Question

question no.5 fast plz...

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Answer 1

triangle AOD and parallelogram ABCD both lie on the same base AD and between the same parallels AD and BC.
so area of triangle = 1/2 area of parallelogram ABCD..........1
similarly area of triangle APD =1/2 the area of parallelogram ABCD.........2

so, by 1 and 2 we get that area of triangles as equal...... hence proved

Answer 2

Given: ABCD is a║gm
           BP=PQ=QC
To Prove:
           arΔAPQ=arΔDPQ= $\frac{1}{6}$ar║gm ABCD
Construction: X and Y on AD, such that AX = XY = YD
Join  XQ, and QY 
Proof: 
In ΔAPQ and ΔDPQ,
they have a common base i.e. PQ
And they lie between the common ║s, i.e. AD║BC
∴ arΔAPQ = arΔDPQ                                       -(i)

Similarly, 
arΔABP = arΔAPQ = arΔQDC (because BP=PQ=QC and AD║BC)
Also,
arΔAQX = arΔXQY = arΔYQD (by construction AX = XY = YD, and they lie between same parallels)

arΔABP + arΔAPQ + arΔQDC + arΔAQX + arΔXQY + arΔYQD = ar(║gm ABCD)
⇒ 3arΔAPQ +3arΔAQX = ar(║gm ABCD)
But, arΔAPQ = arΔAQX because 1/3AB =1/3CD, ∴PQ=AX, and also they lie between same parallels.
∴6arΔAPQ= ar║gmABCD
∴ arΔAPQ = 1/6 ar(ABCD)

∴ arΔAPQ = arΔDPQ = 1/6 ar(ABCD)
Hence proved

P.S: Figure with constructions is attached. Constructions marked in red.