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Lines and Angles
1.
In the given figure, if x: y =1:4, then find the values of x and y, respectively.
Solution.
Given, x : y =1:4
Let x = a and y = 4a
x + y = 180°
[linear pair axiom]
 a+4a= 180°
 5a =180°
1800
 a  360
 a
5
Hence, x – a = 36°
and y = 4a = 4 × 36° = 144°
2.
In the given figure, if AB
x and y.
CD,  APQ =60°and  PRD=137°, then find the values of
Solution.
AB CD
[given]
 x = 60°
[alternate interior angles]
and x + y =137°
(exterior angle of a triangle is equal to the sum of opposite interior angles)
 60° + y =137°
 y =137° – 60°
= 77°
Hence, x = 60° and y = 77°
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4.
In  ABC,  A +  B = 122°and  B +  C =111°, find the measures of  B and  C.
Solution.
Given,
..(i)
 A +  B = 122°
and
..(ii)
 B +  C=111°
In  ABC,
(  A +  B) +  C =180°
[sum of all angles of a triangle is 180°]
122° +  C = 180°
[from Eq. (i)]
  C=180°– 122°
  C = 58°
On putting the value of  C = 58° in Eq, (ii), we get
 B + 58° = 1110
  B = 111° – 58° = 530
Hence,  B = 53° and  C = 58°
In the given figure, AB CD EF. If  CEF = 160° and  BCE=10°, then find  x.
5.
Solution.
In given Figure, AB CD EF and CE is a transversal.
Therefore, sum of two interior angles is 180°
 FEC+  ECD=180°
i.e.,
 160°+  ECD=180° [  FEC =160°, given]
  ECD=20°
Since, AB CD and AB is a transversal
  ABC =  BCD
[alternate angles]
  x=  BCE+  ECD .
= 10° + 20°
[  BCE =10°, given]
= 30°
In the given figure, XOY is a straight line and OQ  XY at O.
3.
Show that 2  QOP =  YOP –  XOP
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Solution.
Given OQ  XY
To prove 2  QOP =  YOP –  XOP
Proof
OQ  XOY
[given]
  XOQ =  QOY = 90°
  QOP +  XOP =  QOY
  QOP =  QOY –  XOP
  QOP +  YOP =  QOP –  XOP
  QOP +  QOP =  YOP –  XOP
 2  QOP =  YOP –  XOP
6.
Hence proved.
In the given figure,  AOC =  ACO and  BDO =  BOD. Show that AC
BD.
Solution.
Given  AOC =  ACO and  BDO =  BOD
To prove AC BD
Proof  AOC =  ACO [given]
 AOC =  BOD
[vertically opposite angles]
  ACO =  BOD
..(i)
Also,
 BOD =  BDO
... (ii) [given]
From Eqs. (i) and (ii), we get
 ACO =  BDO
i.e., angles are alternate interior angles.
 AC BD
Hence proved.
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In the given figure, EF DQ and AB CD. If  FEB = 64° and  PDC=27°, then find
 PDQ,  AED and  DEF.
Solution.
EF DQ and AB CD
  AED =  PDC
[corresponding angles]
 AED = 270
[given]
  AED +  DEF +  EES = 180°
[linear pair axiom]
 27° +  DEF + 64° = 180°
[
 FEB = 64°, given]

91° +  DEP =180°
  DEF =180° – 91°
  DEF = 89°
  PDQ =  DEF
[corresponding angles]
  PDQ = 89°
Hence,
 PDQ = 89°,  AED = 27° and  DEF = 89°.
8.
ABCDE is a regular pentagon and bisector of  BAE meets CD at M. If bisector of  BCD
meets AM at P, then find  CPM.
Solution.
Given, a regular pentagon bisector of  A meets CD at M and bisector of  BCD meets
AM at P.
Since, each angle of a regular pentagon
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=
0
540
 1080
5
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[put n = 5]
So, AM is the bisector of  A.
1
1
 BAM  A   1080  540
2
2
Now, in quadrilateral ABCM,
 BAM +  C +  B +  AMC = 360°
 54° +108° +108° +  AMC – 360°
[
 B =  C =108°]
 270° +  AMC = 360°
  AMC = 360° –270°
  AMC = 90°
  PMC =  AMC = 90°
Since, CP is the bisector of  BCD.
1
1
 PCM   C   1080  540
2
2
Now, in  PCM,
 PCM +  PMC +  CPM = 180°
[by angle sum property of a triangle is 180°]
 54° + 90° +  CPM = 180°
  CPM = 180° – 144
  CPM = 36°.
Hence, CPM is 360.
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9.
In the given figure, AB
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DE, find the value of  BCD.
Solution.
Since, AB DE and DP is a transversal.
  EDF = ZDFB = 130°
[alternate angles] ...(i)
In a figure, DFC is a straight line.
  DPB +  BFC = 180°
[linear pair axiom]
 130° +  BFC=1800
[from Eq. (i)]
  BFC = 50°
...(ii)
In  BPC,
 ABC =  BCF +  BFC
[exterior angle is equal to the sum of two interior angles]
 80° =  BCF + 50°
[from Eq. (ii)]
  BCP = 30°
10. If the sides of an angle are respectively parallel to the sides of another angle, then prove
that these angles are either equal or supplementary.
Solution.
Case I
As
QR BC
 2 = 3
[corresponding angles] ...(i)
and PQ AB, then
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1 = 2
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[corresponding angles]...(ii)
From Eqs. (i) and (ii), we get
1 = 2 = 3
Case II
QR
BC
1  3
[corresponding angles]
But  1 +  2 = 180°, then
 3 +  2 = 180°, supplementary.
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Chapter Test {Lines and Angles}
M: Marks: 40
1.
2.
M: Time: 40 Min.
In the given figure, lines XY and MN intersect at O. If  POY = 900 and a : b = 2 : 3, then find
the value of c.
[4]
Solution.
Given,  POY = 90°
[linear pair axiom]
  POX = 90°
0
  XOM +  MOP = 90
...(i)
 a + b = 90°
Also, given
a : b = 2 :3
Let a = 2k and b = 3k
We have,
a + b = 90°
[from Eq.(i)]
 2k + 3k = 90°
 5k = 90°
900
 k
5
 k = 180
 a = 2 × 180 = 360
and b = 3 × 180 = 540
Now,
[Linear pair axiom]
 MOX +  XON = 180°
 b + c = 180°
[put b = 54°]
 54° + C = 180°
 c = 180° – 54°
 c = 126°
Two plane mirrors m and n are placed perpendicular to each other, as shown in figure. An
incident ray AB to the first mirror is first reflected in the direction of BC and then reflected
by the second mirror in the direction of CD. Prove that AB CD.
[4]
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Solution.
Let BM and CN be the normals to mirrors m and n, respectively and BM and CN intersect
at O.
Now, BM  m, CN  n and m  n
 BM  CN
  BOC = 90°
...(i)
 2 +  3 = 90°
By the laws of reflection, we have
 1 =  2 and  3 =  4
[ angle of incidence = angle of reflection]
 1 + 4 = 2 + 3
(  2 +  3)
 1 + 2 + 3 + 4 = 2
[ adding  2 and  3 both sides]
 (  1 +  2) + (  3 +  4) = 2(90°) = 180°
[ from Eq. (i)]
  ABC +  BCD = 180°
But  ABC and  BCD are consecutive interior angles formed, when the transversal BC
cuts AB and CD.
 AB CD Hence proved.
In the given figure, m and n are two plane mirrors parallel to each other. Show that the
incident ray CA is parallel to the reflected ray BD.
[4]
Solution.
Given, two plane mirrors m and n, such that m n. An incident ray CA after
reflection takes the path AB and then BD. AP and BQ are the normals to the plane
mirrors m and n, respectively.
Since, AP  m, BQ  n and m n
 AP  n, BQ  m =  AP BQ
Thus, AP and BQ are two parallel lines and a transversal AB cuts them at A and B,
respectively.
  2=  3
...(i)
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[alternate interior angles]
But
...(ii)
1 = 2
and
...(iii)
3 = 4
[by the laws of reflection]
1 + 2 = 2 + 2
[adding  2 both sides of Eq. (ii)]
and
3 + 4 = 3 + 3
[adding  3 both sides of Eq. (iii)]
  1 +  2 = 2(  2)
and
 3 +  4 = 2(  3)
  1+  2 =  3 +  4
[  2 =  3  2(  2) = 2(  3)]
  CAB =  ABD
Thus, lines CA and BD are intersected by transversal AB such that  CAB =
 ABD.
i.e., alternate interior angles are equal.
Hence, AC BD
In the figure given below, prove that: x = a + b + c.
[4]
Solution.
In the given figure, join B and D and produce to E.
5.
Here,  1 +  2 = b and  3 +  4 = x
...(i)
Also,  3 =  1 + a
...(ii)
[exterior angle is equal to the sum of opposite interior angles]
4 = 2 + c
and
...(iii)
[exterior angle is equal to the sum of opposite interior angles]
On adding Eqs. (ii) and (iii), we get
3 + 4 = 1 + 2 + a + c
x=b+a+c
[from Eq. (i)]
In the given figure, side BC of  ABC is produced to form ray BD and CE
Show that  ACD =  A+  B. Deduce that:  A +  B +  C = 180°.
BA.
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Solution.
Given, CE BA and AC is the transversal.
...(i)
 4 =1
[alternate interior angles]
Again, CE BA and BD is the transversal.
...(ii)
[corresponding angles]
 5 = 2
On adding Eqs. (i) and (ii), we get
4 + 5 = 1 + 2
Hence,
 ACD =  A +  B
Also,  1 +  2 =  4 +  5
[exterior angle is equal to the sum of the opposite interior angles]
  1 +  2+  3 =  3 +  4 +  5 [adding  3 both sides]
  A +  B +  C = 180°
[  3 +  4 +  5 =  BCD = a straight angle]
6.
In the given figure, AB CD. Find the values of x, y and z.
[4]
Solution.
Since AB CD
(Alternate angles)
 y = 750
0
z + 75 = 1250
(by exterior angle property)
 z = 1250 – 750
z = 500
Also, x + y = 1800
(Linear pair)
0
0
 x + 75 = 180
 x = 1800 – 750
x = 1050
Hence, x = 1050, y = 750 and z = 500
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In the given figure, PQ RS and EF QS. If  PQS = 60°, then find the measure of  RFE. [4]
Solution.
Since PQ RS
  PQS +  RSQ = 1800
(co-interior angles)
 600 +  RSQ = 1800
  RSQ = 1200
Also EF QS
(corresponding angles)
  RFE =  RSQ
0
  RFE = 120
In the given figure, AB CD EF. Also, EA  AB. = 70°, then find the values of p, q and r.
[4]
Solution.
Since AB CD
 p = q (corresponding angles)
Also, EF CD
 q + 700 = 1800
(co-interior angles)
 q = 1800 – 700
 q = 1100
 p = 1100
Since AB CD, CD EF
 AB EF
(co-interior angles)
  EAB +  FEA = 1800
 900 +  FEA = 1800
  FEA = 900
 r + 700 = 900
 r = 200
Hence, p = 1100 , q = 1100 and r = 200.
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In the given figure, l p and m q and lines l and m are intersecting lines. Prove that lines p
and q also intersects.
[4]
Solution.
Given : l p, m q
To prove : p and q intersects
Proof :
Since l and m are intersecting lines and l p
 p and m intersect.
Now, p and m intersect and m q
Hence, p and q intersect.
10. In the given figure, ABCD in a quadrilateral in which  ABC = 73°,  C = 97° and  D =
110°. If AE DC and BE AD and AE intersects BC at F, then find the value of  EBF. [4]
Solution.
Given : AE DC, BE AD,  ABC = 730,  C = 970
To find :  EBF
Since AE DC
 1100  DAE  1800
(Co-interior 's )
  DAF = 700
Also,  BEA =  DCF = 970
(Corresponding B 's )
0
 BEA = 97
Now, in AFB ,
 FAB +  ABF +  AFB = 1800
(Angle sum property)
  FAB + 730 + 970 = 1800
  FAB = 1800 – 1700
= 100
Now, AD BE
(Given)
  DAB +  ABE = 1800
(Co-interior angles)
  DAF +  FAB +  ABF +  EBF = 1800
 700 + 100 + 730 +  EBE = 1800
  FBE = 1800 – 1530
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 FBE =
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