RBSE Solutions Class 9 Maths Chapter 5 Plane Geometry and Line and Angle Exercise 5.2

Chapter 5 Plane Geometry and Line and Angle Ex 5.2

Question 1. In the figure, lines AB, CD and EF are parallel. Find the angles \( \angle x \), \( \angle y \), \( \angle z \) and \( \angle p \).
Answer:In the given figure, we have three parallel lines: \( AB \parallel CD \parallel EF \). First, consider lines CD and EF, and the transversal passing through them. We know that \( \angle y \) and the angle \( 58^\circ \) are interior angles on the same side of the transversal. So, their sum is \( 180^\circ \).
\( \angle y + 58^\circ = 180^\circ \)
\( \implies \angle y = 180^\circ - 58^\circ \)
\( \implies \angle y = 122^\circ \) Next, consider the angles \( \angle x \) and \( \angle y \). They form a linear pair on line AB.
\( \angle x + \angle y = 180^\circ \)
\( \implies \angle x + 122^\circ = 180^\circ \)
\( \implies \angle x = 180^\circ - 122^\circ \)
\( \implies \angle x = 58^\circ \) Now, consider lines AB and CD, and the transversal. The angles \( \angle x \) and \( \angle z \) are alternate interior angles. Since AB is parallel to CD, alternate interior angles are equal.
\( \angle x = \angle z \)
\( \implies \angle z = 58^\circ \) Finally, consider lines CD and EF, and the transversal. The angles \( \angle y \) and \( \angle p \) are corresponding angles. Since CD is parallel to EF, corresponding angles are equal.
\( \angle y = \angle p \)
\( \implies \angle p = 122^\circ \) Therefore, the angles are \( x = 58^\circ \), \( y = 122^\circ \), \( z = 58^\circ \), and \( p = 122^\circ \).
In simple words: We find the angles step-by-step. First, use the rule that interior angles on one side of a line add up to 180 degrees. Then, use the linear pair rule. After that, we use the alternate interior angles rule and the corresponding angles rule to find the rest.

🎯 Exam Tip: Always clearly identify the type of angle pair (linear pair, interior, alternate interior, corresponding) for each step to ensure accuracy.

Question 2. In the given figure, AB || EF. Find \( \angle x \) and \( \angle y \).
Answer:Given that line \( AB \parallel EF \). We need to find \( \angle x \) and \( \angle y \). To do this, we draw a line through point P (the vertex of \( \angle x \)) that is parallel to both \( AB \) and \( EF \). Let this line be PQ. Now, we have \( AB \parallel PQ \). Consider the transversal that cuts AB and PQ. The angles \( 125^\circ \) and \( \angle APQ \) are interior angles on the same side of the transversal. Their sum is \( 180^\circ \).
\( 125^\circ + \angle APQ = 180^\circ \)
\( \implies \angle APQ = 180^\circ - 125^\circ \)
\( \implies \angle APQ = 55^\circ \) Next, consider lines \( PQ \parallel EF \). Consider the transversal that cuts PQ and EF. The angles \( \angle QPE \) and \( 141^\circ \) are also interior angles on the same side of the transversal. Their sum is \( 180^\circ \).
\( \angle QPE + 141^\circ = 180^\circ \)
\( \implies \angle QPE = 180^\circ - 141^\circ \)
\( \implies \angle QPE = 39^\circ \) The angle \( \angle x \) is the sum of \( \angle APQ \) and \( \angle QPE \).
\( \angle x = \angle APQ + \angle QPE \)
\( \implies \angle x = 55^\circ + 39^\circ \)
\( \implies \angle x = 94^\circ \) To find \( \angle y \), we know that angles around a point sum to \( 360^\circ \). So, \( \angle y = 360^\circ - \angle x \)
\( \implies \angle y = 360^\circ - 94^\circ \)
\( \implies \angle y = 266^\circ \) Therefore, \( \angle x = 94^\circ \) and \( \angle y = 266^\circ \).
In simple words: First, draw an extra line through the middle angle point, parallel to the main lines. This splits the middle angle into two parts. Then, use the rule that interior angles on the same side of a transversal add up to 180 degrees for both sets of parallel lines to find these two parts. Add them up to get \( \angle x \). Finally, subtract \( \angle x \) from 360 degrees to find \( \angle y \).

🎯 Exam Tip: When a vertex lies between two parallel lines, constructing an auxiliary line through that vertex, parallel to the given lines, can simplify complex angle problems.

Question 3. In figure, the line \( l \parallel m \). Then find the angles which are equal to \( \angle 1 \).
Answer:Given that line \( l \parallel m \). From the figure, we can identify several angle relationships: 1. \( \angle 1 \) and \( \angle 3 \) are vertically opposite angles. Vertically opposite angles are equal.
\( \implies \angle 1 = \angle 3 \) 2. \( \angle 1 \) and \( \angle 5 \) are corresponding angles. Since \( l \parallel m \), corresponding angles are equal.
\( \implies \angle 1 = \angle 5 \) 3. \( \angle 5 \) and \( \angle 7 \) are vertically opposite angles. Vertically opposite angles are equal.
\( \implies \angle 5 = \angle 7 \) Combining these relationships, we can see that: Since \( \angle 1 = \angle 3 \), and \( \angle 1 = \angle 5 \), and \( \angle 5 = \angle 7 \), it means that \( \angle 1 = \angle 3 = \angle 5 = \angle 7 \). Therefore, the angles that are equal to \( \angle 1 \) are \( \angle 3 \), \( \angle 5 \), and \( \angle 7 \).
In simple words: When two parallel lines are crossed by another line, many angles are equal. The angles that are exactly like \( \angle 1 \) in their position (vertically opposite, corresponding, or vertically opposite to a corresponding angle) are also equal to \( \angle 1 \).

🎯 Exam Tip: Mastering the definitions of vertically opposite, corresponding, alternate interior, and alternate exterior angles is crucial for solving problems involving parallel lines and transversals.

Question 4. In the figure, if \( \angle 1 = 60^\circ \) and \( \angle 6 = 120^\circ \), prove that line \( m \parallel n \).
Answer:Given that \( \angle 1 = 60^\circ \) and \( \angle 6 = 120^\circ \). We need to prove that line \( m \parallel n \). From the figure, \( \angle 1 \) and \( \angle 3 \) are vertically opposite angles. Vertically opposite angles are equal.
\( \implies \angle 3 = \angle 1 \)
\( \implies \angle 3 = 60^\circ \) Now, consider \( \angle 3 \) and \( \angle 6 \). These are interior angles on the same side of the transversal between lines \( m \) and \( n \). Let's find their sum:
\( \angle 3 + \angle 6 = 60^\circ + 120^\circ \)
\( \implies \angle 3 + \angle 6 = 180^\circ \) Since the sum of the interior angles on the same side of the transversal is \( 180^\circ \), the lines \( m \) and \( n \) must be parallel. This is the converse of the interior angles theorem. Therefore, \( m \parallel n \).
In simple words: We are given two angles. We first find the angle that is vertically opposite to \( \angle 1 \). Then, we add this new angle and \( \angle 6 \). If their sum is 180 degrees, it proves that the two lines \( m \) and \( n \) are parallel.

🎯 Exam Tip: To prove that two lines are parallel, demonstrate that either corresponding angles are equal, alternate interior angles are equal, or consecutive interior angles sum to 180 degrees.

Question 5. If a transversal intersects two parallel lines \( l \) and \( m \), then prove that the bisectors AP and BQ of any two alternate interior angles are parallel (i.e., AP || BQ).
Answer:Given: Line \( l \parallel m \), and a transversal (let's call it XY) intersects these lines at points A and B respectively. Let the alternate interior angles be \( \angle XAB \) and \( \angle ABY \). Given that AP is the bisector of \( \angle XAB \), and BQ is the bisector of \( \angle ABY \). To Prove: \( AP \parallel BQ \). Proof: 1. Since \( l \parallel m \) and XY is the transversal, the alternate interior angles are equal.
\( \implies \angle XAB = \angle ABY \) (Alternate interior angles) 2. AP is the bisector of \( \angle XAB \). This means AP divides \( \angle XAB \) into two equal halves.
\( \implies \angle PAB = \frac{1}{2} \angle XAB \) 3. BQ is the bisector of \( \angle ABY \). This means BQ divides \( \angle ABY \) into two equal halves.
\( \implies \angle QBA = \frac{1}{2} \angle ABY \) 4. From step 1, \( \angle XAB = \angle ABY \). If we divide both sides by 2, they will still be equal.
\( \implies \frac{1}{2} \angle XAB = \frac{1}{2} \angle ABY \) 5. Substituting from steps 2 and 3 into step 4:
\( \implies \angle PAB = \angle QBA \) 6. Now consider lines AP and BQ, with AB as the transversal. The angles \( \angle PAB \) and \( \angle QBA \) are alternate interior angles. 7. Since \( \angle PAB = \angle QBA \), and these are alternate interior angles, the lines AP and BQ must be parallel.
\( \implies AP \parallel BQ \) Hence proved.
In simple words: When two parallel lines are crossed by another line, their alternate inside angles are equal. If we draw new lines that cut these equal angles exactly in half, these new "bisector" lines will also be parallel to each other.

🎯 Exam Tip: For proofs, clearly state the "Given," "To Prove," and provide each step of your "Proof" with a valid geometric reason.

Question 6. In figure, BA || ED and BC || EF. Show that \( \angle ABC + \angle DEF = 180^\circ \).
Answer:Given: \( BA \parallel ED \) and \( BC \parallel EF \). We need to show that \( \angle ABC + \angle DEF = 180^\circ \). Let's extend line BC to intersect line ED at a point, say G. 1. Since \( BA \parallel ED \), and BG is a transversal, the corresponding angles are equal.
\( \implies \angle ABC = \angle BGD \) (Corresponding angles) --- (i) 2. Now consider lines \( BC \parallel EF \), and GD is a transversal. The angles \( \angle BGD \) and \( \angle DEF \) are interior angles on the same side of the transversal GD. Their sum is \( 180^\circ \).
\( \implies \angle BGD + \angle DEF = 180^\circ \) --- (ii) 3. Substitute \( \angle ABC \) for \( \angle BGD \) from equation (i) into equation (ii).
\( \implies \angle ABC + \angle DEF = 180^\circ \) Thus, we have shown that \( \angle ABC + \angle DEF = 180^\circ \).
In simple words: If two pairs of lines are parallel in a special way, the angle formed by the first pair and the angle formed by the second pair (when added together) will always make 180 degrees. We prove this by extending one line to create a new transversal and using corresponding and interior angle rules.

🎯 Exam Tip: Sometimes, extending lines or drawing auxiliary lines can help reveal hidden angle relationships needed for a proof.

Question 7. In the figure, DE || QR and AP and BP are the bisectors of \( \angle EAB \) and \( \angle RBA \). Find the value of \( \angle APB \).
Answer:Given that \( DE \parallel QR \). A transversal (let's say AB) intersects these parallel lines. The angles \( \angle EAB \) and \( \angle RBA \) are consecutive interior angles (or interior angles on the same side of the transversal). The sum of consecutive interior angles is \( 180^\circ \).
\( \implies \angle EAB + \angle RBA = 180^\circ \) AP is the bisector of \( \angle EAB \), so it divides the angle into two equal parts:
\( \angle PAB = \frac{1}{2} \angle EAB \) BP is the bisector of \( \angle RBA \), so it divides the angle into two equal parts:
\( \angle PBA = \frac{1}{2} \angle RBA \) Now, divide the equation \( \angle EAB + \angle RBA = 180^\circ \) by 2:
\( \frac{1}{2} \angle EAB + \frac{1}{2} \angle RBA = \frac{180^\circ}{2} \)
\( \implies \angle PAB + \angle PBA = 90^\circ \) --- (i) Now, consider the triangle \( \triangle APB \). The sum of angles in a triangle is \( 180^\circ \).
\( \angle PAB + \angle PBA + \angle APB = 180^\circ \) Substitute the value from equation (i) into this equation:
\( 90^\circ + \angle APB = 180^\circ \)
\( \implies \angle APB = 180^\circ - 90^\circ \)
\( \implies \angle APB = 90^\circ \) Therefore, the value of \( \angle APB \) is \( 90^\circ \).
In simple words: When two parallel lines are cut by a transversal, the two inside angles on the same side add up to 180 degrees. If we then draw lines that cut these two inside angles in half, these cutting lines will meet to form a right angle (90 degrees).

🎯 Exam Tip: Remember that the angle formed by the bisectors of two consecutive interior angles between parallel lines is always a right angle (\( 90^\circ \)).

Question 8. Two lines are perpendicular to two parallel lines respectively. Show that these two lines are also parallel.
Answer:Given: Let \( l \) and \( m \) be two parallel lines, i.e., \( l \parallel m \). Let \( AB \) be a line such that \( AB \perp l \). Let \( CD \) be another line such that \( CD \perp m \). To Prove: \( AB \parallel CD \). Proof: 1. Since line \( l \parallel m \), and line \( AB \) is perpendicular to \( l \), it implies that \( AB \) is also perpendicular to \( m \). This is a property of parallel lines: if a line is perpendicular to one of two parallel lines, it is perpendicular to the other as well.
\( \implies AB \perp m \) 2. We are given that \( CD \perp m \). 3. Now, we have two lines, \( AB \) and \( CD \), which are both perpendicular to the same line \( m \). 4. According to a geometric theorem, if two lines are perpendicular to the same line, then they are parallel to each other.
\( \implies AB \parallel CD \) Hence proved.
In simple words: Imagine two long, straight roads running side-by-side (parallel). If you build a fence straight across the first road (perpendicular), and another fence straight across the second road (perpendicular), then these two fences will also run parallel to each other.

🎯 Exam Tip: A key concept here is that lines perpendicular to the same line are parallel to each other. This is a powerful tool in geometry proofs.

Question 9. In the given figure, if \( AB \parallel CD \parallel EF \), and \( y:z = 3:7 \), find the values of \( x, y, \) and \( z \).
Answer:Given: \( AB \parallel CD \parallel EF \). Also, \( y:z = 3:7 \). 1. First, consider the parallel lines \( CD \) and \( EF \). The transversal intersects them. \( \angle CQR \) and \( \angle QRF \) are alternate interior angles. So, \( \angle CQR = \angle QRF \). From the figure, \( \angle QRF = \angle z \).
\( \implies \angle CQR = \angle z \) --- (i) 2. Next, consider the parallel lines \( AB \) and \( CD \). The transversal intersects them. \( \angle APQ \) and \( \angle CQR \) are corresponding angles. So, \( \angle APQ = \angle CQR \). From the figure, \( \angle APQ = \angle x \).
\( \implies \angle x = \angle CQR \) --- (ii) 3. From (i) and (ii), we can conclude:
\( \angle x = \angle z \) --- (iii) 4. Now, consider the angles \( \angle CQP \) and \( \angle CQR \). They form a linear pair on the line CD.
\( \angle CQP + \angle CQR = 180^\circ \) From the figure, \( \angle CQP = \angle y \).
\( \implies \angle y + \angle CQR = 180^\circ \) Substitute \( \angle CQR = \angle z \) from (i):
\( \implies \angle y + \angle z = 180^\circ \) --- (iv) 5. We are given the ratio \( y:z = 3:7 \). Let \( y = 3k \) and \( z = 7k \) for some constant \( k \). 6. Substitute these values into equation (iv):
\( 3k + 7k = 180^\circ \)
\( \implies 10k = 180^\circ \)
\( \implies k = \frac{180^\circ}{10} \)
\( \implies k = 18^\circ \) 7. Now, find the values of \( y \) and \( z \):
\( y = 3k = 3 \times 18^\circ = 54^\circ \)
\( z = 7k = 7 \times 18^\circ = 126^\circ \) 8. From equation (iii), we know \( \angle x = \angle z \).
\( \implies \angle x = 126^\circ \) Therefore, the values are \( x = 126^\circ \), \( y = 54^\circ \), and \( z = 126^\circ \).
In simple words: First, we use the rules of parallel lines to find out that angles \( x \) and \( z \) are equal. Then, we use the linear pair rule to show that angles \( y \) and \( z \) add up to 180 degrees. Since we know the ratio of \( y \) to \( z \), we can set up an equation to find their exact values. Once we have \( z \), we also know \( x \).

🎯 Exam Tip: When given ratios of angles, express them in terms of a common multiple (e.g., \( 3k, 7k \)) to simplify calculations. Remember that angles forming a linear pair sum to \( 180^\circ \).

Question 10. In the figure, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB || CD.
Answer:Given: Mirrors \( PQ \parallel RS \). Ray AB is incident on PQ at B, reflects as BC. Ray BC is incident on RS at C, reflects as CD. To Prove: \( AB \parallel CD \). Construction: Draw a normal (perpendicular line) BM to PQ at B, and a normal CN to RS at C. Proof: 1. According to the law of reflection, the angle of incidence is equal to the angle of reflection. At mirror PQ: \( \angle ABM = \angle MBC \) (Let's call this \( \angle 1 = \angle 2 \)) --- (i) At mirror RS: \( \angle BCN = \angle NCD \) (Let's call this \( \angle 3 = \angle 4 \)) --- (ii) 2. Since PQ is parallel to RS, and BM is perpendicular to PQ, and CN is perpendicular to RS, then the normals BM and CN are also parallel to each other.
\( \implies BM \parallel CN \) 3. Now, consider \( BM \parallel CN \) and BC as the transversal. The alternate interior angles are equal.
\( \implies \angle MBC = \angle BCN \) (i.e., \( \angle 2 = \angle 3 \)) --- (iii) 4. From (i), \( \angle 1 = \angle 2 \). From (iii), \( \angle 2 = \angle 3 \). From (ii), \( \angle 3 = \angle 4 \). Therefore, by transitivity, \( \angle 1 = \angle 2 = \angle 3 = \angle 4 \). This implies \( \angle 1 = \angle 4 \). 5. Consider the angle \( \angle ABC \). It is formed by \( \angle ABM \) and \( \angle MBC \).
\( \angle ABC = \angle ABM + \angle MBC = \angle 1 + \angle 2 \) Since \( \angle 1 = \angle 2 \), we can write \( \angle ABC = 2 \angle 2 \). 6. Consider the angle \( \angle BCD \). It is formed by \( \angle BCN \) and \( \angle NCD \).
\( \angle BCD = \angle BCN + \angle NCD = \angle 3 + \angle 4 \) Since \( \angle 3 = \angle 4 \), we can write \( \angle BCD = 2 \angle 3 \). 7. From step 4, we know \( \angle 2 = \angle 3 \). Therefore, \( 2 \angle 2 = 2 \angle 3 \).
\( \implies \angle ABC = \angle BCD \) 8. Now, consider lines AB and CD, with BC as the transversal. The angles \( \angle ABC \) and \( \angle BCD \) are alternate interior angles. Since \( \angle ABC = \angle BCD \), the lines AB and CD must be parallel.
\( \implies AB \parallel CD \) Hence proved.
In simple words: To prove the initial and final rays are parallel, we draw lines straight out from the mirrors (normals). The law of reflection says the incoming and outgoing angles are equal. Because the mirrors are parallel, these normal lines are also parallel. This helps us show that the angle of the incoming light ray is the same as the angle of the outgoing light ray when compared to the line joining the mirrors. Since these are alternate inside angles and are equal, the first and last light paths must be parallel.

🎯 Exam Tip: In mirror problems, always draw the 'normal' (a line perpendicular to the mirror surface) at the point of incidence to correctly apply the law of reflection (angle of incidence = angle of reflection).

Question 11. In figure, if PQ || RS, \( \angle MXQ = 135^\circ \) and \( \angle MYR = 40^\circ \) then find \( \angle XMY \).
Answer:Given: \( PQ \parallel RS \), \( \angle MXQ = 135^\circ \), and \( \angle MYR = 40^\circ \). We need to find \( \angle XMY \). Construction: Draw a line TZ through point M, parallel to both PQ and RS.
\( \implies TZ \parallel PQ \) and \( TZ \parallel RS \). Now, consider lines \( PQ \parallel TZ \) and XM as the transversal. The angles \( \angle MXQ \) and \( \angle XMT \) (let's call it \( \angle 1 \)) are interior angles on the same side of the transversal. Their sum is \( 180^\circ \).
\( \angle MXQ + \angle XMT = 180^\circ \)
\( 135^\circ + \angle 1 = 180^\circ \)
\( \implies \angle 1 = 180^\circ - 135^\circ \)
\( \implies \angle 1 = 45^\circ \) --- (i) Next, consider lines \( RS \parallel TZ \) and YM as the transversal. The angles \( \angle MYR \) and \( \angle YMT \) (let's call it \( \angle 2 \)) are alternate interior angles. Alternate interior angles are equal.
\( \implies \angle YMT = \angle MYR \)
\( \implies \angle 2 = 40^\circ \) --- (ii) The angle \( \angle XMY \) is the sum of \( \angle 1 \) and \( \angle 2 \).
\( \angle XMY = \angle 1 + \angle 2 \) Substitute values from (i) and (ii):
\( \angle XMY = 45^\circ + 40^\circ \)
\( \implies \angle XMY = 85^\circ \) Therefore, the value of \( \angle XMY \) is \( 85^\circ \).
In simple words: To solve this, we draw a new line through point M that is parallel to the other two lines. This new line splits the angle \( \angle XMY \) into two smaller angles. For the first part, we use the rule that inside angles on the same side of a transversal add up to 180 degrees. For the second part, we use the rule that alternate interior angles are equal. Finally, we add these two smaller angles to find the total angle \( \angle XMY \).

🎯 Exam Tip: Always look for opportunities to draw auxiliary parallel lines when a vertex is situated between two given parallel lines; this often breaks down a complex problem into simpler, solvable parts.