CLASS-8
PARALLELOGRAM- PROBLEM & SOLUTION

PROBLEM & SOLUTION

Example.1) In the figure ABCD is a parallelogram. Find x & y

Ans.)  The opposite sides of a parallelogram are equal.

So, AB = DC 

=>  8x – 18 = 3x + 7 

=>  5x = 25

=>   x = 5


So, AB = DC => ∠A + ∠B = 180⁰ as these are co-interior angles

                      =>  y + 30 + 25 + 4y + 30 =  180

                      =>  5y = 180 – (30 + 30 + 25)  =  95

                      =>   y = 95/5 =  19

Hence we can get x = 5, and y = 19    (Ans.)



Example.2) In the adjoining figure, ABCD is a rectangle. Find x, y, & z

Ans.) The diagonals of a rectangle are equal and they bisect each other.

OA = OB = OC = OD

In, ∆ OAB, OA = OB  => ∠OBA = ∠OAB = x

The sum of the angles of ∆ OAB = 180⁰


=> x + x + 96⁰ = 180⁰

=>   2x  =  180⁰ - 96⁰ = 84⁰

=>    x  =  42⁰

So, ∠OBA =  42⁰ but ∠ABC = 90⁰

=> ∠OBC = ∠ABC - ∠OBA = 90⁰ - 42⁰ = 48⁰

But, OB = OC

=>  ∠OBC = ∠OCB = y

=>  y = 48⁰

Again,  ∠OAD = ∠DAB - ∠OAB = 90⁰ - x  = 90⁰ - 42⁰ = 48⁰

In, ∆ OAD, OA = OD

=>  ∠ODA = ∠OAD

=>  z  =  48⁰

Hence the result is x = 42⁰, y = 48⁰, & z = 48⁰    (Ans.)



Example.3) In the adjoining figure, ABCD is a square. Find x

Ans.)  the diagonal BD of the square ABCD bisects ∠ABC

So, ∠PBE = ½ ∠ABC

           = ½ X 90⁰ = 45⁰

Also, ∠BPE = opposite ∠DPC = 75⁰

In ∆ PEB, exterior ∠AEP

                  = ∠BPE + ∠PBE

So, x  = 75⁰ + 45⁰ = 120⁰

Hence x = 120⁰  (Ans.)





Example.4) In the adjoining figure, ABCD is a rhombus and  ∠BCD = 80⁰, find x & y

Ans.)  ABCD is a rhombus, so AC bisects ∠BCD

So, ∠PCM = ½ ∠BCD

            = ½ X 80⁰ = 45⁰

In, ∆ PCM, exterior ∠DPC

= x + ∠PCM

=>  110⁰ = x + 40⁰

=>  x = 110⁰ – 40⁰ = 70⁰

Now,  AD || BC  (ABCD being rhombus)

=>  ∠BCD + ∠ADC = 180⁰

=>  80⁰ + ∠ADC = 180⁰

=>  ∠ADC = 180⁰ - 80⁰ = 100⁰

ABCD being a rhombus, the diagonal BD bisects ∠ADC

=> ∠ADB = ½ X ∠ADC

=>     y  = ½ X 100⁰ =  50⁰

Hence the required result is x = 70⁰, y = 50⁰      (Ans.)



Example.5) In the adjoin figure, the bisectors of the two consecutive angles ∠A & ∠B of the parallelogram ABCD meet at point P. Prove that ∠APB = 90⁰

Ans.)  PA & PB are the bisectors of the ∠A & ∠B respectively

                    1                                  1

So,  ∠PAB =  ---------  ∠A  and  ∠PBA =  --------- ∠B 

                    2                                  2

                                                                                                                                   1

Now,  AD || BC  => ∠A + ∠B = 180⁰ => -------- (∠A + ∠B) = 90⁰

                                                 2                                   

        1                   1

=> -------- ∠A +  --------- ∠B  =   90⁰

        2                   2

=>  ∠PAB + ∠PBA = 90⁰

In, ∆ PAB, the sum of angles = 180⁰

=>  ∠PAB + ∠PBA + ∠APB = 180⁰

=> 90⁰ +  ∠APB = 180⁰

=>  ∠APB = 180⁰ - 90⁰ = 90⁰

Hence, ∠APB = 90⁰  (Proved)

Example.6) In the adjoining figure, ABCD is a parallelogram. Prove that a) ∆ ADN ≅ ∆ CBP, and b) AN = CP

Ans.)   In, ∆ ADN and ∆ CBP,  ∠AND = ∠CPB (= 90⁰),

∠ADN = alternate ∠CBP ( AD || BC )

And, AD = BC  (opposite sides of a parallelogram)

∆ ADN ≅ ∆ CBP (A-A-S condition of congruence)


So, the corresponding sides are equal.

Hence,  AN = CP      (Proved)