate
Solution:
In ΔDCP and ΔBPQ,
∠DPC = ∠QPB (Vertically opposite angles)
∠PDC = ∠PQB (Alternate angles)
∴ ΔDCP ≈ ΔBPQ (AA similarity)
⇒
a
r
(
Δ
D
C
P
)
a
r
(
Δ
B
P
Q
)
=
(
P
C
P
B
)
2
⇒
a
r
(
Δ
D
C
P
)
20
=
(
2
)
2
=
4
⇒
a
r
(
Δ
D
C
P
)
=
20
×
4
= 80 square units
© examsnet.com
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