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प्रश्न
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.]
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उत्तर
It can be observed that ΔADX and ΔACX lie on the same base AX and are between the same parallels AB and DC.
∴ Area (ΔADX) = Area (ΔACX) ... (1)
ΔACY and ΔACX lie on the same base AC and are between the same parallels AC and XY.
∴ Area (ΔACY) = Area (ACX) ... (2)
From equations (1) and (2), we obtain
Area (ΔADX) = Area (ΔACY)
संबंधित प्रश्न
In the given figure, AP || BQ || CR. Prove that ar (AQC) = ar (PBR).
P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is the mid-point of AP, show that
(i) ar(PRQ) = 1/2 ar(ARC)
(ii) ar(RQC) = 3/8 ar(ABC)
(iii) ar(PBQ) = ar(ARC)
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)
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