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


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

QUADRILATERALS - EXERCISE 4.2.1 - Class 9

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


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

Solution:

QUADRILATERALS - EXERCISE 4.2.1 - Class 9

 

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

To prove: ABCD is a rhombus

Proof: reason
ABCD is parallelogramgiven
AC ⊥ BD
In ΔAOD and ΔCOD,
OD is common
AO = CODiagonals bisect each other
∠AOD = ∠COD = 90°given
∴ΔAOD ≅ ΔCODSAS
∴AD = CDC.P.C.T
 ∴ABCD is a rhombusadjacent side equal

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

Solution:

QUADRILATERALS - EXERCISE 4.2.1 - Class 9 

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

To prove: ABCD is a rectangle.  

Proof: reason
In ΔDAB and ΔCBAOpposite sides of a parallelogram.
DA = CB
AB = BA
DB =CAgiven
∴ΔDAB ≅ ΔCBASSS
∴∠DAB = ∠CBACPCT
 But ∠DAB + ∠CBA = 180°AD | | BC
∴∠DAB = ∠CBA = 180/2 = 90°angles
 ∴ABCD is a rectanglean angle is a right angles

 

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