P and Q are respectively the midpoints of sides AB and BC of a t

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 Multiple Choice QuestionsShort Answer Type

71. The diagonals of a quadrilateral ABCD intersect at O (see figure). Prove that if BO = OD, then ar(ΔABC) = ar(ΔADC).


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72. In the figure, the vertex A of ΔABC is joined to a point D on the side BC. The mid-point of AD is straight E space Prove space that space ar space left parenthesis increment BEC right parenthesis equals 1 half ar left parenthesis increment ABC right parenthesis.


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73. 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.
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74.

In figure, D and E are two points on BC such that BD = DE = EC. Show that ar(ΔABD) = ar(ΔADE) = ar(ΔAEC). [CBSE 2012 (March)]

Can you now answer the question that you have left in the ‘Introduction’, of this chapter, whether the field of Budhia has been actually divided into three parts of equal area?



[Remark: Note that by taking BD = DE = EC, the triangle ABC is divided into three triangles ABD, ADE and AEC of equal areas. In the same way, by dividing BC into n equal parts and joining the points of division so obtained to the opposite vertex of BC, you can divide ΔABC into n triangles of equal areas.] 

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75. In figure, ABCD, DCFE and ABFE are parallelograms. Show that ar(ΔADE) = ar(ΔBCF).


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76. 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(ΔBPC) = ar(ΔDPQ).



[Hint. Join AC.]
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 Multiple Choice QuestionsLong Answer Type

77. 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: 



left parenthesis straight i right parenthesis space ar left parenthesis increment BDE right parenthesis equals 1 fourth ar left parenthesis increment ABC right parenthesis
left parenthesis ii right parenthesis space ar left parenthesis increment BDE right parenthesis equals 1 half ar space left parenthesis increment BAE right parenthesis
left parenthesis iii right parenthesis space ar left parenthesis increment ABC right parenthesis equals space 2 ar left parenthesis increment BEC right parenthesis
left parenthesis iv right parenthesis space ar left parenthesis increment BFE right parenthesis equals 2 ar left parenthesis increment AFD right parenthesis
left parenthesis straight v right parenthesis space ar left parenthesis increment BFE right parenthesis equals 2 ar left parenthesis increment FED right parenthesis
left parenthesis vi right parenthesis ar left parenthesis increment FED right parenthesis equals 1 over 8 ar left parenthesis AFC right parenthesis.
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 Multiple Choice QuestionsShort Answer Type

78.

Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that ar(ΔAPB) x ar(ΔCPD) = ar(ΔAPD) x ar(ΔBPC).

[Hint From A and C, draw perpendiculars to DD.] 

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 Multiple Choice QuestionsLong Answer Type

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79. P and Q are respectively the midpoints of sides AB and BC of a triangle ABC and R is the mid-point of AP. Show that:

left parenthesis straight i right parenthesis space ar left parenthesis increment PRQ right parenthesis equals 1 half ar left parenthesis increment ARC right parenthesis
left parenthesis II right parenthesis space ar left parenthesis increment RQC right parenthesis equals 3 over 8 ar left parenthesis increment ABC right parenthesis

(iii) ar(ΔPBQ) = ar(ΔARC).


Given: P and Q are respectively the midpoints of sides AB and BC of a triangle ABC and R is the mid-point of AP.


left parenthesis straight i right parenthesis space a r left parenthesis increment P R Q right parenthesis equals 1 half a r left parenthesis increment A R C right parenthesis
left parenthesis I I right parenthesis space a r left parenthesis increment R Q C right parenthesis equals 3 over 8 a r left parenthesis increment A B C right parenthesis

(iii) ar(ΔPBQ) = ar(ΔARC). Construction: Join AQ and CP.


Given: P and Q are respectively the midpoints of sides AB and BC of a

Proof: (i) ar(ΔPRQ) = ar(ΔARQ)
(∵ a median of a A divides it into triangles of equal area)

equals 1 half ar left parenthesis increment APQ right parenthesis equals 1 half ar left parenthesis increment BPQ right parenthesis
equals 1 half ar left parenthesis increment CPQ right parenthesis equals 1 half.1 half ar left parenthesis increment BPC right parenthesis
equals space 1 half ar left parenthesis increment BPC right parenthesis equals 1 fourth.1 half ar left parenthesis increment ABC right parenthesis
equals space 1 half ar left parenthesis increment ABC right parenthesis space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis
1 half a r left parenthesis increment A R C right parenthesis equals 1 half 1 half a r left parenthesis increment A P C right parenthesis
equals space 1 fourth a r left parenthesis increment A P C right parenthesis equals 1 fourth.1 half space a r left parenthesis increment A B C right parenthesis
equals space 1 over 8 a r left parenthesis increment A B C right parenthesis space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis

From (1) and (2), we have

ar left parenthesis increment PRQ right parenthesis equals 1 half ar left parenthesis increment ARC right parenthesis

(ii) ar(ΔRQC) = ar(ΔRBQ) (Δ a median of a A divides it into triangles of equal areas) = ar(ΔPRQ) + ar(ΔBPQ)

equals 1 over 8 ar left parenthesis increment ABC right parenthesis plus 1 half ar left parenthesis increment PBC right parenthesis
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space vertical line space Using space left parenthesis 1 right parenthesis
equals 1 half space ar left parenthesis increment ABC right parenthesis plus 1 half.1 half ar left parenthesis increment ABC right parenthesis
equals space 1 over 8 ar left parenthesis increment ABC right parenthesis plus 1 fourth ar left parenthesis increment ABC right parenthesis
equals 3 over 8 ar left parenthesis increment ABC right parenthesis space space space space space space space space space space space space space space space space space space space space

left parenthesis iii right parenthesis space ar left parenthesis increment PBQ right parenthesis equals 1 fourth ar left parenthesis increment ABC right parenthesis space space space space space space space space space space space space space... left parenthesis 3 right parenthesis
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space vertical line space From space left parenthesis ii right parenthesis
ar space left parenthesis increment ARC right parenthesis equals 1 fourth ar left parenthesis increment ABC right parenthesis space space space space space space space space space space space space space space space space space... left parenthesis 4 right parenthesis
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space vertical line space From space left parenthesis straight i right parenthesis

From (3) and (4),
ar(ΔPBQ) = ar(ΔARC).



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80. 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 ⊥ DE meets BC at Y. Show that:



(i)    ΔMBC ≅ ΔABD
(ii)    ar(BYXD) = 2 ar(ΔMBC)
(iii) ar(BYXD) = ar(ΔABMN)
(iv)    ΔFCB ≅ ΔACE
(v)    ar(CYXE) = 2 ar(ΔFCB)
(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.

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