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प्रश्न
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (AOD) = ar (BOC).
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उत्तर
It can be observed that ΔDAC and ΔDBC lie on the same base DC and between the same parallels AB and CD.
∴ Area (ΔDAC) = Area (ΔDBC)
⇒ Area (ΔDAC) − Area (ΔDOC) = Area (ΔDBC) − Area (ΔDOC)
⇒ Area (ΔAOD) = Area (ΔBOC)
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)
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).
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[Hint : From A and C, draw perpendiculars to BD.]
PQRS is a parallelogram whose area is 180 cm2 and A is any point on the diagonal QS. The area of ∆ASR = 90 cm2.
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).
A point E is taken on the side BC of a parallelogram ABCD. AE and DC are produced to meet at F. Prove that ar (ADF) = ar (ABFC)
The medians BE and CF of a triangle ABC intersect at G. Prove that the area of ∆GBC = area of the quadrilateral AFGE.
In ∆ABC, if L and M are the points on AB and AC, respectively such that LM || BC. Prove that ar (LOB) = ar (MOC)
In the following figure, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. Prove that ar (ABCDE) = ar (APQ)
