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Answer :

We have a parallelogram ABCD and AD = CQ. Let us join AC.

We know that triangles on the same base and between the same parallels are equal in area.

Since, \(\triangle \)QAC and \(\triangle \)QDC are on the same base QC and between the same parallels AD and BQ.

\(\therefore ar(\triangle QAC) = ar(\triangle QDC) \)

Subtracting ar(\(\triangle \) QPC) from both sides, we have

\(ar(\triangle QAQ – ar(\triangle QPC) = ar(\triangle QDC) – ar(\triangle QPC) \)

\(\Rightarrow \)ar(\(\triangle PAQ = ar(\triangle QDP) \) …(1)

Since, \(\triangle \)PAC and \(\triangle \)PBC are on the same base PC and between the same parallels AB and CD.

\(\therefore ar(\triangle PAC) = ar(\triangle PBC) \)…(2)

From (1) and (2), we get

ar(\(\triangle PBC) = ar(\triangle QDP) \)

- Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.
- In figure, D and E are two points on BC such that BD = DE = EC. Show that ar(ABD) = ar(ADE) = ar(AEC).
- In figure, ABCD, DCFE and ABFE are parallelograms. Show that ar(ADE) = ar(BCF).
- In 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) = \(\frac{1}{4} \) ar (ABC) (ii) ar (BDE) = \(\frac{1}{2} \) ar (BAE) (iii) ar (ABC) = 2 ar (BEC) (iv) ar (BFE) = ar (AFD) (v) ar (BFE) = 2 ar (FED) (vi) ar (FED) = \(\frac{1}{8} \) ar (AFC) [Hint : Join EC and AD. Show that BE || AC and DE || AB, etc.]
- Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that ar(APB) x ar(CPD) = ar(APD) x ar(BPC). [Hint From A and C, draw perpendiculars to BD.]
- 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) = \(\frac{1}{2} \) ar(ARC) (ii) ar(PBQ) = ar(ARC) (iii) ar(RQC) = \(\frac{3}{8} \) ar(ABC)
- In 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 \(\perp \) DE meets BC at Y. Show that (i) \(\triangle \)MBC \(\cong \) \(\triangle \)ABD (ii) ar(BYXD) = 2 ar(MBC) (iii) ar(BYXD) = ax(ABMN) (iv) \(\triangle \)FCB \(\cong \) \(\triangle \)ACE (v) ar(CYXE) = 2 ar(FCB) (vi) ar(CYXE) = ar(ACFG) (vii) ar(BCED) = ar(ABMN) + ar(ACFG)

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