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Areas of Parallelograms and Triangles

Short 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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Long 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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Short 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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Long Answer Type

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).

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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.

Given: In figure, ABC is a right angle 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.
To Prove: (i) ΔMBC ≅ ΔABD (ii) ar(BYXD) = 2 ar(ΔMBC)
(iii) ar(BYXD) = ar(ABMN)
(iv)    ΔFCB ≅ ΔACE
(v)    ar(CYXE) = 2 ar(ΔFCB)
(vi)    ar(CYXE) = ar(ACFG)
(vii)    ar(BCED) = ar(ABMN) + ar(ACFG) Proof: (i) In ΔMBC and ΔABD,
MB = AB    ...(1)
| Sides of a square BC = BD    ...(2)
| Sides of a square ∠MBA = ∠CBD    | Each = 90°
⇒ ∠MBA + ∠ABC = ∠CBD + ∠ABC
[adding ∠ABC to both sides] ⇒ ∠MBC = ∠ABD    ...(3)
In view of (1), (2) and (3),
ΔMBC = ΔABD.
| By Congruence SAS Rule
(ii)    ar(BYXD) = 2 ar(ΔABD)
⇒ ar(BYXD) = 2 ar(ΔMBC).
(iii)    ar(BYXD) = 2ar(ΔABD) ar(ABMN) = 2ar(ΔMBC) = 2ar(ΔABD)
| From (i)
∴ ar(BYXD) = ar(ABMN).
(iv)    In ΔFCB and ΔACE,
FC = AC    | Sides of a square
CB = CE    | Sides of a square
∠FCA = ∠BCE    | Each = 90°
⇒ ∠FCA + ∠ACB = ∠BCE + ∠ACB
(Adding the same on both sides) ⇒ ∠FCB = ∠ACE ∴ ΔFCB ≅ ΔACE.
| SAS Congruence Rule
(v)    ar(CYXE) = 2ar(ΔACE) = 2ar(ΔFCB)
| From (iv) ∵ ΔFCB ≅ ΔACE ∴ ar(ΔFCB) = ar(ΔACE) Congruent As have equal areas
(vi) ar(CYXE) = 2ar(ΔACE) = 2ar(ΔFCB) ar(ACFG) = 2ar(ΔFCB) ∴ ar(CYXE) = ar(ACFG).
(vii) ar(BCED) = ar(CYXE) + ar(BYXD)
= ar(ACFG) + ar(ABMN) = ar(ABMN) + ar(ACFG).

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