Question

in the figure ABCD is a parallelogram P is a point on BC such that Bp:pc=1:2 equal produced meets a AB produced at Q given the area of triangle CPQequal to 20 CM square then calculate the area triangle CDP and area of area of parallelogram ABCD

Answer 1

First prove triangle aqb
Then compare it with parallelogram abcd
Then conclude both

Answer 2

Answer:

Step-by-step explanation:

BP:PC=1:2

Area of triangle CPQ=20cm^2

Ar(ΔCPQ)/Ar(ΔBPQ)=2/1

=20/Ar(ΔBPQ)=2/1

=20/2=Ar(ΔBPQ)=10 cm^2

Ar(ΔADC)=Ar(ΔDCQ)

ar(ΔDCQ)=1/2 *ar(parallelogram ABCD)

NOW;

Ar(ΔABC)/Ar(ΔCDP)=(1/2*BC*h)/(1/2*PC*h)

=ar(ΔDCQ)=3/2 Ar (ΔCBP)

=AREA of Δ (DPC+PCQ)=3/2( AREA OF triangle cdp)

=ar(ΔPCQ)=3/2AR(traingle CDP - ΔDpc)

AR(ΔPCQ)=1/2AR(ΔCDP)

2*AR(ΔCPQ)=ar(ΔCDP)

=2×20=AR(Δ CDP)

AR(ΔCDP)=40 cm^2

Ar(ΔCDQ)=ar(ΔDPC)+ar(ΔCPQ)

=40+20 cm^2

AREA OF parallelogram ABCD= 2× ar(Δ CDQ)

=2×60=120 cm^2

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