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प्रश्न
The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed (see the following figure). Show that
ar (ABCD) = ar (PBQR).
[Hint: Join AC and PQ. Now compare area (ACQ) and area (APQ)]
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उत्तर
Let us join AC and PQ.
ΔACQ and ΔAQP are on the same base AQ and between the same parallels AQ and CP.
∴ Area (ΔACQ) = Area (ΔAPQ)
⇒ Area (ΔACQ) − Area (ΔABQ) = Area (ΔAPQ) − Area (ΔABQ)
⇒ Area (ΔABC) = Area (ΔQBP) ... (1)
Since AC and PQ are diagonals of parallelograms ABCD and PBQR respectively,
∴ Area (ΔABC) = 1/2Area (ABCD) ... (2)
Area (ΔQBP) = 1/2Area (PBQR) ... (3)
From equations (1), (2), and (3), we obtain
1/2Area (ABCD) = 1/2Area (PBQR)
Area (ABCD) = Area (PBQR)
संबंधित प्रश्न
In a triangle ABC, E is the mid-point of median AD. Show that ar (BED) = 1/4ar (ABC).
Show that the diagonals of a parallelogram divide it into four triangles of equal area.
In the given figure, ABC and ABD are two triangles on the same base AB. If line-segment CD is bisected by AB at O, show that ar (ABC) = ar (ABD).
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.]
In the following figure, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y. Show that:-
(i) ΔMBC ≅ ΔABD
(ii) ar (BYXD) = 2 ar(MBC)
(iii) ar (BYXD) = ar(ABMN)
(iv) ΔFCB ≅ ΔACE
(v) ar(CYXE) = 2 ar(FCB)
(vi) ar (CYXE) = ar(ACFG)
(vii) ar (BCED) = ar(ABMN) + ar(ACFG)
Note : Result (vii) is the famous Theorem of Pythagoras. You shall learn a simpler proof of this theorem in Class X.
The area of the parallelogram ABCD is 90 cm2 (see figure). Find ar (ΔABD)
The area of the parallelogram ABCD is 90 cm2 (see figure). Find ar (ΔBEF)
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 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).
