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

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 formed when two lines intersect?
Ans. When two lines intersect, they form four angles. The angles are categorized as follows: vertically opposite angles, which are equal, and adjacent angles, which are supplementary (their sum is 180 degrees). The angles can be classified as acute (less than 90 degrees), right (exactly 90 degrees), and obtuse (more than 90 degrees but less than 180 degrees).
2. How do you find the value of an angle when given a pair of parallel lines cut by a transversal?
Ans. When a transversal cuts parallel lines, several pairs of angles are formed. Corresponding angles are equal, alternate interior angles are equal, and same-side interior angles are supplementary. To find the value of an angle, you can use these properties to set up equations based on the given angles and solve for the unknown.
3. What is the relationship between complementary and supplementary angles?
Ans. Complementary angles are two angles whose sum is 90 degrees, while supplementary angles are two angles whose sum is 180 degrees. For example, if one angle measures 30 degrees, its complement would measure 60 degrees, and its supplement would measure 150 degrees.
4. How can you prove that the sum of angles in a triangle is 180 degrees?
Ans. To prove that the sum of angles in a triangle is 180 degrees, you can draw a line parallel to one side of the triangle through the opposite vertex. This will create alternate interior angles with the angles of the triangle. By showing that these angles together form a straight line (180 degrees), you can conclude that the sum of the three angles in the triangle is also 180 degrees.
5. What are some real-life applications of angles and lines?
Ans. Angles and lines have various real-life applications, including architecture, engineering, navigation, and art. For instance, architects use angles to design buildings, engineers apply angles in machinery design, navigators utilize angles for determining direction, and artists use them for creating perspective in their artwork. Understanding lines and angles is essential in these fields for accurate and effective outcomes.
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