Page 8.29
(i) If any transversal intersects two parallel lines in such a way that the ratio of interior angles on the same side is 2 : 7 , then find the measure of the bigger angle in degrees.
(ii) If two parallel lines are intersected by a transversal, what relation will exist between alternate angles ?
(iii) If each of the two lines is parallel to a third line, what relation exists between them ?
(iv) At what angle, in degree measure, all perpendiculars drawn to a line segment are inclined to one another ?
(v) If P be a point not lying on line 'l' then how many lines parallel to 'l' through the point P, can be drawn ?
(vi) If a transversal intersects two parallel lines, then write two conditions for these two lines to be parallel.
(vii) If arms of any two angles are parallel, what relation exists between these angles ?
Sol :
(i) 2:7
2x+7x=180°
9x=180°
$x=\frac{180^{\circ}}{9}$
x=20°
bigger angle-
7x=7×20°
=140°
(ii) will be equal
(iii) they will be Parallel
(iv) 0° or 180°
(v) only one
(vi) Pair of alternate interior angles are equal; pair of corresponding angles are equal
(vii) will be parallel
(i) All perpendiculars drawn on a line segment are __
(ii) If a transversal intersects two lines in such a way that a pair of alternate interior angles are equal, then the two lines are __
(iii) A transversal intersects two parallel lines and one pair of alternate angles x and y are formed , then x and y are __
Sol :
(i) Parallel
(ii) Parallel
(iii) Equal
Sol :
Diagram
Construction: Draw a line RS through P parallel to the AB and CD line.
Given: AB||CD
∠ABP=30°,∠CDP=45°
To prove: ∠DPB=?
PS||AB(from construction)
∠SPB=∠ABP(alternate interior angle)
∵RS||AB
AB||CD(given)
∴RS||CD
∠SPD=∠CDP(alternate interior angles)
∠SPD=45°
∠DPB=∠SPB+∠SPD
=30°+45°
=75°
Type 2
<figure to be added>
[Hint: Draw a line PCQ through C parallel to AB and DE ]
Sol:
Diagram
Given: AB||DE
∠ABC=100°,∠CDE=120°
∠BCD=?
Construction: We can extend the line AB towards the line CD.
Now,
∠ABC+∠CBP=180°(Linear pair)
100°+∠CBP=180°
∠CBP=80°
∠CBP+∠BCP=∠CDE(exterior angle is equal to the sum of opposite interior angles)
80°+∠BCD=120° (∵∠CPT=∠PDE or ∠CPT=∠CDE)
∠BCD=40°
<figure to be added>
Sol :
Diagram
∠5+∠6=180°(linear pair)
∠5+60°=180°
Exercise 8.2
Type 1Question 1
Answer the following questions :(i) If any transversal intersects two parallel lines in such a way that the ratio of interior angles on the same side is 2 : 7 , then find the measure of the bigger angle in degrees.
(ii) If two parallel lines are intersected by a transversal, what relation will exist between alternate angles ?
(iii) If each of the two lines is parallel to a third line, what relation exists between them ?
(iv) At what angle, in degree measure, all perpendiculars drawn to a line segment are inclined to one another ?
(v) If P be a point not lying on line 'l' then how many lines parallel to 'l' through the point P, can be drawn ?
(vi) If a transversal intersects two parallel lines, then write two conditions for these two lines to be parallel.
(vii) If arms of any two angles are parallel, what relation exists between these angles ?
Sol :
(i) 2:7
2x+7x=180°
9x=180°
$x=\frac{180^{\circ}}{9}$
x=20°
bigger angle-
7x=7×20°
=140°
(ii) will be equal
(iii) they will be Parallel
(iv) 0° or 180°
(v) only one
(vi) Pair of alternate interior angles are equal; pair of corresponding angles are equal
(vii) will be parallel
Question 2
Fill up the blanks :(i) All perpendiculars drawn on a line segment are __
(ii) If a transversal intersects two lines in such a way that a pair of alternate interior angles are equal, then the two lines are __
(iii) A transversal intersects two parallel lines and one pair of alternate angles x and y are formed , then x and y are __
Sol :
(i) Parallel
(ii) Parallel
(iii) Equal
Question 3
P is a point between two parallel lines AB and CD. If ∠ABP=30° , ∠CDP=45° , then find the value of ∠DPB.Sol :
Diagram
Construction: Draw a line RS through P parallel to the AB and CD line.
Given: AB||CD
∠ABP=30°,∠CDP=45°
To prove: ∠DPB=?
PS||AB(from construction)
∠SPB=∠ABP(alternate interior angle)
∵RS||AB
AB||CD(given)
∴RS||CD
∠SPD=∠CDP(alternate interior angles)
∠SPD=45°
∠DPB=∠SPB+∠SPD
=30°+45°
=75°
Type 2
Question 4
In the given figure, if AB||DE, then find the value of ∠BCD<figure to be added>
[Hint: Draw a line PCQ through C parallel to AB and DE ]
Sol:
Diagram
Given: AB||DE
∠ABC=100°,∠CDE=120°
∠BCD=?
Construction: We can extend the line AB towards the line CD.
Now,
∠ABC+∠CBP=180°(Linear pair)
100°+∠CBP=180°
∠CBP=80°
∠CBP+∠BCP=∠CDE(exterior angle is equal to the sum of opposite interior angles)
80°+∠BCD=120° (∵∠CPT=∠PDE or ∠CPT=∠CDE)
∠BCD=40°
Question 5
In the given figure, if ∠1=130° and ∠6=60° , is AB||CD ?<figure to be added>
Sol :
Diagram
∠5+∠6=180°(linear pair)
∠5+60°=180°
∠5=120°
∠1≠5
∴AB∦CD
Page 8.30
Question 6
In the given figure , m||n and ∠2:∠3=2:3, then find the values of angles denoted by 1,2,.....7,8.
Sol:
Let ∠2=2x
∠3=3x
∠2+∠3=180°(linear pair)
2x+3x=180°
5x=180°
$x=\frac{180^{\circ}}{5}$
x=36°
∠2=2x
=2×36°
=72°
∠3=3x
=3×36°
=108°
∠4=∠2(V.OA)
∠4=72°
∠1=∠3(V.O.A)
∠1=108°
∠5=∠1(corresponding angles)
∠5=108°
∠7=∠5(V.O.A)
∠7=108°
∠6=∠2(corresponding angles)
∠6=72°
∠8=∠6(V.O.A)
∠8=72°
Question 7
In figures (i) , (ii) , (iii) , (iv) if AB||CD , then find the value of x and y in each of the following cases:(i)
Sol :
∠SRB=∠ARP(V.O.A)
x=65°
∠SRB+∠RSD=180°(Interior angles on same side of transversal are supplementary)
65°+y=180°
y=180°-65°
y=115°
(ii)
Sol :
∠CSR=∠SRB(Alternate interior angle)
x=105°
∠CSR+∠CSQ=180°
105°+y=180°
y=75°
(iii)
(iv)
Sol :
Question 8
Write the pair of parallel lines in each of the following cases shown in figures below:(i)
Sol :
x+60°=180°
x=120°
y=60°(alternate angle)
(ii)
Sol :
x=100°(alternate angle)
y=80°(corresponding angle)
(iii)
Sol :
∠D+∠A
=115°+65°
=180°
∠A+∠B
=115°+65°
=180°
AD||BC
Type 3
Question 9
In the given figure , p is a transversal of the lines m and n and ∠1=60° and $\angle 2=\dfrac{2}{3}$ of a right angle. Prove that m||n.
Sol :
Diagram
Given: P is a transversal of the lines m and n
∠1=60° ,$\angle=\frac{2}{3}$ of right angle
To prove: m∥n
Proof: $\angle 2=\frac{2}{3}$ of right angle
$\angle 2=90^{\circ} \times \frac{2}{3}$
∠2=60°
∠3=∠2(V.O.A)
∠3=60°
∵∠1=60°(∠1 and ∠3 are pair of corresponding angle)
If pair of corresponding angles are equal then the lines are parallel
∴ m∥n
Question 10
In the given figure , m||n and p||q. If ∠1=75° , then prove that $\angle 2=\angle 1+\dfrac{1}{3}$ of right angles
Sol :
[Hint: ∠BAC=∠1=75° [alternate angles]
Now , ∠2+∠BAC=180° [sum of the interior angle son the same side of transversal n]
∴ ∠2=180°-∠BAC
=180°-75°
=105°=75°+30°$=\angle 1+\dfrac{1}{3}$ of right angle ]
Sol :
Given: m||n and p|q , ∠1=75°
To prove: ∠2=∠1+$\frac{1}{3}$ of right angles
Proof: ∵p||q (given)
∠CDB=∠1(corresponding angle)
∠CDB=75°
∵ m|n (given)
∠CDB+∠2=180°
75°+∠2=180°
∠2=180°-75°
∠2=105°
$\angle 2=\angle 1+\frac{1}{3}$ of right angle∠2=105°
∠2=75°+30°
Question 11
In the given figure , if x=y and a=b , prove that r||n
Sol :
Given:x=y or a=b
To prove:r||n
Proof:
x=y(given)
r||m (∵ if corresponding angle of transversal are equal then the given lines is are parallel)
a=b (given)
m||n (∵ if corresponding angle of transversal are equal then the given lines is are parallel)
∵r||m and m||n
∴r||n
Question 12
lf m,n, p are three lines such that p||m and n⊥p, prove that n⊥mSol :
Diagram
Given: P||m or n⟂p
To prove: n⟂m
Proof: n⟂p (given)
∠1=90°
p||m
∠2=∠1 (alternate interior angle)
Question 13
What are the conditions for two lines to be parallel ?Sol :
Alternate angles are equal or corresponding angles are equal or , the co-interior angels are supplementary
Question 14
In the given figure, l and m are two intersecting lines and p||l and q||m. Prove that p and q are also intersecting lines .
Sol:
Given: l and m are intersecting lines p||l and q||m
Let us assume that p and q are not intersecting lines, which means p||q.
We know ,
p||l and q||m
∴It implies that q||p, q||l , l||m [∵p||l] [∵q||m]
But , we are given that l and m intersect each other.
Therefore , our assumption is wrong.
Hence , p and q are intersecting lines.
Question 15
If a line is perpendicular to any one of the two parallel lines , prove that this line will also be perpendicular to the other parallel lineSol :
Diagram
Given: p||q and r⟂p
To prove: r⟂q
Proof: p||q (given)
∠2=∠1 (alternate interior angle)
∠2=90°
∴r⟂q
Question 16
In the given figure, AB||CD and AD||BC, prove that ∠DAB=∠DCB.
Sol :
[Hint: AB||DC and transversal BC meets them]
∴ ∠ABC+∠BCD=180°..(i)
Again AD||BC
∴ ∠DAB+∠ABC=180°..(ii)
From (i) and (ii) , ∠ABC+∠BCD=∠DAB+∠ABC=180°
or ∠DAB=∠BCD=∠DAB ]
Sol :
Given: AB||CD and AD||BC
To prove: ∠DAB=∠DCB
Proof: AB||CD (given)
∠ABC+∠DCB=180°..(i)
AD||BC(given)
∠DAB+∠ABC=180°..(ii)
From equation (i) and (ii)
∠DAB+∠ABC=∠ABC+∠DCB
∠DAB=∠DCB
Question 17
A transversal intersects two given lines m and n in such a way that interior angles on the same side of the transversal are equal , then it is always true that the given lines are parallel ? If not , state the condition under which the two lines will be parallel.[Hint: m||n only when transversal line is perpendicular to both m and n]
Sol :
Diagram
∠AOC+∠BOC=180°
6x+30°+4x=180°
10x+30°=180°
10x=180°-30°
10x=150°
$x=\frac{150^{\circ}}{10}$
x=15°
Question 18
Prove that the two lines that are respectively perpendicular to two intersecting lines intersect each other .
[Hint: As shown in the given figure , lines MN and PQ are perpendicular to intersecting lines AB and CD respectively. If possible , suppose lines MN and PQ do not intersect , then definitely MN||PQ.
But we know that two parallel lines cannot be simultaneously perpendicular to two intersecting lines. So , it is wrong to assume that MN||PQ. Hence MN and PQ are intersecting lines ]
Sol :
Question 19
In the given figure, lines AB and CD are parallel and any point P lies between these lines . Prove that ∠ABP+∠CDO=∠DPB
[Hint: Through point P , draw a line PM||AB and proceed.]
Sol :
Question 20
If a transversal intersects two lines such that the bisectors of a pair of alternate angles are parallel, then prove that the two lines are parallelSol :
[Hint: Proceed as in solved example 20 page 8.27]
Diagram
Given: ∠AOB=∠COD..(i)
∠BOC=∠DOA..(ii)
from diagram
∠AOB+∠BOC+∠COD+∠DOA=360°
⇒2∠COD+2∠DOA=360°
⇒∠COD+∠DOA$=\frac{360^{\circ}}{2}$
⇒∠AOC=180°
Similarly ∠BOD=180°
Sum of all angle at a point on a straight line is 180°
Question 21
Prove that the two lines that are respectively perpendiculars to two parallel lines are parallel to each other.Sol :
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