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प्रश्न
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.]
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उत्तर
Let us draw AM ⊥ BD and CN ⊥ BD
`"Area of a triangle "=1/2xx"Base"xx"Altitude"`
`ar(APB)xxar(CPD)=[1/2xxBPxxAM]xx[1/2xxPDxxCN]`
`=1/4xxBPxxAMxxPDxxCN`
`ar(APD)xxar(BPC)=[1/2xxPDxxAM]xx[1/2xxCNxxBP]`
`=1/4xxPDxxAMxxCNxxBP`
`=1/4xxBPxxAMxxPDxxCN`
∴ ar (APB) × ar (CPD) = ar (APD) × ar (BPC)
संबंधित प्रश्न
D and E are points on sides AB and AC respectively of ΔABC such that
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[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 at O in such a way that ar (AOD) = ar (BOC). Prove that ABCD is a trapezium.
In the given figure, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.
In the following figure, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that
(i) ar (BDE) = 1/4 ar (ABC)
(ii) ar (BDE) = 1/2 ar (BAE)
(iii) ar (ABC) = 2 ar (BEC)
(iv) ar (BFE) = ar (AFD)
(v) ar (BFE) = 2 ar (FED)
(vi) ar (FED) = 1/8 ar (AFC)
[Hint : Join EC and AD. Show that BE || AC and DE || AB, etc.]
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 area of the parallelogram ABCD is 90 cm2 (see figure). Find ar (ΔBEF)
The medians BE and CF of a triangle ABC intersect at G. Prove that the area of ∆GBC = area of the quadrilateral AFGE.
