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प्रश्न
ABCD is a parallelogram a line through A cuts DC at point P and BC produced at Q. Prove that triangle BCP is equal in area to triangle DPQ.
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उत्तर
ΔAPB and parallelogram ABCD are on the same base AB and between the same parallel lines AB and CD.
∴ Ar. ( ΔAPB ) = `1/2` Ar.( parallelogram ABCD ) ......(i)
ΔADQ and parallelogram ABCD are on the same base AD and between the same parallel lines AD and BQ.
∴ Ar.( ΔADQ ) = `1/2` Ar.( parallelogram ABCD ) ......(ii)
Adding equation (i) and (ii), we get
∴ Ar.( ΔAPB ) + Ar.( ΔADQ ) = Ar.(parallelogram ABCD)
Ar.( quad ADQB ) - Ar.(Δ BPQ ) = Ar.(parallelogram ABCD)
Ar.( quad ADQB) - Ar.( ΔBPQ ) = Ar.(quad ADQB ) -Ar.( ΔDCQ )
Ar. ( ΔBPQ ) = Ar. ( ΔDCQ )
Subtracting Ar.ΔPCQ from both sides, we get
Ar. ( ΔBPQ ) - Ar.(ΔPCQ ) = Ar. ( ΔDCQ ) - Ar. ( ΔPCQ)
Ar. ( ΔBCP ) = Ar. ( ΔDPQ )
Hence proved.
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संबंधित प्रश्न
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[ Join B and R ]
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(ii) Area (ΔAQD) = Area (ΔAPD) + Area (ΔCPB)
In the following figure, DE is parallel to BC.
Show that:
(i) Area ( ΔADC ) = Area( ΔAEB ).
(ii) Area ( ΔBOD ) = Area( ΔCOE ).
In the given figure, M and N are the mid-points of the sides DC and AB respectively of the parallelogram ABCD.
If the area of parallelogram ABCD is 48 cm2;
(i) State the area of the triangle BEC.
(ii) Name the parallelogram which is equal in area to the triangle BEC.
In the figure given alongside, squares ABDE and AFGC are drawn on the side AB and the hypotenuse AC of the right triangle ABC.
If BH is perpendicular to FG
prove that:
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find the area of the parallelogram ABCD.
ABCD is a parallelogram. P and Q are the mid-points of sides AB and AD respectively.
Prove that area of triangle APQ = `1/8` of the area of parallelogram ABCD.
In the following figure, BD is parallel to CA, E is mid-point of CA and BD = `1/2`CA
Prove that: ar. ( ΔABC ) = 2 x ar.( ΔDBC )
In ΔABC, E and F are mid-points of sides AB and AC respectively. If BF and CE intersect each other at point O,
prove that the ΔOBC and quadrilateral AEOF are equal in area.
Show that:
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