GSEB Solutions for Class 9 Mathematics – Some Primary Concepts in Geometry : 2

GSEB Solutions for Class 9 Mathematics –  Some Primary Concepts in Geometry : 2 (English Medium)

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Exercise 8:

Solution 1:

Solution 2:

Solution 3:

Solution 4:

Solution 5:

Solution 6:

Given, m ∠XOY : m ∠YOZ = 2 : 3
Let m ∠XOY = 2k, then m ∠YOZ = 3k.
∠XOY and ∠YOZ form a linear pair.
∴ m ∠XOY + m ∠YOZ = 180
∴ 2k + 3k = 180
∴ 5k = 180

∴ k =180/5

∴ k = 36
m ∠XOY = 2k = 2 × 36 = 72
m ∠YOZ = 3k= 3 × 36 = 108
Thus, the measure of ∠XOY is 72 and ∠YOZ that of 108.

Solution 7(1):

c. rays
An angle is a union of rays.

Solution 7(2):

d. 0 and 180
The measure of an angle always lies between 0 and 180.

Solution 7(3):

d. 9
Measure of the complementary angle of ∠A
= 90 – m ∠A
= 90 – 81
= 9

Solution 7(4):

Solution 7(5):

Solution 7(6):

b. skew
If two lines cannot lie in the same plane, they are called skew lines.

Solution 7(7):

c. 113
The measure of the complementary angle of complementary angle of an angle having measure 23 = 90 – 23 = 67
The measure of the supplementary angle of complementary angle of an angle having measure 23 = 180 – 67 = 113

Solution 7(8):

a. -(x – 60)
The measure of the complementary angle of an angle having measure x + 30
= 90 – (x + 30)
= 60 – x – 30
= -(x – 60)

Solution 7(9):

c. obtuse
If one angle of a linear pair is acute, then other angle is obtuse.

Solution 7(10):

a. supplementary
If t is a transversal for two parallel lines l and m, interior angles on the same side of the transversal are supplementary.

Solution 7(11):

b. 15
We know that the sum of measures of angles forming a linear pair is 180.
∴ (6y + 30) + 4y = 180
∴ 10y + 30 = 180
∴ 10y = 180 – 30∴ 10y = 150
∴ y = 15

Solution 7(12):

Exercise 8.1:

Solution 1:

Any one of the following three conditions uniquely determines a plane:

1. Three non-collinear points.
2. A line and a point not lying on it.
3. Two distinct intersecting lines.

Solution 2:

Solution 3:

Solution 4:

• Coplanar and non-coplanar points
  Coplanar points: If there exists a plane containing all the given points, the points are said to be coplanar.
  Non-coplanar points: If there does not exist a plane containing all the given points, we say they are non-coplanar.
• Coplanar and non-coplanar lines
  Coplanar lines: If there exists a plane containing all the given lines, we say the lines are coplanar.
  Non-coplanar points: If there does not exist a plane containing all the given lines, such lines are called non-coplanar lines.

Solution 5:

Solution 6:

Exercise 8.2:

Solution 1:

Solution 2:

1. Coplanar lines: If there exists a plane containing all the given lines, we say the lines are coplanar.
2. Skew lines: The lines which are not coplanar are called skew lines.
3. Coplanar points: If there exists a plane containing all the given points, the points are said to be coplanar.
4. A closed half plane: The union of any half plane formed by line l and the line l itself is a closed half plane.

Solution 3:

Solution 4:

Solution 5:

Exercise 8.3:

Solution 1:

Solution 2:

Pairs of adjacent angles are given below:

1. ∠DOA and ∠DOC
2. ∠DOA and ∠DOB
3. ∠COB and ∠COD
4. ∠COB and ∠COA

Linear pairs of angles are given below:

1. ∠DOA  and ∠DOB
2. ∠COB and ∠COA

Solution 3:

Solution 4:

Solution 5:

Solution 6(1):

Solution 6(2):

Solution 6(3):

The measure of the supplementary angle of the angle having measure 120° = 180° – 120° = 60°.
The measure of the complementary angle of the supplementary angle of the angle having measure 120° = 90° – 60° = 30°.

Solution 6(4):

1. The measure of the complementary angle of the angle with measure 42° = 90° – 42° = 48°.
2. The measure of the complementary angle of the angle with measure 37° = 90° – 37° = 53°.
3. The measure of the complementary angle of the angle with measure (10 + x)° = 90° – (10 + x)° = 90 – 10 – x = (80 – x)°.
4. The measure of the complementary angle of the angle with measure 81° = 90° – 81° = 9°.

Solution 6(5):

1. The measure of the supplementary angle of the angle with measure 100° = 180° – 100° = 80°.
2. The measure of the supplementary angle of the angle with measure 89° = 180° – 89° = 91°.
3. The measure of the supplementary angle of the angle with measure (y – 30)° = 180° – (y – 30)° = 180 – y + 30 = (210 – y)°.
4. The measure of the supplementary angle of the angle with measure 49° = 180° – 49° = 131°.

Exercise 8.4:

Solution 1:

Solution 2:

From the figure, m  ⃦ n. Line t is the transversal of line m and line l.
∴ m∠EFB = m∠DGF (corresponding angles)
But, m∠EFB = 65° …[Given]
∴ m∠DGF = 65° … [∵ m∠EFB = m∠DGF]
Now, ∠DGF and ∠CGF form a linear pair.
∴ m∠DGF + m∠CGF = 180°
∴ 65° + m∠CGF = 180° … [∵ m∠DGF = 65°
∴ m∠CGF = 180° – 65°
∴ m∠CGF = 115°
So, m∠CGF = 115° and m∠DGF = 65°.

Solution 3:

Solution 4:

Solution 5:

From the figure, ∠CQD and ∠PQE are vertically opposite angles.
∴ m∠CQD = m∠PQE
∴ m∠PQE = 85° [∵ m∠CQD = 85°]
m∠APB = 85° …[Given]
∴ m∠APB = m∠PQE
∴ ∠APB @ ∠PQE
∠APB and ∠PQE are corresponding angles formed by transversal t to lines l and m.
Thus, l  ⃦ m.

Exercise 8.5:

Solution 1:

Alternate angles are given below:

1. ∠MQR and ∠QRB
2. ∠NQR and ∠QRA

Corresponding angles are given below:

1. ∠PQM and ∠QRA
2. ∠PQN and ∠QRB
3. ∠MQR and ∠ARS
4. ∠NQR and ∠BRS

Solution 2:

Solution 3:

Solution 4:

Solution 5:

Given: Line t₁ and t₂ are transversals to the lines l, m and n.
m∠ABE = m∠BCD and m∠FEI = m∠GFH
To Prove : l ⃦ n
Proof:
m∠ABE = m∠BCD
∴ ∠ABE ≌ ∠BCD
∠ABE and ∠BCD are corresponding angles formed by transversal t₁ to lines l and m and they are congruent.
∴ l  ⃦ m …[1]
m∠FEI = m∠GFH
∴ ∠FEI ≌ ∠GFH
∠FEI and ∠GFH are corresponding angles formed by transversal t₂ to lines m and n and they are congruent.
∴ m  ⃦ n …[2]
From [1] and [2],
l  ⃦ n.

Solution 6:

Solution 7: