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प्रश्न
In the given figure, AP is parallel to BC, BP is parallel to CQ.
Prove that the area of triangles ABC and BQP are equal.
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उत्तर
Joining PC we get,
ΔABC and ΔBPC are on the same base BC and between the same parallel lines AP and BC.
∴ A( ΔABC ) = A( ΔBPC ) ....(i)
ΔBPC and ΔBQP are on the same base BP and between the same parallel lines BP and CQ.
∴ A( ΔBPC ) = A( ΔBQP ) ....(ii)
From (i) and (ii), we get
∴A( ΔABC ) = A( ΔBQP )
Hence proved.
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Prove that area (ΔAEC) = area (ΔDBF)
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Prove that: Area of ABC = Area of // gm BDEC.
In parallelogram ABCD, P is a point on side AB and Q is a point on side BC.
Prove that:
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(ii) Area (ΔAQD) = Area (ΔAPD) + Area (ΔCPB)
In the following figure, CE is drawn parallel to diagonals DB of the quadrilateral ABCD which meets AB produced at point E.
Prove that ΔADE and quadrilateral ABCD are equal in area.
The given figure shows a pentagon ABCDE. EG drawn parallel to DA meets BA produced at G and CF draw parallel to DB meets AB produced at F.
Prove that the area of pentagon ABCDE is equal to the area of triangle GDF.
In the given figure, the diagonals AC and BD intersect at point O. If OB = OD and AB//DC,
show that:
(i) Area (Δ DOC) = Area (Δ AOB).
(ii) Area (Δ DCB) = Area (Δ ACB).
(iii) ABCD is a parallelogram.
In parallelogram ABCD, E is a point in AB and DE meets diagonal AC at point F. If DF: FE = 5:3 and area of ΔADF is 60 cm2; find
(i) area of ΔADE.
(ii) if AE: EB = 4:5, find the area of ΔADB.
(iii) also, find the area of parallelogram ABCD.
In parallelogram ABCD, P is the mid-point of AB. CP and BD intersect each other at point O. If the area of ΔPOB = 40 cm2, and OP: OC = 1:2, find:
(i) Areas of ΔBOC and ΔPBC
(ii) Areas of ΔABC and parallelogram ABCD.
Show that:
The ratio of the areas of two triangles of the same height is equal to the ratio of their bases.
