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(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.]

Answer :

Let us join EC and AD. Draw EP \(\perp \) BC.

Let AB = BC = CA = a, then

BD = \(\frac{a}{2} \) = DE = BE

(i) area of \(\triangle ABC = \frac{\sqrt{3}}{4} × a^2 \)

And area of \(\triangle BDE = \frac{\sqrt{3}}{4} × \frac{a}{2} ^2 \)

= \(\frac{\sqrt{3}}{4} × \frac{a^2}{4} \)

\(\therefore \) ar (BDE) = \(\frac{1}{4} \) ar (ABC)

(ii) Since, \(\triangle \) ABC and \(\triangle \) BED are equilateral triangles.

\(\Rightarrow \) \(\angle \)ACB = \(\angle \)DBE = 60°

\(\Rightarrow \) BE || AC

\(\triangle \)BAE and \(\triangle \)BEC are on the same base BE and between the same parallels BE and AC.

\(\therefore \) ar(\(\triangle \)BAE) = ar(\(\triangle \)BEC)

\(\Rightarrow \) ar(\(\triangle \)BAE) = 2 ar(\(\triangle \)BDE)

[ DE is median of \(\triangle \)EBC. ]

\(\Rightarrow \) ar(\(\triangle \) BDE) = \(\frac{1}{2} \) ar(\(\triangle \)BAE)

(iii) ar(\(\triangle \)ABC) = 4 ar(\(\triangle \)BDE)

[Proved in (i) part]

ar(\(\triangle \)BEC) = 2 ar(\(\triangle \)BDE)

[ \(\because \)DE is median of \(\triangle \) BEC]

\(\Rightarrow \) ar(\(\triangle \)ABC) = 2 ar(\(\triangle \)BEC)

(iv) Since, \(\triangle \)ABC and \(\triangle \)BDE are equilateral triangles.

\(\Rightarrow \) \(\angle \)ABC = \(\angle \)BDE = 60°

\(\Rightarrow \) AB || DE

\(\triangle \)BED and \(\triangle \)AED are on the same base ED and between the same parallels AB and DE.

\(\therefore \) ar(\(\triangle \)BED) = ar(\(\triangle \)AED)
Subtracting ar(AEFD) from both sides, we get

\(\Rightarrow \) ar(\(\triangle \)BED) – ar(\(\triangle \)EFD) = ar(\(\triangle \)AED) – ar(\(\triangle \)EFD)

\(\Rightarrow \) ar(\(\triangle \)BEE) = ar(\(\triangle \)AFD)

(v) In right angled \(\triangle \) ABD, we get

\( AD^2 = AB^2 - BD^2 = a^2 - (\frac{a}{2} )^2 = \frac{3a^2}{4} \)

AD = \(\frac{\sqrt{3} a}{2} \)

In right triangle PED

Similarly,

EP = \(\frac{\sqrt{3} a}{4} \)

Area of triangle AFD = (\(\frac{1}{2} \) ) × FD × AD

= (\(\frac{1}{2} \) ) × FD × \(\frac{\sqrt{3} a}{2} \).....(i)
And area of triangle EFD = (\(\frac{1}{2} \)) × FD × EP

= (\(\frac{1}{2} \) ) × FD × \(\frac{\sqrt{3} a}{4} \) .....(ii)

From (1) and (2), we get

ar(\(\triangle \) AFD) = 2 ar((\(\triangle \)EFD)

ar(\(\triangle \)AFD) = ar(\(\triangle \)BEF) [From (iv) part]

\(\Rightarrow \) ar(\(\triangle \)BFE) = 2 ar(\(\triangle \)EFD)

(vi) ar(\(\triangle \)AFC) = ar(\(\triangle \)AFD) + ar(\(\triangle \)ADC)

= ar(\(\triangle \)BFE) + \(\frac{1}{2} \) ar(\(\triangle \)ABC) [From (iv) part]

= ar(\(\triangle \)BFE) + (\(\frac{1}{2} \) ) x 4 x ar(\(\triangle \)BDE) [From (i) part]

= ar(\(\triangle \)BFE) + 2ar(\(\triangle \)BDE)

= 2ar(\(\triangle \)FED) + 2[ar(\(\triangle \)BFE) + ar(\(\triangle \)FED)]

= 2ar(\(\triangle \)FED) + 2[2ar(\(\triangle \)FED) + ar(\(\triangle \)FED)]

[From (v) part]

= 2ar(\(\triangle \)FED) + 2[3ar(\(\triangle \)FED)]

= 2ar(\(\triangle \)FED) + 6ar(\(\triangle \)FED)

= 8ar(\(\triangle \)FED)

\(\Rightarrow \) ar(\(\triangle \)FED) = \(\frac{1}{8} \) ar(\(\triangle \)AFC)

- 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.]
- 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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