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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 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 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 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.
In Fig. below, ABC and BDE are two equilateral triangles such that D is the mid-point of
BC. AE intersects BC in F. Prove that
(1) ar (Δ BDE) = `1/2` ar (ΔABC)
(2) Area ( ΔBDE) `= 1/2 ` ar (ΔBAE)
(3) ar (BEF) = ar (ΔAFD)
(4) area (Δ ABC) = 2 area (ΔBEC)
(5) ar (ΔFED) `= 1/8` ar (ΔAFC)
(6) ar (Δ BFE) = 2 ar (ΔEFD)
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)
PQRS is a parallelogram whose area is 180 cm2 and A is any point on the diagonal QS. The area of ∆ASR = 90 cm2.
The area of the parallelogram ABCD is 90 cm2 (see figure). Find ar (ΔABD)
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)
In the following figure, CD || AE and CY || BA. Prove that ar (CBX) = ar (AXY).
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)
