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प्रश्न
In the given figure, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.
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उत्तर
It is given that
Area (ΔDRC) = Area (ΔDPC)
As ΔDRC and ΔDPC lie on the same base DC and have equal areas, therefore, they must lie between the same parallel lines.
∴ DC || RP
Therefore, DCPR is a trapezium.
It is also given that
Area (ΔBDP) = Area (ΔARC)
⇒ Area (BDP) − Area (ΔDPC) = Area (ΔARC) − Area (ΔDRC)
⇒ Area (ΔBDC) = Area (ΔADC)
Since ΔBDC and ΔADC are on the same base CD and have equal areas, they must lie between the same parallel lines.
∴ AB || CD
Therefore, ABCD is a trapezium.
संबंधित प्रश्न
In the given figure, E is any point on median AD of a ΔABC. Show that ar (ABE) = ar (ACE)
D and E are points on sides AB and AC respectively of ΔABC such that
ar (DBC) = ar (EBC). Prove that DE || BC.
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (AOD) = ar (BOC).
In the given figure, AP || BQ || CR. Prove that ar (AQC) = ar (PBR).
If a triangle and a parallelogram are on the same base and between same parallels, then the ratio of the area of the triangle to the area of parallelogram is ______.
The area of the parallelogram ABCD is 90 cm2 (see figure). Find
- ar (ΔABEF)
- ar (ΔABD)
- ar (ΔBEF)
The area of the parallelogram ABCD is 90 cm2 (see figure). Find ar (ΔABD)
In ∆ABC, D is the mid-point of AB and P is any point on BC. If CQ || PD meets AB in Q (Figure), then prove that ar (BPQ) = `1/2` ar (∆ABC).
The medians BE and CF of a triangle ABC intersect at G. Prove that the area of ∆GBC = area of the quadrilateral AFGE.
In the following figure, X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. Prove that ar (ABP) = ar (ACQ).
