RBSE Solutions Class 9 Maths Chapter 5 Plane Geometry and Line and Angle Important Questions

Multiple Choice Questions

Question 1. In the figure, PQ || RS, ∠QPR = 70°, ∠ROT = 20° then the value of x is:

(a) 20°
(b) 70°
(c) 110°
(d) 50°
Answer: (d) 50°
In simple words: Imagine drawing a line through point T that is parallel to PQ and RS. The angle formed between PT and this new line (on the left side) would be 70 degrees, just like angle QPR, because they are alternate interior angles. Similarly, the angle formed between TO and this new line (on the right side) would be 20 degrees, just like angle ROT. Since x is the angle between PT and TO, we subtract these two angles (70 - 20) to get 50 degrees.

🎯 Exam Tip: For problems involving angles between parallel lines and a zig-zag transversal, draw an auxiliary line through the vertex of the zig-zag, parallel to the given parallel lines, to create simpler alternate interior angles.

Question 2. Two angles measures (30 – a)° and (125 + 2a)°. If each one is the supplement of the other, then 'a' is:
(a) 35°
(b) 25°
(c) 65°
(d) 45°
Answer: (b) 25°
In simple words: When two angles are supplementary, it means they add up to 180 degrees. So, we add the two given expressions for the angles and set the total equal to 180. Then, we solve the simple equation to find the value of 'a'.

🎯 Exam Tip: Remember that "complementary angles" add up to 90 degrees, while "supplementary angles" add up to 180 degrees. Be careful not to confuse the two terms during calculations.

Question 5. In the given figure, what value of x would make AOB a straight line?

(a) 12°
(b) 14°
(c) 16°
(d) 18°
Answer: (c) 16°
In simple words: If AOB is a straight line, it means the angles on that line add up to 180 degrees. So, we add the two angle expressions, set them equal to 180, and then solve for x.

🎯 Exam Tip: Always look for keywords like "straight line" or "linear pair" as they indicate that angles add up to 180 degrees. This helps set up the correct equation for solving.

Question 7. In the given figure, if l || m then the value of x is:

(a) 30°
(b) 40°
(c) 35°
(d) 45°
Answer: (a) 30°
In simple words: Since lines l and m are parallel, the angles on the same side of the transversal (also called consecutive interior angles) add up to 180 degrees. So, we add 110 degrees and (3x - 20) degrees, set the sum equal to 180, and solve for x.

🎯 Exam Tip: Always remember the properties of parallel lines and transversals: alternate interior angles are equal, corresponding angles are equal, and consecutive interior angles are supplementary (add to 180°).

Very Short Answer Type Questions

Question 1. If the angles (3x - 20)° and (2x - 40)° are complementary angles. Find x.
Answer: When two angles are complementary, their sum is 90°.
\( (3x - 20)^\circ + (2x - 40)^\circ = 90^\circ \)
\( 5x - 60^\circ = 90^\circ \)
\( 5x = 90^\circ + 60^\circ \)
\( 5x = 150^\circ \)
\( x = \frac{150^\circ}{5} \)
\( x = 30^\circ \)
In simple words: Complementary angles always add up to 90 degrees. So, we add the two given angle expressions together and set the total equal to 90. Then we solve the equation to find x.

🎯 Exam Tip: Clearly write out the definition of complementary angles first, then substitute the given expressions to form the equation. This shows your understanding and helps avoid calculation errors.

Question 2. An angle is 25° less than its complement. What is its measure?
Answer: Let the angle be \( x \). The complement of this angle will be \( 90^\circ - x \). According to the problem, the angle is 25° less than its complement:
\( x = (90^\circ - x) - 25^\circ \)
\( x = 90^\circ - x - 25^\circ \)
\( x = 65^\circ - x \)
\( x + x = 65^\circ \)
\( 2x = 65^\circ \)
\( x = \frac{65^\circ}{2} \)
\( x = 32.5^\circ \) or \( 32\frac{1}{2}^\circ \). Therefore, the measure of the required angle is \( 32.5^\circ \).
In simple words: We call the unknown angle 'x'. Its complement is (90-x). The problem says 'x' is 25 less than (90-x). We write this as an equation: x = (90-x) - 25, and then solve to find x.

🎯 Exam Tip: When dealing with relationships between an angle and its complement or supplement, always start by defining the angle as 'x' and its complement/supplement in terms of 'x'. This makes setting up the equation much clearer.

Question 3. In the given figure, POS is a line, find the measure of ∠QOR.

Answer: We are given that POS is a straight line. The sum of angles on a straight line is 180°.
\( \angle POQ + \angle QOR + \angle ROS = 180^\circ \) (Angles on a straight line)
\( 60^\circ + (5x - 20^\circ) + 40^\circ = 180^\circ \)
\( 5x + 80^\circ = 180^\circ \)
\( 5x = 180^\circ - 80^\circ \)
\( 5x = 100^\circ \)
\( x = \frac{100^\circ}{5} \)
\( x = 20^\circ \) Now we find the measure of \( \angle QOR \):
\( \angle QOR = (5x - 20^\circ) = (5 \times 20^\circ - 20^\circ) = (100^\circ - 20^\circ) = 80^\circ \)
In simple words: Because POS is a straight line, all the angles along it add up to 180 degrees. We add the three given angle expressions (60, 5x-20, and 40) and set the total to 180 to find x. Once we have x, we plug it back into the expression for angle QOR to get its value.

🎯 Exam Tip: Clearly state the reason for your first step, such as "angles on a straight line" or "linear pair," as this demonstrates proper geometric reasoning.

Question 5. In the given figure, find the value of x.

Answer: In the given figure, BCD is a straight line. So, \( \angle ACB + \angle ACD = 180^\circ \) (Linear pair of angles)
\( 5y + 7y = 180^\circ \)
\( 12y = 180^\circ \)
\( y = \frac{180^\circ}{12} \)
\( y = 15^\circ \) Now, using the exterior angle property for triangle ABC: The exterior angle is equal to the sum of the two opposite interior angles.
\( \angle ACD = \angle BAC + \angle ABC \)
\( 7y = 3y + x \)
\( x = 7y - 3y \)
\( x = 4y \) Substitute the value of \( y = 15^\circ \) into the equation for x:
\( x = 4 \times 15^\circ \)
\( x = 60^\circ \)
In simple words: First, we use the fact that angles on a straight line add up to 180 degrees. This helps us find the value of 'y' from the angles 5y and 7y. Then, we use the rule that an exterior angle of a triangle (7y) is equal to the sum of the two opposite interior angles (3y and x). We substitute the value of 'y' to find 'x'.

🎯 Exam Tip: Always clearly identify linear pairs and use the exterior angle property of a triangle. Labeling all known and unknown angles on the diagram can simplify complex problems.

Question 6. In the given figure, find x and also find ∠POS.

Answer: In the given figure, POQ is a straight line. The sum of angles on a straight line is 180°.
\( \angle POS + \angle SOR + \angle ROQ = 180^\circ \) (Angles on a straight line)
\( (x + 45^\circ) + x + (x + 30^\circ) = 180^\circ \)
\( 3x + 75^\circ = 180^\circ \)
\( 3x = 180^\circ - 75^\circ \)
\( 3x = 105^\circ \)
\( x = \frac{105^\circ}{3} \)
\( x = 35^\circ \) Now we find the measure of \( \angle POS \):
\( \angle POS = (x + 45^\circ) = (35^\circ + 45^\circ) = 80^\circ \)
In simple words: Since POQ is a straight line, all the angles around the point O, on one side of the line, must add up to 180 degrees. We add the three given angle expressions (x+45, x, and x+30) and set them equal to 180 to solve for x. Then, we substitute the value of x into the expression for angle POS to find its measure.

🎯 Exam Tip: Remember that if multiple angles share a common vertex and lie on a straight line, their sum is 180°. This is a fundamental concept for solving many geometry problems.

Question 8. In the figure, ∠POM and ∠QOM form a linear pair. If x – 2y = 30°, find x and y.

Answer: Given that \( \angle POM \) and \( \angle QOM \) form a linear pair. This means their sum is 180°.
\( x + y = 180^\circ \) ...(i) We are also given the equation:
\( x - 2y = 30^\circ \) ...(ii) Now we solve this system of two linear equations. From equation (i), we can write \( x = 180^\circ - y \). Substitute this expression for x into equation (ii):
\( (180^\circ - y) - 2y = 30^\circ \)
\( 180^\circ - 3y = 30^\circ \)
\( 180^\circ - 30^\circ = 3y \)
\( 150^\circ = 3y \)
\( y = \frac{150^\circ}{3} \)
\( y = 50^\circ \) Now, substitute the value of y back into equation (i) to find x:
\( x + 50^\circ = 180^\circ \)
\( x = 180^\circ - 50^\circ \)
\( x = 130^\circ \) So, the values are \( x = 130^\circ \) and \( y = 50^\circ \).
In simple words: A linear pair means two angles add up to 180 degrees, so x + y = 180. We also have another equation, x - 2y = 30. We use these two equations together to find both x and y. One way is to find x from the first equation (x = 180 - y) and put it into the second equation, then solve for y. After finding y, we can easily find x.

🎯 Exam Tip: When given two equations with two variables (like x and y), use substitution or elimination methods to solve them. Clearly label your equations to keep track of your steps.

Question 9. In the figure, a : b : c = 4 : 3 : 5. If AOB is a straight line, find the value of a, b and c.

Answer: We are given that the ratio of the angles is a : b : c = 4 : 3 : 5. We can represent the angles using a common multiplier, say k. So, \( \angle a = 4k \), \( \angle b = 3k \), and \( \angle c = 5k \). Given that AOB is a straight line. This means the sum of the angles on the straight line is 180°.
\( \angle a + \angle b + \angle c = 180^\circ \) (Angles on a straight line)
\( 4k + 3k + 5k = 180^\circ \)
\( 12k = 180^\circ \)
\( k = \frac{180^\circ}{12} \)
\( k = 15^\circ \) Now we can find the measure of each angle:
\( \angle a = 4k = 4 \times 15^\circ = 60^\circ \)
\( \angle b = 3k = 3 \times 15^\circ = 45^\circ \)
\( \angle c = 5k = 5 \times 15^\circ = 75^\circ \)
In simple words: Since AOB is a straight line, all the angles along it add up to 180 degrees. The angles are given as a ratio, so we write them as 4k, 3k, and 5k. We add these up and set them equal to 180 to find k. Then, we multiply k by each part of the ratio to find the actual size of each angle.

🎯 Exam Tip: When angles are given in a ratio, introduce a common multiplier (like k or x) to represent their actual measures. Remember that angles on a straight line sum to 180 degrees.

Short Answer Type Questions

Question 1. In the given figure, AC ⊥ AB. If ∠BAP = x, ∠PAQ = (x + 5°) and ∠CAQ = (2x + 5°), find the value of x and reflex ∠PAQ.

Answer: Given that AC ⊥ AB, this means the angle \( \angle CAB \) is a right angle, so \( \angle CAB = 90^\circ \). From the figure, the sum of the angles \( \angle BAP \), \( \angle PAQ \), and \( \angle CAQ \) makes up the total angle \( \angle CAB \).
\( \angle BAP + \angle PAQ + \angle CAQ = \angle CAB \)
\( x + (x + 5^\circ) + (2x + 5^\circ) = 90^\circ \) Combine like terms:
\( (x + x + 2x) + (5^\circ + 5^\circ) = 90^\circ \)
\( 4x + 10^\circ = 90^\circ \)
\( 4x = 90^\circ - 10^\circ \)
\( 4x = 80^\circ \)
\( x = \frac{80^\circ}{4} \)
\( x = 20^\circ \) Now we find the measure of \( \angle PAQ \):
\( \angle PAQ = (x + 5^\circ) = (20^\circ + 5^\circ) = 25^\circ \) To find the reflex angle of \( \angle PAQ \), we subtract it from 360°:
Reflex \( \angle PAQ = 360^\circ - \angle PAQ \)
Reflex \( \angle PAQ = 360^\circ - 25^\circ \)
Reflex \( \angle PAQ = 335^\circ \)
In simple words: Since AC is perpendicular to AB, angle CAB is 90 degrees. The three smaller angles (x, x+5, and 2x+5) add up to this 90-degree angle. We combine them to form an equation and solve for x. Once we find x, we can calculate angle PAQ. To find the reflex angle, we subtract PAQ from 360 degrees.

🎯 Exam Tip: Always remember that perpendicular lines form a 90° angle. For reflex angles, the formula is 360° minus the angle itself.

Question 2. In the given figure, AB || CD, if ∠OAB = 58° and ∠OCD = 45° then find the value of x.

Answer: Given that AB || CD. We have \( \angle OAB = 58^\circ \) and \( \angle OCD = 45^\circ \). We need to find the value of x, which is the reflex angle \( \angle AOC \). Let's draw a line PQ through point O, such that PQ is parallel to AB and CD.
\( \implies \) Since PQ || AB, \( \angle AOQ = \angle OAB = 58^\circ \) (These are alternate interior angles).
\( \implies \) Since PQ || CD, \( \angle COQ = \angle OCD = 45^\circ \) (These are alternate interior angles). Now, the angle \( \angle AOC \) is the sum of \( \angle AOQ \) and \( \angle COQ \):
\( \angle AOC = \angle AOQ + \angle COQ \)
\( \angle AOC = 58^\circ + 45^\circ \)
\( \angle AOC = 103^\circ \) The value of x is the reflex angle \( \angle AOC \).
\( x = \text{Reflex } \angle AOC = 360^\circ - \angle AOC \)
\( x = 360^\circ - 103^\circ \)
\( x = 257^\circ \)
In simple words: To solve this, we draw an imaginary line through point O that is parallel to both AB and CD. Because of this, the angle OAB (58°) becomes an alternate interior angle to the upper part of angle AOC. Similarly, angle OCD (45°) becomes an alternate interior angle to the lower part of angle AOC. We add these two parts to find angle AOC. Since x is the reflex angle of AOC, we subtract angle AOC from 360 degrees.

🎯 Exam Tip: When a vertex lies between two parallel lines, always consider drawing an auxiliary line through that vertex, parallel to the given lines. This simplifies the problem into separate sets of alternate interior angles.

Question 4. "If a transversal intersects two parallel lines, then each pair of interior angles are supplementary”. Prove it.

Answer:Given: AB and CD are two parallel lines (AB || CD), and EF is a transversal that intersects them at points M and N respectively. The transversal forms two pairs of interior angles: \( \angle 1, \angle 4 \) and \( \angle 2, \angle 3 \). (Using the common interpretation for interior angles, where \( \angle 1 \) is AMN, \( \angle 2 \) is BMN, \( \angle 3 \) is CNM, \( \angle 4 \) is DNM, as per image's numerical labeling.)To Prove: Each pair of interior angles is supplementary. That means:
(i) \( \angle 1 + \angle 3 = 180^\circ \)
(ii) \( \angle 2 + \angle 4 = 180^\circ \)Proof:Consider the line CD and ray NE standing on it.
\( \angle CNM + \angle MNF = 180^\circ \) (Angles forming a linear pair). Let \( \angle MNF \) be \( \angle 5 \).
\( \implies \angle 3 + \angle 5 = 180^\circ \) ...(i) Now, since AB || CD and EF is the transversal:
\( \angle BMN = \angle CNF \) (These are corresponding angles). Let \( \angle CNF \) be \( \angle 5 \).
\( \implies \angle 2 = \angle 5 \) ...(ii) From equations (i) and (ii), substitute \( \angle 2 \) for \( \angle 5 \):
\( \angle 3 + \angle 2 = 180^\circ \) Thus, one pair of interior angles, \( \angle 2 \) and \( \angle 3 \), are supplementary. Similarly, we can prove the other pair. Consider the line AB and ray ME standing on it.
\( \angle AMN + \angle AMF = 180^\circ \) (Linear pair). Let \( \angle AMF \) be \( \angle 6 \).
\( \implies \angle 1 + \angle 6 = 180^\circ \) ...(iii) Also, since AB || CD and EF is the transversal:
\( \angle CNM = \angle AMF \) (These are corresponding angles). Let \( \angle AMF \) be \( \angle 6 \).
\( \implies \angle 3 = \angle 6 \) ...(iv) From equations (iii) and (iv), substitute \( \angle 3 \) for \( \angle 6 \):
\( \angle 1 + \angle 3 = 180^\circ \) Thus, the other pair of interior angles, \( \angle 1 \) and \( \angle 3 \), are also supplementary. Hence, if a transversal intersects two parallel lines, each pair of interior angles is supplementary.
In simple words: When a line cuts across two parallel lines, it creates special angle pairs. For any two interior angles on the same side of this cutting line, they will always add up to 180 degrees. We prove this by using the idea of a linear pair (angles on a straight line add to 180) and corresponding angles (which are equal when lines are parallel).

🎯 Exam Tip: Always clearly state the geometric reasons for each step in a proof (e.g., "Linear pair," "Corresponding angles"). Drawing and labeling the diagram accurately is crucial for understanding and presenting the proof.