RBSE Solutions Class 9 Maths Chapter 5 Plane Geometry and Line and Angle More Ques

Multiple Choice Questions (Q1 to Q7)

Question 1. In the adjoining figure, if AB || CD || EF, PQ || RS, ∠RQD = 25° and ∠CQP = 60°, then the value of ∠QRS.
(A) 85°
(B) 135°
(C) 145°
(D) 110°
Answer: (A) 85°
In simple words: First, find the angle between PQ and QR inside the parallel lines. Since the total angle on a straight line at Q is 180 degrees, and two parts are 60 and 25 degrees, the middle part is 180 - (60 + 25) = 95 degrees. Then, because PQ and RS are parallel, the interior angles ∠PQR and ∠QRS must add up to 180 degrees. So, 180 - 95 = 85 degrees. This property is very useful when dealing with parallel lines cut by transversals, as consecutive interior angles always add up to 180 degrees.

🎯 Exam Tip: When dealing with multiple parallel lines and transversals, always identify pairs of alternate interior, corresponding, or consecutive interior angles to find unknown values.

Question 2. In figure, for what value of x, POQ is a line?
Answer: For POQ to be a straight line, the sum of the angles on it must be 180 degrees.
\[ 4x + 40 + 3x = 180 \]
\[ 7x + 40 = 180 \]
\[ 7x = 180 - 40 \]
\[ 7x = 140 \]
\[ x = \frac{140}{7} \]
\[ x = 20 \]Thus, for POQ to be a line, the value of x must be 20 degrees. A straight line always forms a 180-degree angle, which is why these angles add up to that specific sum.
In simple words: When angles are on a straight line, they always add up to 180 degrees. So, add up all the given angles and set them equal to 180, then solve for x.

🎯 Exam Tip: Remember that angles on a straight line form a linear pair and always sum to 180 degrees. This is a basic but essential geometric principle.

Question 3. In figure, if OP || RS, ∠OPQ =110° and ∠QRS = 130° then ∠PQR is equal to
(A) 40°
(B) 50°
(C) 60°
(D) 70°
Answer: (C) 60°
In simple words: Draw a line through Q parallel to OP and RS. The angle formed by PQ with this new line (on the same side as P) is 180 - 110 = 70 degrees. The angle formed by QR with this new line (on the same side as R) is 180 - 130 = 50 degrees. Subtract the smaller angle from the larger supplementary angle to get 70 - 10 = 60. In problems with parallel lines and zigzag transversals, creating a third parallel line through the vertex of the unknown angle often simplifies the calculation.

🎯 Exam Tip: For zigzag angles between parallel lines, drawing an auxiliary line parallel to the given lines through the "bend" point is a standard technique to solve for the unknown angle.

Question 4. In figure, the reflex ∠AOB is equal to:
(A) 60°
(B) 120°
(C) 300°
(D) 360°
Answer: (C) 300°
In simple words: The diagram shows that angle AOB is 60 degrees. To find the reflex angle, subtract this 60 degrees from the full circle angle of 360 degrees. A reflex angle measures the 'larger' angle around a point, encompassing all the space not covered by the smaller angle.

🎯 Exam Tip: Remember that a complete angle around a point is 360 degrees. A reflex angle is always greater than 180 degrees but less than 360 degrees.

Question 6. In figure, which of the following pairs of angles are not corresponding angles:
(A) ∠1, ∠5
(B) ∠2, ∠6
(C) ∠3, ∠7
(D) ∠3, ∠5
Answer: (D) ∠3, ∠5
In simple words: Corresponding angles are found in the same position at each intersection where a transversal crosses parallel lines. Pairs like (1,5), (2,6), (3,7), and (4,8) are corresponding. Angles 3 and 5 are on the same side of the transversal and between the parallel lines, which means they are consecutive interior angles, not corresponding angles. Identifying types of angle pairs, like consecutive interior or alternate interior angles, is crucial for determining if lines are parallel.

🎯 Exam Tip: Clearly differentiate between corresponding angles (same relative position), alternate interior angles (opposite sides, between lines), and consecutive interior angles (same side, between lines) to avoid common mistakes.

Question 9. In figure, AB || CD, find ∠x and ∠y from the angles given in the figure.
Answer: First, consider the angles formed by parallel lines. The sum of angle \( \angle x \) and 80 degrees is equal to angle \( \angle BCD \) because they are alternate interior angles.
\( \implies \angle x + 80° = 116° \)
Subtracting 80 degrees, we get:
\( \implies \angle x = 116° - 80° \)
\( \implies \angle x = 36° \)
Next, at point B, the angles \( \angle y \), \( \angle x \), and 80 degrees form a straight line, so their sum is 180 degrees. Using the value of \( \angle x \) (which is 36 degrees), and the given 80 degrees, we find:
\( \implies \angle y = 180° - (36° + 80°) = 180° - 116° \)
\( \implies \angle y = 64° \)
Thus, the values are \( \angle x = 36° \) and \( \angle y = 64° \). Understanding how angles relate, such as alternate interior angles or angles on a straight line, is fundamental in geometry.
In simple words: Use the property of alternate interior angles to find x by subtracting. Then, use the property that angles on a straight line add up to 180 degrees to find y.

🎯 Exam Tip: Always clearly state the geometric reason (e.g., alternate angles, linear pair) for each step in your solution to ensure full marks.

Question 14. In figure, find which of the lines l, m, n, p, q and r are parallel and why? Give reasons?
Answer: Let's examine the pairs of lines:
First, consider lines l and m and transversal n. The sum of the interior angles \( \angle PAC = 75° \) and \( \angle QCA = 105° \) is \( 75° + 105° = 180° \). Since the sum of consecutive interior angles is 180 degrees, lines l and m are parallel to each other. Therefore, n || p (following the provided solution's conclusion). Identifying types of angle pairs, like consecutive interior or alternate interior angles, is crucial for determining if lines are parallel.
Next, consider lines m and r and transversal q. The alternate interior angles \( \angle ACD \) and \( \angle CDS \) are both 114 degrees. Since alternate interior angles are equal, lines m and r are parallel to each other. So, m || r.
In simple words: If two lines are cut by a transversal and their consecutive interior angles add up to 180 degrees, or if their alternate interior angles are equal, then the two lines are parallel. We apply these rules to find the parallel lines in the figure.

🎯 Exam Tip: When proving lines are parallel, always explicitly state the pair of angles you are using (e.g., alternate interior, corresponding, consecutive interior) and the property that confirms parallelism.

Question 15. In figure, the two lines intersect each other. If ∠1 + ∠2 + ∠3 = 230°, then find the value of ∠1 and ∠4.
Answer: We know that the sum of all angles formed around a point is 360 degrees. So:
\( \angle 1 + \angle 2 + \angle 3 + \angle 4 = 360° \)
We are given that \( \angle 1 + \angle 2 + \angle 3 = 230° \). Substitute this into the equation:
\( 230° + \angle 4 = 360° \)
Now, subtract 230 degrees from both sides to find \( \angle 4 \):
\( \implies \angle 4 = 360° - 230° \)
\( \implies \angle 4 = 130° \)
We also know that \( \angle 1 \) and \( \angle 2 \) form a linear pair (angles on a straight line), so \( \angle 1 + \angle 2 = 180° \). Similarly, \( \angle 2 \) and \( \angle 4 \) are vertically opposite angles, so \( \angle 2 = \angle 4 \).
Since \( \angle 4 = 130° \), then \( \angle 2 = 130° \).
Now, using the linear pair property with \( \angle 1 \) and \( \angle 2 \):
\( \angle 1 + \angle 2 = 180° \)
\( \angle 1 + 130° = 180° \)
\( \implies \angle 1 = 180° - 130° \)
\( \implies \angle 1 = 50° \)
Thus, \( \angle 1 = 50° \) and \( \angle 4 = 130° \). Understanding vertically opposite angles and angles on a straight line is key to solving problems involving intersecting lines.
In simple words: All angles around a point add up to 360 degrees. Use the given sum to find angle 4. Then, use the fact that opposite angles are equal (vertically opposite) and angles on a straight line add up to 180 degrees to find angle 1.

🎯 Exam Tip: Remember to utilize all angle relationships (vertically opposite, linear pair, angles around a point) to solve for all unknown angles systematically.

Question 16. Two plane mirrors PQ and QR are joint together at Q with an angle of 30°. The incident ray AB parallel to the mirror RQ, then find the value of ∠BCQ, ∠CBQ and ∠BDC.
Answer: Given that the angle between the two mirrors PQ and QR is \( \angle PQR = 30° \). The incident ray AB is parallel to mirror QR.
1. **Finding ∠CBQ:**
Since the incident ray AB is parallel to mirror QR, the angle it makes with the mirror PQ is equal to the angle between the mirrors. Thus, the angle between incident ray AB and mirror PQ is 30 degrees. This means \( \angle AB\text{(point on PQ)} = 30° \). If we draw a normal (perpendicular line) to mirror PQ at point B, the angle of incidence will be \( 90° - 30° = 60° \). According to the law of reflection, the angle of reflection is also 60 degrees. Therefore, the reflected ray BC makes an angle of \( 90° - 60° = 30° \) with mirror PQ. So, \( \angle CBQ = 30° \).
2. **Finding ∠BCQ:**
Now consider the triangle formed by the two mirrors and the ray BC, which is \( \triangle BQC \). We know \( \angle BQC = 30° \) (the angle between the mirrors) and we just found \( \angle CBQ = 30° \). The sum of angles in a triangle is 180 degrees, so:
\( \angle BCQ = 180° - (\angle BQC + \angle CBQ) \)
\( \angle BCQ = 180° - (30° + 30°) \)
\( \angle BCQ = 180° - 60° \)
\( \implies \angle BCQ = 120° \)
3. **Finding ∠BDC (or ∠DCQ if D is on QR):**
The ray BC strikes mirror QR at point C. The angle between the ray BC and mirror QR is not \( \angle BCQ \) (which is 120 degrees and an interior angle of \( \triangle BQC \)), but rather the angle formed by BC with the line QR. If \( \angle BCQ = 120° \), then the angle made by BC with the line QR (assuming C is on QR) is \( 180° - 120° = 60° \). This means at point C, the angle of incidence with the normal to mirror QR is \( 90° - 60° = 30° \). The angle of reflection is also 30 degrees. Therefore, the reflected ray CD makes an angle of \( 90° - 30° = 60° \) with mirror QR. If D is a point on the mirror line QR, then \( \angle BDC \) refers to \( \angle DCQ \) in the context of the reflected ray and mirror. So, \( \angle DCQ = 60° \).
The laws of reflection, where the angle of incidence equals the angle of reflection, are fundamental for understanding how light behaves in mirrors.
In simple words: First, use the fact that the incident ray is parallel to one mirror and the angle between mirrors is 30 degrees to find the reflection angle at the first mirror (BCQ = 30). Then, use the triangle angle sum to find BCQ. For the second reflection, calculate the angle the ray makes with the mirror and use reflection laws again to find the final angle.

🎯 Exam Tip: Always draw normals at the points of reflection. The angle of incidence equals the angle of reflection, measured from the normal, not the mirror surface directly. Also, remember standard triangle properties.