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Page 1
Question:29
In the given figure, l || m and a transversal t cuts them. If ?1 = 120°, find the measure of each of the remaining marked angles.
Solution:
We have, ?1 = 120°
. Then,
?1 = ?5 [Corresponding angles] ? ?5 = 120° ?1 = ?3 [Vertically-opposite angles] ? ?3 = 120°
?5 = ?7 [Vertically-opposite angles] ? ?7 = 120° ?1 + ?2 = 180° [Since AFB is a straight line] ? 120°+ ?2 = 180°
? ?2 = 60° ?2 = ?4 [Vertically-opposite angles] ? ?4 = 60° ?2 = ?6 [Corresponding angles]
? ?6 = 60° ?6 = ?8 [Vertically-opposite angles] ? ?8 = 60° ? ?1 = 120°, ?2 = 60°, ?3 = 120°, ?4 = 60°, ?5 = 120°, ?6 = 60°, ?7 = 120° and ?8 = 60°
Question:30
In the given figure, l || m and a transversal t cuts them. If ?7 = 80°, find the measure of each of the remaining marked angles.
Solution:
In the given figure, ?7 and ?8 form a linear pair.
? ?7 + ?8 = 180º
? 80º + ?8 = 180º
? ?8 = 180º - 80º = 100º
Now,
?6 = ?8 = 100º Verticallyoppositeangles
?5 = ?7 = 80º Verticallyoppositeangles
It is given that, l || m and t is a transversal.
? ?1 = ?5 = 80º Pairofcorrespondingangles
?2 = ?6 = 100º Pairofcorrespondingangles
?3 = ?7 = 80º Pairofcorrespondingangles
?4 = ?8 = 100º Pairofcorrespondingangles
Question:31
In the given figure, l || m and a transversal t cuts them. If ?1 : ?2 = 2 : 3, find the measure of each of the marked angles.
Page 2
Question:29
In the given figure, l || m and a transversal t cuts them. If ?1 = 120°, find the measure of each of the remaining marked angles.
Solution:
We have, ?1 = 120°
. Then,
?1 = ?5 [Corresponding angles] ? ?5 = 120° ?1 = ?3 [Vertically-opposite angles] ? ?3 = 120°
?5 = ?7 [Vertically-opposite angles] ? ?7 = 120° ?1 + ?2 = 180° [Since AFB is a straight line] ? 120°+ ?2 = 180°
? ?2 = 60° ?2 = ?4 [Vertically-opposite angles] ? ?4 = 60° ?2 = ?6 [Corresponding angles]
? ?6 = 60° ?6 = ?8 [Vertically-opposite angles] ? ?8 = 60° ? ?1 = 120°, ?2 = 60°, ?3 = 120°, ?4 = 60°, ?5 = 120°, ?6 = 60°, ?7 = 120° and ?8 = 60°
Question:30
In the given figure, l || m and a transversal t cuts them. If ?7 = 80°, find the measure of each of the remaining marked angles.
Solution:
In the given figure, ?7 and ?8 form a linear pair.
? ?7 + ?8 = 180º
? 80º + ?8 = 180º
? ?8 = 180º - 80º = 100º
Now,
?6 = ?8 = 100º Verticallyoppositeangles
?5 = ?7 = 80º Verticallyoppositeangles
It is given that, l || m and t is a transversal.
? ?1 = ?5 = 80º Pairofcorrespondingangles
?2 = ?6 = 100º Pairofcorrespondingangles
?3 = ?7 = 80º Pairofcorrespondingangles
?4 = ?8 = 100º Pairofcorrespondingangles
Question:31
In the given figure, l || m and a transversal t cuts them. If ?1 : ?2 = 2 : 3, find the measure of each of the marked angles.
Solution:
Let ?1 = 2k and ?2 = 3k, where k is some constant.
Now, ?1 and ?2 form a linear pair.
? ?1 + ?2 = 180º
? 2k + 3k = 180º
? 5k = 180º
? k = 36º
? ?1 = 2k = 2 × 36º = 72º
?2 = 3k = 3 × 36º = 108º
Now,
?3 = ?1 = 72º Verticallyoppositeangles
?4 = ?2 = 108º Verticallyoppositeangles
It is given that, l || m and t is a transversal.
? ?5 = ?1 = 72º Pairofcorrespondingangles
?6 = ?2 = 108º Pairofcorrespondingangles
?7 = ?1 = 72º Pairofalternateexteriorangles
?8 = ?2 = 108º Pairofalternateexteriorangles
Question:32
For what value of x will the line l and m be parallel to each other?
Solution:
For the lines l and m to be parallel
? 3x -20 = 2x +10 [Corresponding Angles] ? x = 30
Question:33
For what value of x will the lines l and m be parallel to each other?
Solution:
? 3x +5 +4x = 180 [Consecutive Interior Angles] ? 7x = 175 ? x = 25
Question:34
In the given figure, AB || CD and BC || ED. Find the value of x.
Solution:
BC ? ED
Page 3
Question:29
In the given figure, l || m and a transversal t cuts them. If ?1 = 120°, find the measure of each of the remaining marked angles.
Solution:
We have, ?1 = 120°
. Then,
?1 = ?5 [Corresponding angles] ? ?5 = 120° ?1 = ?3 [Vertically-opposite angles] ? ?3 = 120°
?5 = ?7 [Vertically-opposite angles] ? ?7 = 120° ?1 + ?2 = 180° [Since AFB is a straight line] ? 120°+ ?2 = 180°
? ?2 = 60° ?2 = ?4 [Vertically-opposite angles] ? ?4 = 60° ?2 = ?6 [Corresponding angles]
? ?6 = 60° ?6 = ?8 [Vertically-opposite angles] ? ?8 = 60° ? ?1 = 120°, ?2 = 60°, ?3 = 120°, ?4 = 60°, ?5 = 120°, ?6 = 60°, ?7 = 120° and ?8 = 60°
Question:30
In the given figure, l || m and a transversal t cuts them. If ?7 = 80°, find the measure of each of the remaining marked angles.
Solution:
In the given figure, ?7 and ?8 form a linear pair.
? ?7 + ?8 = 180º
? 80º + ?8 = 180º
? ?8 = 180º - 80º = 100º
Now,
?6 = ?8 = 100º Verticallyoppositeangles
?5 = ?7 = 80º Verticallyoppositeangles
It is given that, l || m and t is a transversal.
? ?1 = ?5 = 80º Pairofcorrespondingangles
?2 = ?6 = 100º Pairofcorrespondingangles
?3 = ?7 = 80º Pairofcorrespondingangles
?4 = ?8 = 100º Pairofcorrespondingangles
Question:31
In the given figure, l || m and a transversal t cuts them. If ?1 : ?2 = 2 : 3, find the measure of each of the marked angles.
Solution:
Let ?1 = 2k and ?2 = 3k, where k is some constant.
Now, ?1 and ?2 form a linear pair.
? ?1 + ?2 = 180º
? 2k + 3k = 180º
? 5k = 180º
? k = 36º
? ?1 = 2k = 2 × 36º = 72º
?2 = 3k = 3 × 36º = 108º
Now,
?3 = ?1 = 72º Verticallyoppositeangles
?4 = ?2 = 108º Verticallyoppositeangles
It is given that, l || m and t is a transversal.
? ?5 = ?1 = 72º Pairofcorrespondingangles
?6 = ?2 = 108º Pairofcorrespondingangles
?7 = ?1 = 72º Pairofalternateexteriorangles
?8 = ?2 = 108º Pairofalternateexteriorangles
Question:32
For what value of x will the line l and m be parallel to each other?
Solution:
For the lines l and m to be parallel
? 3x -20 = 2x +10 [Corresponding Angles] ? x = 30
Question:33
For what value of x will the lines l and m be parallel to each other?
Solution:
? 3x +5 +4x = 180 [Consecutive Interior Angles] ? 7x = 175 ? x = 25
Question:34
In the given figure, AB || CD and BC || ED. Find the value of x.
Solution:
BC ? ED
and CD is the transversal.
Then,
?BCD + ?CDE = 180° [Angles on the same side of a transversal line are supplementary] ? ?BCD +75 = 180 ? ?BCD = 105°
AB ? CD
and BC is the transversal.
?ABC = ?BCD (alternate angles) ? x° = 105° ? x = 105
Question:35
In the given figure, AB || CD || EF. Find the value of x.
Solution:
EF ? CD
and CE is the transversal.
Then,
?ECD + ?CEF = 180° [Consecutive Interior Angles] ? ?ECD +130° = 180° ? ?ECD = 50°
Again, AB ? CD
and BC is the transversal.
Then,
?ABC = ?BCD [Alternate Interior Angles] ? 70° = x +50° [ ? ?BCD = ?BCE + ?ECD] ? x = 20°
Question:36
In the given figure, AB || CD. Find the values of x, y and z.
Solution:
AB ? CD
and let EF and EG be the transversals.
Now, AB ? CD
and EF is the transversal.
Then,
?AEF = ?EFG [Alternate Angles] ? y° = 75° ? y = 75
Also,
?EFC + ?EFD = 180° [Since CFD is a straight line] ? x +y = 180 ? x +75 = 180 ? x = 105
And,
?EGF+ ?EGD = 180° [Since CFGD is a straight line] ? ?EGF+125 = 180 ? ?EGF = 55°
We know that the sum of angles of a triangle is 180°
?EFG+ ?GEF+ ?EGF = 180° ? y +z +55 = 180 ? 75 +z +55 = 180 ? z = 50 ? x = 105, y = 75 and z = 50
Question:37
In each of the figures given below, AB || CD. Find the value of x in each case.
Solution:
i
Draw EF ? AB ? CD
.
Now, AB ? EF
and BE is the transversal.
Then,
?ABE = ?BEF [Alternate Interior Angles] ? ?BEF = 35°
Page 4
Question:29
In the given figure, l || m and a transversal t cuts them. If ?1 = 120°, find the measure of each of the remaining marked angles.
Solution:
We have, ?1 = 120°
. Then,
?1 = ?5 [Corresponding angles] ? ?5 = 120° ?1 = ?3 [Vertically-opposite angles] ? ?3 = 120°
?5 = ?7 [Vertically-opposite angles] ? ?7 = 120° ?1 + ?2 = 180° [Since AFB is a straight line] ? 120°+ ?2 = 180°
? ?2 = 60° ?2 = ?4 [Vertically-opposite angles] ? ?4 = 60° ?2 = ?6 [Corresponding angles]
? ?6 = 60° ?6 = ?8 [Vertically-opposite angles] ? ?8 = 60° ? ?1 = 120°, ?2 = 60°, ?3 = 120°, ?4 = 60°, ?5 = 120°, ?6 = 60°, ?7 = 120° and ?8 = 60°
Question:30
In the given figure, l || m and a transversal t cuts them. If ?7 = 80°, find the measure of each of the remaining marked angles.
Solution:
In the given figure, ?7 and ?8 form a linear pair.
? ?7 + ?8 = 180º
? 80º + ?8 = 180º
? ?8 = 180º - 80º = 100º
Now,
?6 = ?8 = 100º Verticallyoppositeangles
?5 = ?7 = 80º Verticallyoppositeangles
It is given that, l || m and t is a transversal.
? ?1 = ?5 = 80º Pairofcorrespondingangles
?2 = ?6 = 100º Pairofcorrespondingangles
?3 = ?7 = 80º Pairofcorrespondingangles
?4 = ?8 = 100º Pairofcorrespondingangles
Question:31
In the given figure, l || m and a transversal t cuts them. If ?1 : ?2 = 2 : 3, find the measure of each of the marked angles.
Solution:
Let ?1 = 2k and ?2 = 3k, where k is some constant.
Now, ?1 and ?2 form a linear pair.
? ?1 + ?2 = 180º
? 2k + 3k = 180º
? 5k = 180º
? k = 36º
? ?1 = 2k = 2 × 36º = 72º
?2 = 3k = 3 × 36º = 108º
Now,
?3 = ?1 = 72º Verticallyoppositeangles
?4 = ?2 = 108º Verticallyoppositeangles
It is given that, l || m and t is a transversal.
? ?5 = ?1 = 72º Pairofcorrespondingangles
?6 = ?2 = 108º Pairofcorrespondingangles
?7 = ?1 = 72º Pairofalternateexteriorangles
?8 = ?2 = 108º Pairofalternateexteriorangles
Question:32
For what value of x will the line l and m be parallel to each other?
Solution:
For the lines l and m to be parallel
? 3x -20 = 2x +10 [Corresponding Angles] ? x = 30
Question:33
For what value of x will the lines l and m be parallel to each other?
Solution:
? 3x +5 +4x = 180 [Consecutive Interior Angles] ? 7x = 175 ? x = 25
Question:34
In the given figure, AB || CD and BC || ED. Find the value of x.
Solution:
BC ? ED
and CD is the transversal.
Then,
?BCD + ?CDE = 180° [Angles on the same side of a transversal line are supplementary] ? ?BCD +75 = 180 ? ?BCD = 105°
AB ? CD
and BC is the transversal.
?ABC = ?BCD (alternate angles) ? x° = 105° ? x = 105
Question:35
In the given figure, AB || CD || EF. Find the value of x.
Solution:
EF ? CD
and CE is the transversal.
Then,
?ECD + ?CEF = 180° [Consecutive Interior Angles] ? ?ECD +130° = 180° ? ?ECD = 50°
Again, AB ? CD
and BC is the transversal.
Then,
?ABC = ?BCD [Alternate Interior Angles] ? 70° = x +50° [ ? ?BCD = ?BCE + ?ECD] ? x = 20°
Question:36
In the given figure, AB || CD. Find the values of x, y and z.
Solution:
AB ? CD
and let EF and EG be the transversals.
Now, AB ? CD
and EF is the transversal.
Then,
?AEF = ?EFG [Alternate Angles] ? y° = 75° ? y = 75
Also,
?EFC + ?EFD = 180° [Since CFD is a straight line] ? x +y = 180 ? x +75 = 180 ? x = 105
And,
?EGF+ ?EGD = 180° [Since CFGD is a straight line] ? ?EGF+125 = 180 ? ?EGF = 55°
We know that the sum of angles of a triangle is 180°
?EFG+ ?GEF+ ?EGF = 180° ? y +z +55 = 180 ? 75 +z +55 = 180 ? z = 50 ? x = 105, y = 75 and z = 50
Question:37
In each of the figures given below, AB || CD. Find the value of x in each case.
Solution:
i
Draw EF ? AB ? CD
.
Now, AB ? EF
and BE is the transversal.
Then,
?ABE = ?BEF [Alternate Interior Angles] ? ?BEF = 35°
Again, EF ? CD
and DE is the transversal.
Then,
?DEF = ?FED ? ?FED = 65° ? x° = ?BEF+ ?FED = (35 +65)° = 100°or, x = 100
ii
Draw EO ? AB ? CD
.
Then, ?EOB + ?EOD = x°
Now, EO ? AB
and BO is the transversal.
? ?EOB + ?ABO = 180° [Consecutive Interior Angles] ? ?EOB +55° = 180° ? ?EOB = 125°
Again, EO ? CD
and DO is the transversal.
? ?EOD + ?CDO = 180° [Consecutive Interior Angles] ? ?EOD +25° = 180° ? ?EOD = 155°
Therefore,
x° = ?EOB + ?EOD = (125 +155)° = 280°or, x = 280
iii
Draw EF ? AB ? CD
.
Then, ?AEF+ ?CEF = x°
Now, EF ? AB
and AE is the transversal.
? ?AEF+ ?BAE = 180° [Consecutive Interior Angles] ? ?AEF+116 = 180 ? ?AEF = 64°
Again, EF ? CD
and CE is the transversal.
?CEF+ ?ECD = 180° [Consecutive Interior Angles] ? ?CEF+124 = 180 ? ?CEF = 56°
Therefore,
x° = ?AEF+ ?CEF = (64 +56)° = 120°or, x = 120
Question:38
In the given figures, AB || CD. Find the value of x.
Solution:
Draw EF ? AB ? CD
.
EF ? CD
and CE is the transversal.
Then,
?ECD + ?CEF = 180° [Angles on the same side of a transversal line are supplementary] ? 130°+ ?CEF = 180° ? ?CEF = 50°
Again, EF ? AB
and AE is the transversal.
Then,
?BAE + ?AEF = 180° [Angles on the same side of a transversal line are supplementary] ? x°+20°+50° = 180° [ ?AEF = ?AEC + ?CEF] ? x°+70° = 180° ? x° = 110° ? x = 110
Question:39
In the given figure, AB || PQ. Find the values of x and y.
Page 5
Question:29
In the given figure, l || m and a transversal t cuts them. If ?1 = 120°, find the measure of each of the remaining marked angles.
Solution:
We have, ?1 = 120°
. Then,
?1 = ?5 [Corresponding angles] ? ?5 = 120° ?1 = ?3 [Vertically-opposite angles] ? ?3 = 120°
?5 = ?7 [Vertically-opposite angles] ? ?7 = 120° ?1 + ?2 = 180° [Since AFB is a straight line] ? 120°+ ?2 = 180°
? ?2 = 60° ?2 = ?4 [Vertically-opposite angles] ? ?4 = 60° ?2 = ?6 [Corresponding angles]
? ?6 = 60° ?6 = ?8 [Vertically-opposite angles] ? ?8 = 60° ? ?1 = 120°, ?2 = 60°, ?3 = 120°, ?4 = 60°, ?5 = 120°, ?6 = 60°, ?7 = 120° and ?8 = 60°
Question:30
In the given figure, l || m and a transversal t cuts them. If ?7 = 80°, find the measure of each of the remaining marked angles.
Solution:
In the given figure, ?7 and ?8 form a linear pair.
? ?7 + ?8 = 180º
? 80º + ?8 = 180º
? ?8 = 180º - 80º = 100º
Now,
?6 = ?8 = 100º Verticallyoppositeangles
?5 = ?7 = 80º Verticallyoppositeangles
It is given that, l || m and t is a transversal.
? ?1 = ?5 = 80º Pairofcorrespondingangles
?2 = ?6 = 100º Pairofcorrespondingangles
?3 = ?7 = 80º Pairofcorrespondingangles
?4 = ?8 = 100º Pairofcorrespondingangles
Question:31
In the given figure, l || m and a transversal t cuts them. If ?1 : ?2 = 2 : 3, find the measure of each of the marked angles.
Solution:
Let ?1 = 2k and ?2 = 3k, where k is some constant.
Now, ?1 and ?2 form a linear pair.
? ?1 + ?2 = 180º
? 2k + 3k = 180º
? 5k = 180º
? k = 36º
? ?1 = 2k = 2 × 36º = 72º
?2 = 3k = 3 × 36º = 108º
Now,
?3 = ?1 = 72º Verticallyoppositeangles
?4 = ?2 = 108º Verticallyoppositeangles
It is given that, l || m and t is a transversal.
? ?5 = ?1 = 72º Pairofcorrespondingangles
?6 = ?2 = 108º Pairofcorrespondingangles
?7 = ?1 = 72º Pairofalternateexteriorangles
?8 = ?2 = 108º Pairofalternateexteriorangles
Question:32
For what value of x will the line l and m be parallel to each other?
Solution:
For the lines l and m to be parallel
? 3x -20 = 2x +10 [Corresponding Angles] ? x = 30
Question:33
For what value of x will the lines l and m be parallel to each other?
Solution:
? 3x +5 +4x = 180 [Consecutive Interior Angles] ? 7x = 175 ? x = 25
Question:34
In the given figure, AB || CD and BC || ED. Find the value of x.
Solution:
BC ? ED
and CD is the transversal.
Then,
?BCD + ?CDE = 180° [Angles on the same side of a transversal line are supplementary] ? ?BCD +75 = 180 ? ?BCD = 105°
AB ? CD
and BC is the transversal.
?ABC = ?BCD (alternate angles) ? x° = 105° ? x = 105
Question:35
In the given figure, AB || CD || EF. Find the value of x.
Solution:
EF ? CD
and CE is the transversal.
Then,
?ECD + ?CEF = 180° [Consecutive Interior Angles] ? ?ECD +130° = 180° ? ?ECD = 50°
Again, AB ? CD
and BC is the transversal.
Then,
?ABC = ?BCD [Alternate Interior Angles] ? 70° = x +50° [ ? ?BCD = ?BCE + ?ECD] ? x = 20°
Question:36
In the given figure, AB || CD. Find the values of x, y and z.
Solution:
AB ? CD
and let EF and EG be the transversals.
Now, AB ? CD
and EF is the transversal.
Then,
?AEF = ?EFG [Alternate Angles] ? y° = 75° ? y = 75
Also,
?EFC + ?EFD = 180° [Since CFD is a straight line] ? x +y = 180 ? x +75 = 180 ? x = 105
And,
?EGF+ ?EGD = 180° [Since CFGD is a straight line] ? ?EGF+125 = 180 ? ?EGF = 55°
We know that the sum of angles of a triangle is 180°
?EFG+ ?GEF+ ?EGF = 180° ? y +z +55 = 180 ? 75 +z +55 = 180 ? z = 50 ? x = 105, y = 75 and z = 50
Question:37
In each of the figures given below, AB || CD. Find the value of x in each case.
Solution:
i
Draw EF ? AB ? CD
.
Now, AB ? EF
and BE is the transversal.
Then,
?ABE = ?BEF [Alternate Interior Angles] ? ?BEF = 35°
Again, EF ? CD
and DE is the transversal.
Then,
?DEF = ?FED ? ?FED = 65° ? x° = ?BEF+ ?FED = (35 +65)° = 100°or, x = 100
ii
Draw EO ? AB ? CD
.
Then, ?EOB + ?EOD = x°
Now, EO ? AB
and BO is the transversal.
? ?EOB + ?ABO = 180° [Consecutive Interior Angles] ? ?EOB +55° = 180° ? ?EOB = 125°
Again, EO ? CD
and DO is the transversal.
? ?EOD + ?CDO = 180° [Consecutive Interior Angles] ? ?EOD +25° = 180° ? ?EOD = 155°
Therefore,
x° = ?EOB + ?EOD = (125 +155)° = 280°or, x = 280
iii
Draw EF ? AB ? CD
.
Then, ?AEF+ ?CEF = x°
Now, EF ? AB
and AE is the transversal.
? ?AEF+ ?BAE = 180° [Consecutive Interior Angles] ? ?AEF+116 = 180 ? ?AEF = 64°
Again, EF ? CD
and CE is the transversal.
?CEF+ ?ECD = 180° [Consecutive Interior Angles] ? ?CEF+124 = 180 ? ?CEF = 56°
Therefore,
x° = ?AEF+ ?CEF = (64 +56)° = 120°or, x = 120
Question:38
In the given figures, AB || CD. Find the value of x.
Solution:
Draw EF ? AB ? CD
.
EF ? CD
and CE is the transversal.
Then,
?ECD + ?CEF = 180° [Angles on the same side of a transversal line are supplementary] ? 130°+ ?CEF = 180° ? ?CEF = 50°
Again, EF ? AB
and AE is the transversal.
Then,
?BAE + ?AEF = 180° [Angles on the same side of a transversal line are supplementary] ? x°+20°+50° = 180° [ ?AEF = ?AEC + ?CEF] ? x°+70° = 180° ? x° = 110° ? x = 110
Question:39
In the given figure, AB || PQ. Find the values of x and y.
Solution:
Given, AB ? PQ
.
Let CD be the transversal cutting AB and PQ at E and F, respectively.
Then,
?CEB + ?BEG+ ?GEF = 180° [Since CD is a straight line] ? 75°+20°+ ?GEF = 180° ? ?GEF = 85°
We know that the sum of angles of a triangle is 180°
.
? ?GEF+ ?EGF+ ?EFG = 180 ? 85°+x +25° = 180° ? 110°+x = 180° ? x = 70°
And
?FEG+ ?BEG = ?DFQ [Corresponding Angles] ? 85°+20° = ?DFQ ? ?DFQ = 105° ?EFG+ ?GFQ+ ?DFQ = 180° [Since CD is a straight line] ? 25°+y +105° = 180° ? y = 50° ? x =
Question:40
In the given figure, AB || CD. Find the value of x.
Solution:
AB ? CD
and AC is the transversal.
Then,
?BAC + ?ACD = 180° [Consecutive Interior Angles] ? 75 + ?ACD = 180 ? ?ACD = 105°
And,
?ACD = ?ECF [Vertically-Opposite Angles] ? ?ECF = 105°
We know that the sum of the angles of a triangle is 180°
.
?ECF+ ?CFE + ?CEF = 180° ? 105°+30°+x = 180° ? 135°+x = 180° ? x = 45°
Question:41
In the given figure, AB || CD. Find the value of x.
Solution:
AB ? CD
and PQ is the transversal.
Then,
?PEF = ?EGH [Corresponding Angles] ? ?EGH = 85°
And,
?EGH + ?QGH = 180° [Since PQ is a straight line] ? 85°+ ?QGH = 180° ? ?QGH = 95°
Also,
?CHQ+ ?GHQ = 180° [Since CD is a straight line] ? 115°+ ?GHQ = 180° ? ?GHQ = 65°
We know that the sum of angles of a triangle is 180°
.
? ?QGH + ?GHQ+ ?GQH = 180° ? 95°+65°+x = 180° ? x = 20° ? x = 20°
Question:42
In the given figure, AB || CD. Find the values of x, y and z.
Solution:
?ADC = ?DAB [Alternate Interior Angles] ? z = 75° ?ABC = ?BCD [Alternate Interior Angles] ? x = 35°
We know that the sum of the angles of a triangle is 180°
.
? 35°+y +75° = 180° ? y = 70° ? x = 35°, y = 70° and z = 75°.
Question:43
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Additional Information about Lines and Angles - Exercise 7C for Class 9 Preparation
Lines and Angles - Exercise 7C Free PDF Download
The Lines and Angles - Exercise 7C is an invaluable resource that delves deep into the core of the Class 9 exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the Lines and Angles - Exercise 7C now and kickstart your journey towards success in the Class 9 exam.
Importance of Lines and Angles - Exercise 7C
The importance of Lines and Angles - Exercise 7C cannot be overstated, especially for Class 9 aspirants.
This document holds the key to success in the Class 9 exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
Lines and Angles - Exercise 7C Notes
Lines and Angles - Exercise 7C Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to Lines and Angles - Exercise 7C.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, Lines and Angles - Exercise 7C Notes on EduRev are your ultimate resource for success.
Lines and Angles - Exercise 7C Class 9 Questions
The "Lines and Angles - Exercise 7C Class 9 Questions" guide is a valuable resource for all aspiring students preparing for the
Class 9 exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study Lines and Angles - Exercise 7C on the App
Students of Class 9 can study Lines and Angles - Exercise 7C alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the Lines and Angles - Exercise 7C,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of Lines and Angles - Exercise 7C is prepared as per the latest Class 9 syllabus.
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