Answer :

Since, ABCD is a parallelogram [Given]

\(\therefore \)Its opposite sides are parallel and equal.

\(\Rightarrow \) AD = BC …(1)

Now, \(\triangle \)ADE and \(\triangle \)BCF are on equal bases AD = BC [from (1)] and between the same parallels AB and EF.

So, ar(\(\triangle \)ADE) = ar(\(\triangle \)BCF).

- 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 is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that ar(\(\triangle \)BPC) = ar(\(\triangle \)DPQ).[Hint Join AC.]
- 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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