Samacheer Kalvi Class 7 Maths Term 1 Chapter 5 Geometry Solutions
Welcome to NCTB Solutions. Here with this post we are going to help 7th class students by providing Solutions for Samacheer Kalvi Class 7 Maths Term 1 chapter 5 Geometry. Here students can easily find all the solutions for Geometry Exercise 5.1, 5.2, 5.3, 5.4, 5.5 and 5.6. Also here our Expert Maths Teacher’s solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 5 solutions. Here all the solutions are based on Tamil Nadu State Board latest syllabus.
Geometry Exercise 5.1 Solutions :
(1) Name the pairs of adjacent angles
Solution :
The pairs of adjacent angle are ∠ACF and ∠FCE , ∠ACF and ∠ECD, ∠BCF and ∠FCE, ∠FCE and ∠ECD, ∠ABG and ∠GBC.
(2) Find the angle ∠JIL from the given figure.
Solution :
From the fig,
∠JIL = ∠JIK + ∠KIL
= 38° + 27°
= 65°
∴ ∠JIL = 65°
(3) Find the angle ∠GEH from the given figure.
Solution :
From the fig,
∠HEF = ∠GEH + ∠GEF
120° = ∠GEH + 34°
Subtract both sides by 34°, we get
= 120° – 34°
= ∠GEH + 34° – 34°
= ∠GEH = 86°
Therefore, ∠GEH = 86°
(4) Given that AB is a straight line. Calculate the value of x° in the following cases.
Solution :
Figure – (i)
Since the angles are linear pair,
∠AOC + ∠COB = 180°
72° + x° = 180°
Subtract both sides by 72°, we get
72° + x° – 72°= 180° – 72°
x° = 108°
Figure – (ii)
Since the angles are linear pair,
∠AOC + ∠COB = 180°
3x° + 42° = 180°
Subtract both sides by 42°, we get
3x° + 42° – 42°= 180° – 42°
3x° = 138°
Dividing both sides by 3, we get
3x°/3 = 138°/3
x° = 46°
Figure – (iii)
Since the angles are linear pair,
∠AOC + ∠COB = 180°
4x° + 2x° = 180°
6x° = 180°
Dividing both sides by 6, we get
6x°/6 = 180°/6
x° = 30°
(5) One angle of a linear pair is a right angle. What can you say about the other angle?
Solution :
We know, sum of linear pair is 180°
Since one angle of a linear pair is a right angle
Let other is x
Thus, 90° + x°
= 180°
Therefore,
x° = 180° – 90°
= 90°
Hence, we can say that other angel is also right angle.
(6) If the three angles at a point are in the ratio 1 : 4 : 7, find the value of each angle?
Solution :
Let, the angles be x, 4x and 7x
Since they are angles at a point, their sum is 360°
x + 4x + 7x = 360°
12x = 360°
x = 360°/12
x = 30°
The angels are, x = 30°
4x = 4 × 30 = 120°
7x = 7 × 30 = 210°
(7) There are six angles at a point. One of them is 45° and the other five angles are all equal. What is the measure of all the five angles?
Solution :
Given, there are six angles at a point.
One of the angel = 45°
Let x be equal angel.
Since they are angles at a point, their sum is 360°
Thus, 45° + x + x + x + x + x = 360°
5x = 360° – 45°
5x = 315°
x = 315°/5
x = 63°
Hence, the measure of all five angel is 63°
(8) In the given figure, identify
(i) any two pairs of adjacent angles.
(ii) two pairs of vertically opposite angles.
Solution :
(i) Adjacent angles : ∠TQS and ∠SQR ; ∠RQU and ∠PQU
(ii) Opposite pairs : ∠PQU and ∠TQR ; ∠PQT and ∠UQR
(9) The angles at a point are x°, 2x°, 3x°, 4x° and 5x°. Find the value of the largest angle?
Solution :
Given, the angles at a point are x°, 2x°, 3x°, 4x° and 5x°.
Since, they are angles at a point, their sum is 360°
x° + 2x° + 3x° + 4x° + 5x° = 360°
15x° = 360°
x° = 360°/15
x° = 24°
Now, 5x° = 5 × 24
= 120°
Hence, the value of the largest angle is 120°
(10) From the given figure, find the missing angle.
Solution :
From fig,
Since, ∠ROQ and ∠SOP are vertical opposite angel and they are equal.
Therefore, x = 105°
(11) Find the angles x° and y° in the figure shown.
Solution :
From fig, ∠3x and ∠x are liner pair.
Their sum is 180°
Thus, 3x° + x° = 180°
4x° = 180°
x° = 180° /4
x° = 45°
Also since, 3x and y are vertical opposite angel and they are equal
Therefore,
y = 3x = 3 × 45
y = 135°
(12) Using the figure, answer the following questions.
(i) What is the measure of angle x°?
(ii) What is the measure of angle y°?
Solution :
(i) From fig,
Angles 125° and x° are vertical opposite angel and they are equal
Thus, x° = 125°
(ii) From fig, angles 125° and y°are liner pair.
Their sum is 180°
125° + y° = 180°
y° = 180° – 125°
y° = 55°
Objective Type Question Solutions :
(13) Adjacent angles have
(i) no common interior, no common arm, no common vertex
(ii) one common vertex, one common arm, common interior
(iii) one common arm, one common vertex, no common interior
(iv) one common arm, no common vertex, no common interior
Solution :
Correct Option → (iii)
Adjacent angles have one common arm, one common vertex, no common interior.
(14) In the given figure the angles ∠1 and ∠2 are
(i) opposite angles
(ii) Adjacent angles
(iii) Linear pair
(iv) Supplementary angles
Solution :
Correct Option → (iii)
In the given figure the angles ∠1 and ∠2 are Linear pair.
(15) Vertically opposite angles are
(i) not equal in measure
(ii) complementary
(iii) supplementary
(iv) equal in measure
Solution :
Correct Option → (iv)
Vertically opposite angles are equal in measure.
(16) The sum of all angles at a point is
(i) 360°
(ii) 180°
(iii) 90°
(iv) 0°
Solution :
Correct Option → (i)
The sum of all angles at a point is 360°
(17) The measure of ∠BOC is
(i) 90°
(ii) 180°
(iii) 80°
(iv) 100°
Solution :
Correct Option → (iii) 80° is the correct answer to this question.
Geometry Exercise 5.2 Solutions :
(1) From the figures name the marked pair of angles.
Solution :
Figure – (i)
The angles are exterior angles on the same side of the transversal.
Figure – (ii)
The angles are alternate exterior angles.
Figure – (iii)
The angles are corresponding angles.
Figure – (iv)
The angles are interior angles on the same side of the transversal.
Figure – (v)
The angles are alternate interior angles.
Figure – (vi)
The angles are corresponding angles.
(2) Find the measure of angle x in each of the following figures.
Solution :
Figure – (i)
Given n is parallel to m and l is transversal to n and m.
We get, x = 35° [since, corresponding angles are equal]
Figure – (ii)
Given n is parallel to m and l is transversal to n and m.
We get, x = 65° [since, corresponding angles are equal]
Figure – (iii)
Given n is parallel to m and l is transversal to n and m.
We get, x = 145° [since, corresponding angles are equal]
Figure – (iv)
Given n is parallel to m and l is transversal to n and m.
We get, x = 135° [since, corresponding angles are equal]
Figure – (v)
Given n is parallel to m and l is perpendicular to n and m.
We get, x = 90° [since, line l is perpendicular to line m]
(3) Find the measure of angle y in each of the following figures.
Solution :
Figure – (i)
Given n is parallel to m and l is transversal to n and m.
We get, y = 28° [since, alternate interior angles are equal]
Figure – (ii)
Given n is parallel to m and l is transversal to n and m.
We get, y = 58° [since, alternate exterior angles are equal ]
Figure – (iii)
Given n is parallel to m and l is transversal to n and m.
We get, y = 123° [since, alternate interior angles are equal]
Figure – (iv)
Given n is parallel to m and l is transversal to n and m.
We get, y = 108° [since, alternate exterior angles are equal ]
(4) Find the measure of angle z in each of the following figures.
Solution :
Figure – (i)
Given n is parallel to m and l is transversal to n and m.
Now, z° + 31° = 180° …[sum of interior angles that lie on the same side of the transversal]
z° = 180° – 31°
z° = 149°
Figure – (ii)
Given n is parallel to m and l is transversal to n and m.
Now, z° + 135° = 180° …[sum of interior angles that lie on the same side of the transversal]
z° = 180° – 135°
z° = 45°
(5) Find the value of angle a in each of the following figures.
Solution :
Figure – (i)
Given n is parallel to m and l is transversal to n and m.
We get, 3a = 126° [since, corresponding angles are equal]
a = 126/3
a = 42
Figure – (iii)
Given n is parallel to m and l is transversal to n and m.
We get, 8a + 29 = 45° [since, alternate interior angles are equal]
8a = 45 – 29
8a = 16
a = 16/8
a = 2
Figure – (iv)
Given, line l is perpendicular to line n.
Thus, 6a = 90
a = 90/6
a = 15°
(6) Find the value of angle x in both the figures.
Solution :
Figure – (i)
Given n is parallel to m and l is transversal to n and m.
We get, 2x + 15 = 3x – 40 [since, alternate interior angles are equal]
3x -2x = 40 + 15
x = 55°
(7) Anbu has marked the angles as shown below in (i) and (ii). Check whether both of them are correct. Give reasons.
Solution :
Figure – (i)
= No.
Reason : Since in a fig the sum of interior angles on the same side of the transversal are not equal to 180°
Figure – (ii)
= No.
Reason : Since in a fig corresponding angles are not equal.
Objective Type Questions Solutions :
(10) A line which intersects two or more lines in different points is known as
(i) parallel lines
(ii) transversal
(iii) non-parallel lines
(iv) intersecting line
Solution :
The correct option is – (ii)
A line which intersects two or more lines in different points is known as Transversal.
(11) In the given figure, angles a and b are
(i) alternate exterior angles
(ii) corresponding angles
(iii) alternate interior angles
(iv) vertically opposite angles
Solution :
The appropriate option is – (iii)
In the given figure, angles a and b are alternate exterior angles.
(12) Which of the following statements is ALWAYS TRUE when parallel lines are cut by a transversal
(i) corresponding angles supplementary.
(ii) alternate interior angles supplementary.
(iii) alternate exterior angles supplementary.
(iv) interior angles on the same side of the transversal are supplementary.
Solution :
Alternative (iv) interior angles on the same side of the transversal are supplementary is the correct answer of this question.
(13) In the diagram, what is the value of angle x?
(i) 43°
(ii) 44°
(iii) 132°
(iv) 134°
Solution :
Alternative (ii) 44° is the correct answer to this question.
Geometry Exercise 5.6 Solutions :
(1) Find the value of x if ∠AOB is a right angle.
Solution :
From fig,
Now, 2x + 3x = ∠AOB
5x = 90° …( Since, ∠AOB is a right angle)
x = 90/5
x = 18
∴ The value of x is 18.
(2) In the given figure, find the value of x.
Solution :
From fig,
Since, angel 2x + 23 and 3x – 48 is liner angel pair.
Therefore, 2x + 23 + 3x – 48 = 180°
5x – 25° = 180°
5x = 180° – 25°
5x = 155°
x = 155/5
x = 31°
Therefore, the value of x will be 31°
(3) Find the value of x, y and z
Solution :
From fig, ∠DOB and ∠ BOC are liner pair.
Thus, ∠DOB + ∠ BOC = 180°
x + 3x+40 = 180
4x = 180 – 40
4x = 140
x = 140/4
x = 35°
Since, ∠DOB and ∠AOC are vertical opposite angle
Thus, ∠DOB = ∠AOC
35 = z + 10
z = 35 – 10
z = 25°
Now, ∠ BOC and ∠ DOA are vertical opposite angle
Therefore, ∠ BOC = ∠ DOA
3x + 40 = y+30
3*35 + 40 = y + 30
105 + 40 = y + 30
145 = y + 30
y = 145 – 30
y = 115°
(4) Two angles are in the ratio 11: 25. If they are linear pair, find the angles.
Solution :
Suppose, this angles are 11x and 25 x and it’s liner pair
Thus , 11x + 25x = 180
36x = 180
x = 180/36
x = 5
Thus, 11x = 11 * 5 = 55°
25x = 25*5 = 125°
Therefore, this angles are 55° and 125°.
(5) Using the figure, answer the following questions and justify your answer.
(i) Is ∠1 adjacent to ∠2?
(ii) Is ∠AOB adjacent to ∠BOE?
(iii) Does ∠BOC and ∠BOD form a linear pair?
(iv) Are the angles ∠COD and ∠BOD supplementary?
(v) Is ∠3 vertically opposite to ∠1?
Solution :
(i) Yes. They have a common vertex, a common arm and their interiors do not overlap.
(ii) No. They have overlapping interiors.
(iii) No. Since ∠BOCis a straight angle, their sum will exceed 180°.
(iv) Yes. They are linear pair.
(v) No. They are not formed by intersecting lines.
(6) In the figure POQ, ROS and TOU are straight lines. Find the x°.
Solution :
Given, ∠TOU are straight angle.
Therefore, 36 + 47 + 45 + x = 180°
128 + x = 180°
x = 180° – 128°
x = 52°
(7) In the figure AB is parallel to DC. Find the value of ∠1 and ∠2. Justify your answer.
Solution :
From fig, AB is parallel to DC
Since, ∠2 and 80 alternate internal angel and it’s equal.
∠2 = 80
Now, also since, ∠1 and 30 alternate internal angel and it’s equal.
∠1 = 30
(8) In the figure AB is parallel to CD. Find x, y and z.
Solution :
From fig, ∠C, 42 and 63 is angles of triangle
We know, sum of angles of triangle is 180
Thus, ∠C + 42 + 63 = 180
∠C + 105 = 180
∠C = 180 – 105
∠C = 75
Since, AB is parallel to CD
Angles x and C are alternate internal angel and it’s equal.
x = ∠C = 75
Also, angles z and 48 are alternate internal angel and it’s equal.
z = 48
Now, angles y and z are corresponding angel and it’s equal.
y = z = 48
(10) A plumber must install pipe 2 parallel to pipe 1. He knows that ∠1 is 53. What is the measure of ∠2?
Solution :
Given, a plumber must install pipe 2 parallel to pipe 1.
Thus, ∠1 and ∠2 are interior angles that lie on the same side of the transversal
Therefore, ∠1 + ∠2 = 180
53 + ∠2 = 180 ….. (Given, ∠1 = )
∠2 = 180 – 53
∠2 = 127
Challenge Problem Solutions :
(11) Find the value of y.
Solution :
From fig,
Since, ∠POQ is straight angle
Thus, 60 + 3y – 20 + y + y + 10 = 180
50 + 5y = 180
5y = 180 – 50
5y = 130
y = 130/5
y = 26
Therefore, the value of y will be 26
(12) Find the value of z.
Solution :
From fig,
Since, 4z – 25 = 3z + 2z-5 + z+10
4z – 25 = 6z +5
6z – 4z = -25 – 5
2z = -30
z = -30/2
z = -15
Hence, the value of z will be -15
(15) In the figure, the lines GH and IJ are parallel. If ∠1=108° and ∠2 = 123°, find the value of x, y and z.
Solution :
From fig,
Since, ∠1 and ∠IGH are liner pair
Thus, ∠1 + ∠IGH = 180
108 + ∠IGH = 180 …. ( given, ∠1=108)
∠IGH = 180 – 108
∠IGH = 72
the lines GH and IJ are parallel and line KG is transversal.
Therefore, ∠IGH and ∠x are corresponding angels
So, ∠x = 72
Similarly, we can find, ∠y = 57
Now, ΔIKJ is triangle and sum of angels is 180
Thus, x+y+z = 180
72 + 57 + z = 180
129 + z = 180
z = 180 – 129
z = 51
(18) In the figure AB is parallel to CD. Find x ̊, y ̊ and z ̊.
Solution :
Given, AB is parallel to CD .
As AD is transversal to line AB || line CD
Angle x and 48 is alternate internal angle and there are equal.
Thus, x = 48
Similarly, as CB is transversal to line AB || line CD
Angel y and 60 is alternate internal angle and there are equal.
Thus, y = 60
Now, since form fig ,
z = 60 + x
z = 60 + 48
z = 108
(20) In the given figure, ∠8 = 107 ̊, what is the sum of the ∠2 and ∠4?
Solution :
From fig, ∠8 and ∠4 are corresponding angels and there are equal.
Also, ∠2 and ∠4 are opposite angels and there are equal.
Therefore, ∠4 = ∠8 = 107
And ∠2 = ∠4 = 107
Now, ∠2 + ∠4 = 107 + 107
∠2 + ∠4 = 214
Next Chapter Solution :
