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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
D, E and F are respectively the mid-points of the sides BC, CA and AB of a ΔABC. Show that
(i) BDEF is a parallelogram.
(ii) ar (DEF) = 1/4ar (ABC)
(iii) ar (BDEF) = 1/2ar (ABC)
In the given figure, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that:
(i) ar (DOC) = ar (AOB)
(ii) ar (DCB) = ar (ACB)
(iii) DA || CB or ABCD is a parallelogram.
[Hint: From D and B, draw perpendiculars to AC.]
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, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that
(i) ar (ACB) = ar (ACF)
(ii) ar (AEDF) = ar (ABCDE)
ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (ADX) = ar (ACY).
[Hint: Join CX.]
In the given figure, AP || BQ || CR. Prove that ar (AQC) = ar (PBR).
Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that ar (APB) × ar (CPD) = ar (APD) × ar (BPC).
[Hint : From A and C, draw perpendiculars to BD.]
In the below fig. D and E are two points on BC such that BD = DE = EC. Show that ar
(ΔABD) = ar (ΔADE) = ar (ΔAEC).
In a ΔABC, if L and M are points on AB and AC respectively such that LM || BC. Prove
that:
(1) ar (ΔLCM ) = ar (ΔLBM )
(2) ar (ΔLBC) = ar (ΔMBC)
(3) ar (ΔABM) ar (ΔACL)
(4) ar (ΔLOB) ar (ΔMOC)
The medians BE and CF of a triangle ABC intersect at G. Prove that the area of ∆GBC = area of the quadrilateral AFGE.
