QUADRILATERALS – EXERCISE 4.2.1 – Class 9

1. Is a parallelogram a rectangle? Can you call a rectangle a parallelogram?

Solution:

 A parallelogram becomes a rectangle when its angles measure 90° each. Therefore a parallelogram need not be a rectangle. But a rectangle is a parallelogram

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2. Prove that bisectors of two opposite angles of parallelogram are parallel.

Given: ABCD is a | | gm AE and CF bisect DAB and DCB respectively

To prove: AE | | CF

Proof:

Statement Reason

ABCD is | |gm given

∴ DC | | AB

∠CDA + ∠DAB = 180° Co-interior angles

∴ ∠DAB = 180° – ∠CDA are supplementary

But ∠DAE = ¹/₂ ∠DAB ∵ AE bisects DAB

∴ ∠DAE = ¹/₂ [180° – ∠CDA]

∠DAE = 90° – 12 [180 – ∠CDA] ……. (1)

In Δle ADE

∠DEA = 180° – (∠DAE + ∠ADE )

= 180° – [90°– 12 ∵ ∠ ADE = ∠CDA + ∠ADE ∠CDA from equation(1)

= 180° – [90°– ¹/₂ ∠CDA + ∠CDA] [∵∠CDA – ¹/₂ ∠CDA = ∠CDA]

= 180° – [90°– ¹/₂ ∠CDA]

= 180° – 90°– ¹/₂ ∠CDA

∠DAE = 90° – ¹/₂ ∠CDA …….. (2)

From (1) and (2)

∠DAE = ∠DEA …….. (3)

Further, ∠ECF = ¹/₂ ∠DCB, But ∠DCB = ∠DAB

∠ECF = ¹/₂ ∠DEA

ECF = ∠DAF But from (3) DAE = DEA

∠ECF = ∠DEA

But these are corresponding angles

∴ AE | | FC

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3. Prove that if the diagonals of a parallelogram are perpendicular to each other, the parallelogram is a rhombus.

Solution:

 

Given: ABCD is a parallelogram. Diagonals AC and BD intersect at right angles.

To prove: ABCD is a rhombus

Proof:	reason
ABCD is parallelogram	given
AC ⊥ BD
In ΔAOD and ΔCOD,
OD is common
AO = CO	Diagonals bisect each other
∠AOD = ∠COD = 90°	given
∴ΔAOD ≅ ΔCOD	SAS
∴AD = CD	C.P.C.T
∴ABCD is a rhombus	adjacent side equal

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4. Prove that if the diagonals of a parallelogram are equal then it is a rectangle.

Solution:

 

Given: ABCD is a parallelogram. Diagonals AC and BD are equal.

To prove: ABCD is a rectangle.  

Proof:	reason
In ΔDAB and ΔCBA	Opposite sides of a parallelogram.
DA = CB
AB = BA
DB =CA	given
∴ΔDAB ≅ ΔCBA	SSS
∴∠DAB = ∠CBA	CPCT
But ∠DAB + ∠CBA = 180°	AD | | BC
∴∠DAB = ∠CBA = 180/2 = 90°	angles
∴ABCD is a rectangle	an angle is a right angles

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