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Properties Of Parallel Lines And Their Transversal

Class 8th Mathematics West Bengal Board Solution
Lets Work Out 8.1
  1. Chandra took a ruled paper. She drew a transversal between two lines. As a result 4…
  2. Let’s see and write from the figure given below which are corresponding angles,…
  3. If AB||CD then let’s write the measurement of the angles given below:…
  4. If AB||CD then let’s write the measurement of the angles given below:…
  5. If AB||CD then let’s write the measurement of the angles given below:…
  6. Lets write the measurement of 7 angles from the figure beside if XY||PQ.…
  7. Examining the measurement of the angles given below let’s conclude logically ABCD…
  8. Examining the measurement of the angles given below let’s conclude logically ABCD…
  9. Examining the measurement of the angles given below let’s conclude logically ABCD…
  10. In the figure AB||CD and∠EGB=50°; then write the measurement of ∠AGE, ∠AGH, ∠BGH, ∠GHV,…
  11. In the figure beside AB||CD; let’s write the measurement of ∠PQR.…
  12. In the figure beside PQ||RS, ∠BPQ=40°, ∠BPR=155° and ∠CRS=70°; let’s write the…
  13. O is any point inside two parallel lines AB and CD. OP and OQ are two perpendiculars on…
  14. Each pair of arms of two angles is parallel to each other. Let’s prove either these…
  15. The diagonal AC of the parallelogram ABCD bisects ∠BAD. Let’s prove that the diagonal…
  16. Let’s prove if one angle of a parallelogram is right angle, then all other angles of…
Lets Work Out 8.1
Question 1.

Chandra took a ruled paper. She drew a transversal between two lines. As a result 4 corresponding angles, two pairs of alternate angles and two pairs of interior angles are formed. Let’s find out and number them. Verify by measuring with protractor whether

(i) corresponding angles are equal in measurement

(ii) alternate angles are equal in measurement and

(iii) the interior angles on the same side of a transversal are supplementary.
Answer:


Given that AB||CD and EF is the transversal.


So, if two lines are parallel and cut by a transversal then the corresponding angles are equal,


Alternate interior angles are equal


Alternate exterior angles are equal,


Sum of co-interior angles is 180°.


We can say that,


Corresponding angles are (2,6), (1,5),(4,8)and (3,7)


Alternate angles are (2,8),(1,7),(3,5)and (4,6)


Interior angles on same side of transversal are (3,6) and(4,5).
Question 2.

Let’s see and write from the figure given below which are corresponding angles, alternate angles and of them interior angles in the same side of the transversal
Answer:

From the given figure,

We can say that,


Corresponding angles are (2,6), (1,5),(4,8)and (3,7)


Alternate angles are (2,8),(1,7),(3,5)and (4,6)


Interior angles on same side of transversal are (3,6) and(4,5).
Question 3.

If AB||CD then let’s write the measurement of the angles given below:
Answer:


Given that AB||CD and EF is the transversal.


So, if two lines are parallel and cut by a transversal then the corresponding angles are equal,


Alternate interior angles are equal


Alternate exterior angles are equal,


Sum of co-interior angles is 180°.


∠2=55°, ∠3=x and ∠8=y


∠3=∠2(vertically opposite angles)


⇒ x=55°


∠6=∠3(alternate interior angles)


⇒ ∠6=x=55°


Now, as EF is a straight line,


∠8+∠6=180° (linear pair)


⇒ y+∠6=180° ⇒ y+55° =180°


⇒ y=180° -55°


⇒ y=125°


Hence, x=55° and y=125°
Question 4.

If AB||CD then let’s write the measurement of the angles given below:
Answer:


Given that AB||CD and EF is the transversal.


So, if two lines are parallel and cut by a transversal then the corresponding angles are equal,


Alternate interior angles are equal


Alternate exterior angles are equal,


Sum of co-interior angles is 180°.


∠6=68° , ∠1=x


∠5=∠1=x (corresponding angles)……………(1)


Now, as CD is a straight line,


∠5+∠6=180° (linear pair)


⇒ x+68° =180°


⇒ x=180°-68°


⇒ x=112°


Hence, x=112°
Question 5.

If AB||CD then let’s write the measurement of the angles given below:
Answer:


Given that AB||CD and EF is the transversal.


So, if two lines are parallel and cut by a transversal then the corresponding angles are equal,


Alternate interior angles are equal


Alternate exterior angles are equal,


Sum of co-interior angles is 180°.


∠4=100° , ∠7=x


∠6=∠7=x (vertically opposite angles)……………(1)


Now, as AB||CD,


∠6+∠4 =180°


⇒ x+100° =180° (co-interior angles)


⇒ x=180°-100°


⇒ x=80° (from (1))


Hence, x=80°
Question 6.

Lets write the measurement of 7 angles from the figure beside if XY||PQ.
Answer:

Given that XY||PQ and cut by a transversal.

So, if two lines are parallel and cut by a transversal then the corresponding angles are equal,


Alternate interior angles are equal


Alternate exterior angles are equal,


Sum of co-interior angles is 180°.


∠2=50° (vertically opposite angles)


Now, as XY is a straight line,


∠2+∠3=180° (linear pair)


⇒ ∠3=180° -∠2


⇒ ∠3=180° -50°


⇒ ∠3=130°


∠1=∠3=130° (vertically opposite angles)


Also, ∠6=∠3=130° (corresponding angles)


∠7=∠2=50° (corresponding angles)


∠4=∠3=130° (alternate interior angles)


∠5=∠2=50° (alternate exterior angles)


Hence, ∠1=130°, ∠2=50°, ∠3=130°, ∠4=130°, ∠5=50°, ∠6=130° and ∠7=50°
Question 7.

Examining the measurement of the angles given below let’s conclude logically AB&CD are parallel.
Answer:

As we know that if if two lines are parallel and cut by a transversal then the corresponding angles are equal,

Alternate interior angles are equal


Alternate exterior angles are equal,


Sum of co-interior angles is 180°.


So, the reverse will also be true.


Now, from the figure we have,


Sum of co-interior angles = 125° +30° =155° ≠180°


Hence, AB and CD are not parallel.
Question 8.

Examining the measurement of the angles given below let’s conclude logically AB&CD are parallel.
Answer:


As we know that if two lines are parallel and cut by a transversal then the corresponding angles are equal,


Alternate interior angles are equal


Alternate exterior angles are equal,


Sum of co-interior angles is 180°.


∠6=∠3=120° (alternate interior angles)


∠4=∠8=60° (alternate interior angles)


Now, Sum of co-interior angles = 120° +60° =180°


Hence, AB||CD
Question 9.

Examining the measurement of the angles given below let’s conclude logically AB&CD are parallel.
Answer:


As we know that if two lines are parallel and cut by a transversal then the corresponding angles are equal,


Alternate interior angles are equal


Alternate exterior angles are equal,


Sum of co-interior angles is 180°.


∠6=∠2=75° (corresponding angles)


∠4=∠8=95° (corresponding angles)


Now,


Sum of co-interior angles = 75° +95° =170° ≠180°


So, AB is not parallel to CD.
Question 10.

In the figure AB||CD and

∠EGB=50°; then write the measurement of ∠AGE, ∠AGH, ∠BGH, ∠GHV, ∠GHD,

∠CGF and ∠DHF.
Answer:

Given that ∠EGB=50°

Since, AB||CD and EF is a transversal,


∠GHD=∠EGB=50° (corresponding angles) ……….(1)


∠CHF=∠EGB=50° (alternate exterior angles)


∠AGH=∠GHD =50° (alternate interior angles)


Now, ∠BGH+∠GHD=180° (co-interior angles)


⇒ ∠BGH+50° =180° (from (1))


⇒ ∠BGH=180°-50°


⇒ ∠BGH=130°


∠GHC=∠BGH=130° (alternate interior angles)


∠AGE=∠GHC=130° (corresponding angles)


∠DHF=∠AGE=130° (alternate exterior angles)


Hence, ∠GHD=50°, ∠CHF=50°, ∠AGH=50°, ∠BGH=130°, ∠GHC=130°, ∠AGE=130° and ∠DHF=130°
Question 11.

In the figure beside AB||CD; let’s write the measurement of ∠PQR.
Answer:


Given that AB||CD,


Draw a line through Q parallel to both AB and CD such that AB||XY||CD


Also, ∠APQ=30° and ∠CRQ=40°


Now, as AB||XY,


∠PQY=∠APQ=30° (alternate interior angles)………..(1)


Also, XY||CD,


∠YQR=∠CRQ=40° (alternate interior angles)…………(2)


Now, ∠PQR=∠PQY+∠YQR


⇒ ∠PQR=30°+40° (from (1) and (2))


⇒ ∠PQR=70°


Hence, ∠PQR=70°
Question 12.

In the figure beside PQ||RS, ∠BPQ=40°, ∠BPR=155° and ∠CRS=70°; let’s write the measurement of all the angles of APR
Answer:


Given that PQ||RS and ∠BPQ=40° and ∠BPR=155°


Also, ∠CRS=70°


Draw a line through A such that PQ||AY||RS


Since AB is a straight line ,


∠APB=180° (straight angle)


⇒ ∠BPR+∠APR=180°


⇒ 155° +∠APR=180°


⇒ ∠APR=180°-155°


⇒ ∠APR=25°…………..(1)


Since PQ||AY


So, ∠PAY=∠BPQ=40° (corresponding angles)


Similarly, AY||RS


So, ∠RAY=∠CRS=70° (corresponding angles)


Now, ∠PAR=∠PAY+∠RAY=40°+70°=110°


Now, in ∆APR,


∠PAR+∠APR+∠ARS=180°


⇒ 110°+25° +∠ARS=180°


⇒ 135° +∠ARS=180°


⇒ ∠ARS=180°-135°


⇒ ∠ARS=45°


Hence, ∠ARS=45°, ∠PAR=110° and ∠APR=25°
Question 13.

O is any point inside two parallel lines AB and CD. OP and OQ are two perpendiculars on AB and CD respectively. Let’s prove that P,O and Q are collinear.
Answer:


Given that AB||CD and O is a point inside AB and CD.


Also, P and Q are two points on AB and CD respectively such that


OPꞱAB and OQꞱCD.


So, ∠OPA=∠OPB=90°


And ∠OQC=∠OQD=90°


Draw XY through O such that AB||XY||CD


Since, AB||XY


So, ∠YOQ=∠BPO=90° (corresponding angles)


And ∠YOP=∠DQO=90° (corresponding angles)


∴ ∠POQ=∠YOP+∠YOQ=90°+90°=180°


So, PQ is s straight line.


∴ P,O and Q are collinear.


Proved.
Question 14.

Each pair of arms of two angles is parallel to each other. Let’s prove either these angles are equal in measurement or they are supplementary.
Answer:


Given that BC||DE.


∠ABC+∠CBD=180° (straight angle)


⇒ ∠ABC=180°-∠CBD ……………….(1)


Also, draw FG||BD,


∴ BFGD is a parallelogram


So, ∠CBD+∠BDE=180° (co-interior angles)


⇒ ∠BDE =180°-∠CBD ……………….(2)


From (1) and (2)


⇒ ∠ABC=∠BDE


Hence Proved.
Question 15.

The diagonal AC of the parallelogram ABCD bisects ∠BAD. Let’s prove that the diagonal AC also bisects ∠BCD.
Answer:


Given ABCD is a parallelogram with diagonals AC and BD.


Also, AC bisects ∠BAD


So, ∠DAC=∠BAC …………..(1)


Since, CD||AB,


∠DCA=∠BAC (alternate interior angles)


And ∠BCA=∠DAC (alternate interior angles)


From (1),


∠DAC=∠BAC


⇒ ∠BCA=∠DCA


Hence, AC bisects ∠BCD
Question 16.

Let’s prove if one angle of a parallelogram is right angle, then all other angles of the parallelogram are also right angles.
Answer:


Given ABCD is a parallelogram in which AB||CD and BC||AD.


∠A=90°


Since, DC||AB,


∠D+∠A=180° (co-interior angles)


⇒ ∠D+90°=180°


⇒ ∠D=180°-90°


⇒ ∠D=90°…………..(1)


Since, AD||BC,


∠B+∠A=180° (co-interior angles)


⇒ ∠B+90°=180°


⇒ ∠B=180°-90°


⇒ ∠B=90°…………..(2)


Also, ∠D+∠C=180° (co-interior angles)


⇒ ∠C+90°=180° (from(1))


⇒ ∠C=180°-90°


⇒ ∠C=90°…………..(3)


Hence, if each angle of a parallelogram is 90° then all the other angles of the parallelogram are 90°.