(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)

Answer :

We have a right \(\triangle \)ABC such that BCED, ACFG and ABMN are squares on its sides BC, CA and AB respectively. Line segment AX \(\perp \) DE is also drawn such that it meets BC at Y.

(i) \(\angle \)CBD = \(\angle \)MBA [Each90°]

\(\therefore \angle CBD + \angle ABC = \angle MBA + \angle ABC \)

(By adding \(\angle \)ABC on both sides)

\(\Rightarrow \) \(\angle \)ABD = \(\angle \)MBC

In \(\triangle \)ABD and \(\triangle \)MBC, we have

AB = MB [Sides of a square]

BD = BC

\(\angle \)ABD = \(\angle \)MBC [Proved above]

\(\therefore \) \(\triangle \)ABD \(\cong \) \(\triangle \)MBC [By SAS congruency]

(ii) Since parallelogram BYXD and \(\triangle \)ABD are on the same base BD and between the same parallels BD and AX.

\(\therefore \) ar(\(\triangle \) ABD) = \(\frac{1}{2} \) ar(||gm BYXD)

But \(\triangle \)ABD \(\cong \) \(\triangle \)MBC [From (i) part]

[Since, congruent triangles have equal
areas.]

\(\therefore \) ar(\(\triangle \)MBC) = \(\frac{1}{2} \) ar(||gm BYXD)

\(\Rightarrow \) ar(||gm BYXD) = 2ar (\(\triangle \)MBC)

(iii) Since, ar(||gm BYXD) = 2ar(\(\triangle \)MBC) …(1) [From (ii) part]

and ar(\(\square \) ABMN) = 2ar(\(\triangle \)MBC) …(2)

[ABMN and AMBC are on the same base MB and between the same parallels MB and NC]

From (1) and (2), we have

ar(BYXD) = ar(ABMN) .

(iv) \(\angle \)FCA = \(\angle \)BCE (Each 90°)

\(\Rightarrow \) \(\angle \)FCA+ \(\angle \)ACB = \(\angle \)BCE+ \(\angle \)ACB

[By adding \(\triangle \)ACB on both sides]

\(\Rightarrow \) \(\angle \)FCB = \(\angle \)ACE

In \(\triangle \)FCB and \(\triangle \)ACE, we have

FC = AC [Sides of a square]

CB = CE [Sides of a square]

\(\angle \)FCB = \(\angle \)ACE [Proved above]

\(\Rightarrow \) \(\triangle \)FCB \(\cong \) \(\triangle \)ACE [By SAS congruency]

(v) Since, ||gm CYXE and \(\triangle \)ACE are on the same base CE and between the same parallels CE and AX.

\(\therefore \) ar(||gm CYXE) = 2ar((\(\triangle \)ACE)

But \(\triangle \)ACE \(\cong \) \(\triangle \)FCB [From (iv) part]

[Since, congruent triangles are equal in areas.]

\(\therefore \) ar (||gm CYXE) = 2ar(\(\triangle \)FCB)

(vi) Since, ar(||gm CYXE) = 2ar (\(\triangle \)FCB) …(3)
[From (v) part]

Also (quad. ACFG) and \(\triangle \) FCB are on the same base FC and between the same parallels FC and BG.

\(\Rightarrow \) ar(quad. ACFG) = 2ar((\(\triangle \)FCB) …(4)

From (3) and (4), we get

ar( CYXE) = ar(ACFG) …(5)

(vii) We have ar(BCED) = ar(CYXE) + ar(BYXD)

= ar(CYXE) + ar(ABMN)

[From (iii) part]

Thus, ar (BCED) = ar( ABMN) + ar(ACFG)

[From (vi) part]

- 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, 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)

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