Class 9 Exam  >  Class 9 Notes  >  Mathematics (Maths) Class 9  >  HOTS Questions: Lines & Angles

Class 9 Maths Chapter 6 HOTS Questions - Lines and Angles

Q1. If a ray CD stands on a line AB, then prove that
 ∠ACD + ∠BCD = 180º.

 Solution: Let us draw CE ⊥ AB.
∴ ∠ACE = 90º and
∠BCE = 90º 

Class 9 Maths Chapter 6 HOTS Questions - Lines and Angles

Now, ∠ACD = ∠ACE + ∠ECD = 90º + ∠ECD …(1)
∠BCD = ∠BCE - ∠ECD = 90º - ∠ECD …(2)
Adding (1) and (2), we have: ∠ACD + ∠BCD
= [90º + ∠ECD] + [90º - (∠ECD)]
= 90º + 90º + ∠ECD - ∠ECD
= 180º

Note: Above example may be stated as: “The sum of the angles of linear pair is 180°.”
OR
“Sum of all the angles formed on the same side of a line at a given point on the line is 180°.”

 Q2. Two lines AB and CD intersect each other at a point O such that ∠AOC : ∠AOD = 5 : 7. Find all the angles.

Class 9 Maths Chapter 6 HOTS Questions - Lines and Angles

Solution: Let ∠AOC = 5k and ∠AOD = 7k, where k is some constant.
Here, ∠AOC and ∠AOD form a linear pair.
∴ ∠AOC + ∠AOD = 180º
⇒ 5k + 7k = 180º
⇒ 12k = 180º
⇒ k = 15º
∴ ∠AOC = 5k = 5 × 15º = 75º
∠AOD = 7k = 7 × 15º = 105º
Now, ∠BOD = ∠AOC = 75º (Vertically opposite angles)
∠BOC = ∠AOD = 105º (Vertically opposite angles)

Q3. In the following figure, AOB is a straight line. Find ∠AOC and ∠BOD.

Class 9 Maths Chapter 6 HOTS Questions - Lines and Angles

Solution: Since AOB is a straight line.
∴ The sum of all the angles on the same side of AOB at a point on it is 180º.
∠AOC + ∠COD + ∠DOB = 180º
∴ x + 60º + (2x - 15)º = 180º
or 3x +  60º - 15º = 180º
or 3x = 180º - 60º + 15º = 135º
or x =  (135°/3) = 45º
Now 2x - 15 = 2(45) - 15 = 75º
or ∠AOC = 45º and ∠BOD = 75º 

Q4. In the following figure, p: q: r = 2: 3: 4. If AOB is a straight line, then find the values of p, q and r.

Class 9 Maths Chapter 6 HOTS Questions - Lines and Angles

Solution: Let p = 2x, q = 3x and r = 4x                         [∵ p : q : r = 2 : 3 : 4]
∵ AOB is a straight line.
∴ ∠AOC + ∠COD + ∠DOB = 180º
or 2x + 3x + 4x = 180º
or 9x = 180º
or x = (180°/9) = 20º
So, 2x = 2 x 20º = 40º
∴ 3x = 3 × 20° = 60°
4x = 4 × 20° = 80°
Thus, p = 40°, q = 60°, r = 80°

Q5. In the figure, AB || CD. EG and FH are bisectors of ∠PEB and ∠EFD respectively.
 Show that EG || FH.

Class 9 Maths Chapter 6 HOTS Questions - Lines and Angles

OR

If two parallel lines are intersected by a transversal then prove that the bisectors of any pair of corresponding angles are parallel.

Solution: ∵ AB || CD and EF is a transversal.
∴ ∠BEP = ∠EFD                   [corresponding angles]
⇒ (1/2) ∠BEP = (1/2)∠EFD
⇒ ∠PEG = ∠EFH                   [∵ EG and FH are the angle bisectors of ∠BEP and ∠EFD respectively]
But they form pair of corresponding angles.
∴ EG || FH.

Q6. Prove that the sum of the angles of a triangle is 180º.

Solution: Let us consider ΔABC and through A draw DE || BC.

Class 9 Maths Chapter 6 HOTS Questions - Lines and Angles

∵ BC || DE and AB is a transversal.
∴ ∠4= ∠1                  [Interior alternate angles]                   …(1)

Again, BC || DF and AC is a transversal.
∴ ∠5= ∠2                  [Interior alternate angles]                  …(2)
Adding (1) and (2), we have ∠4 + ∠5= ∠1 + ∠2
Adding ∠3 on both sides, we have ∠4 + ∠5 + ∠3 = ∠1 + ∠2 + ∠3
Since, ΔAF is a straight line,
∴ ∠4 + ∠3 + ∠5 = 180º  ,∠1 + ∠2 + ∠3 = 180°
i.e. ∠ABC + ∠BCA + ∠BAC = 180º
∴ The sum of angles of a triangle is 180º.

Q7. Prove the following statement: “If a side of a triangle is produced, then the exterior angle so formed is equal to the sum of the two interior opposite angles.” 

Solution: Let us consider a triangle ABC such that its side BC is produced to D, forming exterior ∠ACD.

Class 9 Maths Chapter 6 HOTS Questions - Lines and Angles

∵ Sum of the angles of a triangle = 180º
∴ ∠1 + ∠2 + ∠3 = 180º                  …(1)
Since BCD is a straight line.
∴ ∠2 + ∠4 = 180º                  …(2)
From (1) and (2), we have ∠2 + ∠4= ∠1 + ∠2 + ∠3
⇒ ∠4= ∠1 + ∠3 i.e.
Exterior ∠ACD = [Sum of the interior opposite angles]

Remember:
An exterior angle of a triangle is greater than either of the interior opposite angles.

Q8. In a triangle, the bisectors of ∠B and ∠C intersect each other at a point O. Prove that ∠BOC = 90º + 1/2∠A.

Solution: In a ΔABC, we have: ∠A + ∠B + ∠C = 180º                  [By angle sum property]

Class 9 Maths Chapter 6 HOTS Questions - Lines and Angles

Class 9 Maths Chapter 6 HOTS Questions - Lines and Angles

Again, in ΔOBC, we have ∠1 + ∠2 + ∠BOC = 180º                   [By angle sum property]
⇒ (∠1 + ∠2) + ∠BOC = 180º

Class 9 Maths Chapter 6 HOTS Questions - Lines and Angles

Q9. In a ΔABC, if ∠A + ∠B = 150º and ∠B + ∠C = 100º. Find the measure of each angle of the triangle.

 Solution: We have ∠ A + ∠ B = 150º                   …(1)
∠B + ∠C = 100º                   …(2)
Adding (1) and (2), we get
∴ ∠A + 2∠B + ∠C = 150 + 100º = 250º
⇒ (∠A + ∠B + ∠C) + ∠B = 250º
⇒ 180º + ∠B = 250º                   [∵ ∠ A + ∠B + ∠ C = 180º]
∴ ∠B = 250º - 180º = 70º.
Now, ∠A + ∠B = 150º
⇒ ∠A = 150º - ∠B = 150º - 70º = 80º
Also ∠B + ∠C = 100º
⇒ ∠C = 100 - ∠B = 100 - 70º = 30º
Thus, ∠A = 80º, ∠B = 70º and ∠C = 30º.

Q10. In a triangle, if ∠A = 2∠B = 6∠C, find the measures of ∠A, ∠B and ∠C.

Solution: Let ∠A = 2∠B = 6∠C = x
∴ ∠A= x 2∠B= x
⇒ ∠B = x 2
6∠C = x
⇒ ∠C = x 6
We know that ∠A + ∠B + ∠C = 180º (using angle sum property)

Class 9 Maths Chapter 6 HOTS Questions - Lines and Angles

or  6xº + 3xº + xº = 6 x 180º 

⇒ 10xº = 6 x 180º
Class 9 Maths Chapter 6 HOTS Questions - Lines and Angles

Q11. If the arms of an angle are respectively parallel to the arms of another angle, then show that the two angles are either equal or supplementary.

Solution: We have two angles ∠ABC and ∠DEF such that BA || ED and BC || EF in the same sense or in the opposite sense.
∴ We can have the following three cases:
Case I:                   [Both pairs of arms are parallel in the same sense.]

Class 9 Maths Chapter 6 HOTS Questions - Lines and Angles

∵ BA || ED and BC is a transversal,
∴ ∠1= ∠2                   [Corresponding angles]                   …(1)
Again, BC || EF and DE is a transversal,
∴ ∠3= ∠2                   [Corresponding angles]                   …(2)
From (1) and (2), we have ∠1= ∠3
i.e. ∠ABC = ∠DEF.

Case II:                   [Both pairs of arms are parallel in the opposite sense.]

Class 9 Maths Chapter 6 HOTS Questions - Lines and Angles

∵ BA || ED and BC is a transversal,
∴ ∠1= ∠2                   [Corresponding angles]                   …(1)
Again BC || EF and ED is a transversal,                   [Alternate interior angles]
∴ ∠3= ∠2 …(2)
From (1) and (2), we have ∠1= ∠3
i.e. ∠ABC = ∠DEF
Case III:                           [One pair of arms are parallel in the same sense and another pair parallel in an opposite sense.]

Class 9 Maths Chapter 6 HOTS Questions - Lines and Angles

∵ BA || ED and BC is a transversal.
∴ ∠1= ∠2                   [Interior alternate angles]                   …(1)
Again BC || EF and DE is a transversal.
∴ ∠3 + ∠2 = 180º                   [Sum of interior opposite angles]
⇒ ∠3 + ∠1 = 180º                  [From (1)]
i.e. ∠DEF + ∠ABC = 180º
Hence, ∠ABC and ∠DEF are supplementary.

The document Class 9 Maths Chapter 6 HOTS Questions - Lines and Angles is a part of the Class 9 Course Mathematics (Maths) Class 9.
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FAQs on Class 9 Maths Chapter 6 HOTS Questions - Lines and Angles

1. What are the different types of angles in geometry?
Ans. In geometry, angles are classified into several types based on their measures: 1. Acute Angle: An angle less than 90 degrees. 2. Right Angle: An angle exactly equal to 90 degrees. 3. Obtuse Angle: An angle greater than 90 degrees but less than 180 degrees. 4. Straight Angle: An angle exactly equal to 180 degrees. 5. Reflex Angle: An angle greater than 180 degrees but less than 360 degrees. 6. Complete Angle: An angle equal to 360 degrees.
2. How do you calculate the measure of an angle formed by two intersecting lines?
Ans. To calculate the measure of an angle formed by two intersecting lines, you can use the following properties: 1. Vertically Opposite Angles: When two lines intersect, the opposite angles are equal. 2. Adjacent Angles: The sum of angles on a straight line is 180 degrees. For example, if one angle measures 50 degrees, the adjacent angle will measure 130 degrees (180 - 50 = 130).
3. What are complementary angles and how do you find them?
Ans. Complementary angles are two angles whose measures add up to 90 degrees. To find a complementary angle, subtract the measure of the given angle from 90 degrees. For example, if you have an angle measuring 30 degrees, its complement is 90 - 30 = 60 degrees.
4. What is the relationship between parallel lines and angles formed by a transversal?
Ans. When a transversal intersects two parallel lines, several pairs of angles are formed. The key relationships include: 1. Corresponding Angles: They are equal. 2. Alternate Interior Angles: They are equal. 3. Alternate Exterior Angles: They are equal. 4. Consecutive Interior Angles: Their measures add up to 180 degrees. These relationships are crucial for solving problems involving parallel lines and angles.
5. How can you prove that two angles are supplementary?
Ans. To prove that two angles are supplementary (sum to 180 degrees), you can use the following steps: 1. Measure both angles using a protractor. 2. Add their measures together. 3. If the sum equals 180 degrees, then the angles are supplementary. Alternatively, if the angles are adjacent and form a straight line, they are automatically supplementary.
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