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In Fig. 9.32, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that ar (BPC) = ar (
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In fig, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ . If AQ intersect DC at P , show that ar(BPC) = ar(DPQ) .
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SOLVED: As shown in the figure, a pair of rings of radius R are separated by distance AR, where A = 4, and are aligned with their symmetry axes along the z-axis.
In trapezium ABCD, AB ∥ DC and L is the mid - point of BC. Through L, a line PQ ∥ AD has been drawn which meets AB is P and DC
In Fig. 9.32, ABCD is a parallelogram and BC is produced to a point Q such
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ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ ( Fig. 9.10). If AQ intersects DC at P, show that ar (BPC) = ar (DPQ)
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SOLVED: #AR (=18,100.04) A4. As is shown in the figure, a moving point charge q get through an unique magnetic field. We assume the velocity of the charge is V, the magnitude
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In △ ABC, D is the mid-point of AB and P is any point on BC. If CQ || PD meets AB in Q (Fig.), then prove that ar (BPQ) = ½
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In the given figure, `AP*AR=AS+AQ` prove that `angle P= angle S and angle Q= angle R`. - YouTube
