RS Aggarwal Solutions for Class 6 Chapter 19 Three Dimensional Shapes

Exercise 19.1

Question 1. Construct line segments whose lengths are:
(i) 4.8 cm
(ii) 12 cm 5 mm
(iii) 7.6 cm
Answer:
(i) Start by drawing a line L and placing point A on it. Obtain a compass and position its metal tip at the zero mark on the ruler. Adjust the compass opening so the pencil reaches the 4.8 cm mark on the ruler. Now transfer the compass to line L with the metal point on A. Create a small mark at B where the pencil touches the line. The segment AB measures 4.8 cm.

(ii) Create a line L and mark a point A on it. Take a compass and set its metal point at the zero mark of the ruler. Set the compass width so the pencil aligns with the point midway between the 12 cm and 13 cm marks (which is 5 mm from 12 cm). Move the compass to line L with the metal point at A. Make a small mark at B on the line L where the pencil point lands. The segment AB measures 12 cm 5 mm.

(iii) Draw a line L and place point A on it. Take a compass and set its metal point at the zero mark on the ruler. Adjust the compass so the pencil sits at the point halfway between the 7 cm and 8 cm marks (which is 6 mm from 7 cm). Position the compass on line L with the metal point on A. Mark a small point at B on line L where the pencil touches. The segment AB measures 7.6 cm.

Exam Tip: Always check that the compass opening remains fixed while transferring it to the line, and ensure the metal point stays exactly on the starting point A.

Question 2. Construct two segments of lengths 4.3 cm and 3.2 cm. Construct a segment whose length is equal to the sum of the lengths of these segments.
Answer:
Using compass and ruler, first create two segments AB and CD with lengths 4.3 cm and 3.2 cm, respectively.

Draw a line L and mark point P on it. Place the compass with its metal tip at A and adjust it so the pencil reaches B. Transfer the compass to line L, positioning the metal point at P, and mark point Q where the pencil touches the line.

Now reset the compass by placing the metal tip at C and adjusting the pencil to reach D. Move the compass to line L with the metal point at Q. Make a mark at point R where the pencil touches. The segment PR represents the required length, equal to the combined length of both segments.

Exam Tip: Keep the compass setting unchanged when moving it from one segment to another to maintain accurate lengths.

Exercise 19.2

Question 1. How many lines can be drawn which are perpendicular to a given line and pass through a given point lying
(i) outside it?
(ii) on it?
Answer:
(i) The perpendicular from any point to a line represents the shortest distance between them. Since only one shortest distance exists, just one perpendicular line can be drawn from an outside point to a given line.

(ii) At any location on a line, we can construct only one perpendicular. Therefore, when a point lies on the given line, only one perpendicular line can be drawn through that point.

Exam Tip: Remember that perpendicularity is unique - there is exactly one shortest path from any point to a line.

Question 2. Draw a line PQ. Take a point R on it. Draw a line perpendicular to PQ and passing through R.
(i) Using ruler and set square.
(ii) ruler and compasses.
Answer:
(i) Create a line PQ and choose a point R on it. Position a set-square so that one arm of the right angle aligns with line PQ. Without moving the set-square, place a ruler along its other arm. Carefully remove the set-square while keeping the ruler fixed, then draw line MN through point R. This line MN is perpendicular to line PQ at point R.

(ii) Draw line PQ and select point R on it. Using R as the center and an appropriate radius, draw an arc that crosses PQ at two points, A and B. With A and B as centers and a radius greater than AR, draw two more arcs that intersect at point S on the opposite side of PQ. Connect R to S and extend the line in both directions. The line RS is perpendicular to PQ and passes through R.

Exam Tip: For the set-square method, ensure the ruler stays aligned with the set-square edge throughout. For the compass method, always use a radius greater than half the distance AB.

Question 3. Draw a line I, take a point A not lying on I. Draw a line m such and passing through A. Using
(i) ruler and set square
(ii) ruler and compass.
Answer:
(i) Draw line L and select point A outside it. Position a set-square PQR so one arm PQ of the right angle sits along line L. Without moving the set-square, align a ruler along its edge PR. Now, while keeping the ruler fixed, slide the set-square along the ruler until arm QR reaches point A. Draw line m. Line m is perpendicular to line L.

(ii) Using A as the center, draw an arc PQ that intersects line L at points P and Q. With P and Q as centers and a radius exceeding half the distance PQ, construct two arcs that intersect at point B. Connect points A and B and extend in both directions. This is the required line perpendicular to line L.

Exam Tip: In method (i), maintain pressure on both the ruler and set-square to prevent slipping. In method (ii), ensure the radius is more than half PQ to guarantee intersection of the arcs.

Question 4. Draw a line AB and take two points C and E on opposite sides of AB. Through C, draw CD⊥AB and through E draw EF⊥AB. Using
(i) ruler and set square
(ii) ruler and compass.
Answer:
(i) Create a line AB and place points C and E on opposite sides. On the side of C, position a set-square PQR so one arm PQ of the right angle aligns with line AB. Without moving the set-square, place a ruler along its edge PR. Keeping the ruler stationary, slide the set-square until arm QR reaches point C. Draw line CD where D is a point on AB. CD is perpendicular to AB. Repeat this same procedure starting with the set-square on the side of E to obtain line EF perpendicular to AB.

(ii) Create a line AB with points C and E on opposite sides. Using C as the center, draw an arc PQ that crosses AB at points P and Q. With P and Q as centers and a radius exceeding distance AR, construct two arcs that meet at point H. Connect C to H and cross AB at D. The line CD is perpendicular to AB. Similarly, using E as the center, draw arc RS intersecting AB at points R and S. With R and S as centers and a radius exceeding distance RS divided by 2, create two arcs intersecting at point G. Connect E to G, which crosses AB at F. The line EF is perpendicular to AB.

Exam Tip: Ensure both perpendiculars are constructed using identical methods for consistency. Always verify that the radius exceeds half the segment length.

Question 5. Draw a line segment AB of length 10 cm. Mark a point P on AB such that AP = 4 cm. Mark a point P on AB such that AP = 4 cm. Draw a line through P perpendicular to AB.
Answer:
Draw line L and place point A on it. Using ruler and compass, mark point B on line L such that the distance AB equals 10 cm. This is the required segment of 10 cm. Next, measure 4 cm from A toward B and mark this position as P. Using P as the center with a radius of 4 cm, construct an arc that touches line L at two points, A and E. With A and E as centers and applying a radius of 6 cm, draw two arcs that cross each other at point R. Extend PR in both directions. This line PR is perpendicular to segment AB.

Exam Tip: The key is to use P as center with its distance from A as the radius, ensuring the arc intersects the line at equal distances on either side of P.

Question 6. Draw a line segment PQ of length of length 12 cm. Mark a point O outside this segment. Draw a line through O perpendicular to PQ.
Answer:
Create line L and mark point P on it. Using ruler and compass, identify point Q on line L such that PQ measures 12 cm. Select an external point O. With O as the center and an appropriate radius, draw an arc that intersects line L at points A and B. Using A and B as centers, construct two arcs with identical radius such that they meet at point C. Connect O and C. The line OC is perpendicular to segment PQ.

Exam Tip: Choose the arc radius centered at O large enough so the arc cuts the segment at two distinct points, allowing you to find their midpoint.

Question 7. Using a protractor, draw ∠BAC of measure 70°. On side AC, take a point P, such that AP = 2cm. From P draw a line perpendicular to AB.
Answer:
Create line segment AC on a line L. Position a protractor on segment AC so the segment aligns with the protractor's diameter line and the midpoint of this line aligns with point A. Measuring from the right side, mark point B at the 70° position on the protractor and draw segment AB. This creates angle BAC of 70°. From point A along segment AC, measure 2 cm and mark point P. Using P as the center, draw an arc that crosses line AB at points E and F. With E and F as centers and using the same radius, construct two arcs that intersect at point G on the opposite side. Connect P to G. Line PG is perpendicular to segment AB.

Exam Tip: When using the protractor, ensure the baseline aligns perfectly with the segment and read angles carefully from the appropriate side scale.

Question 8. Draw a line segment AB of length 8 cm. At each end of this line segment, draw a line perpendicular to AB. Are these two lines parallel?
Answer:
Using convenient radius and A as center, construct an arc crossing the line at points W and X. With W and X as centers and radius exceeding distance AW, draw two arcs intersecting at M. Connect A to M and extend in both directions toward P and Q. This line is perpendicular to AB at A. Similarly, using convenient radius and B as center, construct an arc crossing the line at points Y and Z. With Y and Z as centers and radius exceeding distance YB, draw two arcs intersecting at N. Connect B to N and extend in both directions toward S and R. This line is perpendicular to AB at B. Let the perpendiculars at A and B be PQ and RS, respectively. Since angle QAB equals 90° and angle ABR equals 90°, we have angle QAB equal to angle ABR. When two parallel lines are cut by a transversal, alternate interior angles are equal. Since angle QAB equals angle ABR, lines PQ and RS are parallel.

Exam Tip: Two perpendiculars to the same line are always parallel to each other - this is a fundamental property worth remembering for geometry problems.

Question 9. Using a protractor, draw ∠BAC of measure 45°. Take a point P in the interior of ∠BAC. From P draw line segments PM and PN such that PM⊥AB and PN⊥AC. Measure ∠MPN.
Answer: Draw the angle BAC with a measure of 45° using your protractor. Select a point P inside this angle. From P, construct a perpendicular line to ray AB, meeting it at M. Similarly, draw a perpendicular from P to ray AC, meeting it at N. When you measure the angle MPN using your protractor, you will find it equals 135°.
In simple words: When you draw perpendiculars from a point inside an angle to both sides of the angle, the angle formed between these two perpendiculars always adds up with the original angle to make 180°. So if your angle is 45°, the angle between the perpendiculars will be 135°.

Exam Tip: Always verify that PM and PN are truly perpendicular to AB and AC respectively - use the right angle symbol (90°) to mark these. The relationship ∠MPN = 180° - ∠BAC is key.

Question 10. Draw an angle and label it as ∠BAC. Draw a bisector ray AX and take point P on it. From P draw line segments PM and PN, such that PM⊥AB and PN⊥ AC, where M and N are respectively, points on rays AB and AC. Measure PM and PN. Are the two lengths equal?
Answer: Construct angle BAC and draw its bisector ray AX. Select any point P on this bisector. From P, drop perpendiculars to both AB and AC, creating points M and N where these perpendiculars meet the rays. Using a ruler, measure the lengths PM and PN. You will discover that PM and PN have the same length.
In simple words: Any point lying on the angle bisector is equally distant from both sides of the angle. This is why the two perpendicular segments have equal length.

Exam Tip: This demonstrates a fundamental property of angle bisectors - always check that your perpendiculars are marked with right angle symbols and measure carefully to show PM = PN as evidence of this property.

Exercise 19.3

Question 1. Draw a line segment of length 8.6 cm. Bisect it and measure the length of each part.
Answer: Construct a line segment AB measuring 8.6 cm using your ruler. With A as the center and a radius greater than half of AB, draw arcs on both sides of the segment. Using the same radius and B as the center, draw arcs that intersect the previous arcs at points E and F. Connect E and F with a line; this line crosses AB at point C, which is the midpoint. When you measure AC and BC with your ruler, you find that both equal 4.3 cm.
In simple words: To cut a line in half, use the compass and straightedge method to find the exact middle point, then measure each half to confirm they are equal.

Exam Tip: Always ensure the arc radius is greater than half the segment length, and mark the intersection points E and F clearly before drawing the perpendicular bisector.

Question 2. Draw a line segment AB of length 5.8 cm. Draw the perpendicular bisector of this line segment.
Answer: Using your ruler, draw line segment AB measuring 5.8 cm. Set your compass to a radius larger than the midpoint of AB. With A as the center, mark arcs on both the top and bottom sides of the segment. Without changing the compass width, place the compass point at B and draw arcs that cross the earlier arcs at two points, L and M. Now connect L to M with a straight line. This line LM is the perpendicular bisector of AB - it crosses AB at a right angle and divides it into two equal parts.
In simple words: A perpendicular bisector is a line that cuts a segment exactly in half and meets it at a 90° angle. You can construct it using compass arcs.

Exam Tip: Mark the right angle symbol where LM crosses AB, and label the intersection point clearly to show that the perpendicular bisector passes through the midpoint.

Question 3. Draw a circle with centre at point O and radius 5 cm. Draw its chord AB, draw the perpendicular bisector of line segment Ab. Does it pass through the centre of the circle?
Answer: Draw a point O and construct a circle with center O and radius 5 cm. Select two arbitrary points A and B on the circle's circumference and connect them to form chord AB. To find the perpendicular bisector of AB, use the compass method: with A as center and a radius exceeding half of AB, draw arcs on both sides. Using the same radius and B as center, draw arcs that intersect the previous ones at points E and F. Draw the line through E and F. This perpendicular bisector passes through the center O of the circle.
In simple words: The perpendicular bisector of any chord in a circle always goes through the circle's center. This is a key property of circles.

Exam Tip: Verify that the line EF crosses AB at right angles (mark the right angle) and passes through center O to demonstrate this circle property clearly.

Question 4. Draw a circle with centre at point O. Draw its two chords AB and CD such that AB is not parallel to CD. Draw the perpendicular bisector of AB and CD. At what point do they intersect?
Answer: Construct a circle centered at O. Draw two non-parallel chords, AB and CD. For chord AB, construct its perpendicular bisector using the standard compass technique: with A and B as centers and equal radii (greater than half AB), draw intersecting arcs to locate two points, then draw the line through them. Repeat this process for chord CD with centers C and D. When you draw both perpendicular bisectors, they will intersect at point O, the center of the circle.
In simple words: The perpendicular bisectors of any two chords in a circle always meet at the circle's center. This is true regardless of whether the chords are parallel or not.

Exam Tip: Mark the perpendicular bisectors clearly with dashed lines and show that they meet precisely at O. This reinforces the fundamental property that all perpendicular bisectors of a circle's chords pass through its center.

Question 5. Draw a line segment of length 10 cm and bisect it. Further bisect one of the equal parts and measure its length.
Answer: Create a line segment AB with a length of 10 cm. Using the compass bisection method - with A and B as centers and equal radii exceeding half of AB - construct the perpendicular bisector to find the midpoint C. This divides AB into two 5 cm parts. Now bisect one of these parts, say AC, using the same technique. With A and C as centers, draw intersecting arcs to find the midpoint D of AC. When you measure AD with your ruler, it equals 2.5 cm.
In simple words: Bisecting a line means cutting it in half. If you bisect one of those halves, you get a quarter of the original. So 10 cm divided in half is 5 cm, and half of that is 2.5 cm.

Exam Tip: Show each bisection step clearly with labeled points and verify measurements using a ruler. Demonstrate that the final segment is exactly one-quarter of the original.

Question 6. Draw a line segment AB and bisect it. Bisect one of the equal parts to obtain a line segment of length 1/2(AB).
Answer: Draw line segment AB using your ruler. Locate its midpoint C by using the standard compass bisection method - create arcs centered at A and B with equal radii, then join the intersection points. This produces two segments AC and CB, each measuring 1/2(AB). To get a segment of length 1/2(AB), bisect AC using the same process. Create arcs centered at A and C with equal radii to find point D, the midpoint of AC. The segment AD now has length 1/4(AB). Now bisect DC the same way to find point E, making DE equal to 1/8(AB). The segment AC itself equals 1/2(AB), so you have achieved the target.
In simple words: To get 1/2(AB), simply bisect the original segment once to create two equal halves. Each half is 1/2(AB).

Exam Tip: Be careful with notation - 1/2(AB) means half the original length, not 1/4. Use clear labels and measurements to show each bisection step and confirm final lengths match the required fractions.

Question 7. Draw a line segment AB and by ruler and compasses, obtain a line segment of length 3/4(AB).
Answer: Using your ruler, draw segment AB. Bisect it using compass arcs centered at A and B to locate the midpoint C. This gives AC = CB = 1/2(AB). Next, bisect segment AC by creating arcs centered at A and C; let D be the intersection point, so AD = 1/4(AB). Now, the segment DB consists of DC + CB. Since DC = 1/4(AB) and CB = 1/2(AB), you have DB = 1/4(AB) + 1/2(AB) = 3/4(AB). Measure DB with your ruler to verify.
In simple words: To build 3/4 of a segment, first find the midpoint (1/2), then find a quarter (1/4), and combine them: 1/4 + 1/2 = 3/4.

Exam Tip: Label all intermediate points (C and D) clearly and show the arithmetic breakdown: DB = 1/4(AB) + 1/2(AB) = 3/4(AB) to demonstrate the construction logic.

Exercise 19.4

Question 1. Construct the following angles with the help of protractor: 45°, 67°, 38°, 110°, 179°, 98°, 84°.
Answer: For each angle, follow this method: Draw a ray OA. Position your protractor so its center aligns with point O and its baseline sits along ray OA. Locate the desired angle measurement on the protractor scale (counting from the right side for standard positioning) and mark a point B at that mark. Remove the protractor and connect O to B with a straight line. Ray OB, together with ray OA, forms the required angle. Repeat this process for all seven angles: 45°, 67°, 38°, 110°, 179°, 98°, and 84°, creating a separate diagram for each.
In simple words: To draw any angle with a protractor, start with a ray, line up the protractor center at the ray's starting point, find your angle on the scale, mark the spot, and connect back to create the angle.

Exam Tip: Always ensure the protractor's center point is exactly at vertex O and the baseline aligns with ray OA. Use both angle scales on the protractor correctly - if measuring from the right, read the outer scale; if from the left, use the inner scale.

Question 2. Draw two rays PQ and RS as shown in figure. Using the protractor, construct angles of 15° and 138° with one arm PQ and RS respectively.
Answer: To create a 15° angle with arm PQ, position the protractor so its center aligns with point P and the baseline matches PQ. Mark point B at the 15° mark on the protractor. Remove the protractor and join P to B; angle QPB equals 15°. For the 138° angle with arm RS, place the protractor's center at point R with its baseline along RS. Mark point T at the 138° position. Take away the protractor and draw line RT; angle SRT equals 138°.

Exam Tip: Always ensure the protractor's center point coincides exactly with the vertex and the baseline aligns perfectly with one arm - misalignment leads to incorrect angle measures.

Exercise 19.5

Question 1. Draw an angle and label it as ∠BAC. Construct another angle, equal to ∠BAC
Answer: Start by drawing angle ∠BAC and a separate ray OP. Using a suitable radius with A as the center, draw an arc that intersects sides AB and AC at points X and Y respectively. Using the same radius and O as the center, create an arc that meets ray OP at point M. Measure the distance XY with a compass. With M as center and radius equal to XY, draw another arc to intersect the arc from O at point N. Join O and N, extending the line to Q. The angle ∠POQ is the required angle equal to ∠BAC.
In simple words: Copy the angle by taking equal arc measurements. The arc distance between the two points where the sides meet the arc stays the same in both angles.

Exam Tip: Maintain consistent compass radius throughout the construction - any variation will make the copied angle unequal to the original.

Question 2. Draw an obtuse angle. Bisect it. Measure each of the angle obtained.
Answer: An obtuse angle is one that measures more than 90° but less than 180°. Draw an obtuse angle ∠BAC on your paper. With center at A and a suitable radius, draw an arc intersecting sides AB and AC at points P and Q respectively. With center P and a radius exceeding half of PQ, sketch an arc. Using the same radius and Q as center, draw another arc intersecting the previous arc at R. Join A and R, extending it to X. The ray AX bisects ∠BAC into two equal angles. Measure both resulting angles using a protractor to verify they are equal and sum to the original obtuse angle.

Exam Tip: The bisecting ray should pass exactly through the intersection point of the two arcs - any deviation makes the angles unequal.

Question 3. Using protractor, draw an angle of measure 108°. With this angle as given, draw an angle of 54°.
Answer: Draw a ray OA using a ruler. Using a protractor, construct angle ∠AOB measuring 108° at point O. Since 108° ÷ 2 = 54°, bisecting the 108° angle will yield 54°. With O as center and a suitable radius, draw an arc cutting rays OA and OB at points P and Q respectively. With center P and radius exceeding half of PQ, draw an arc. Using the same radius and Q as center, draw another arc meeting the first arc at R. Connect O and R, extending it to X. The angle ∠AOX measures exactly 54°.
In simple words: The angle of 54° is exactly half of 108°. By finding the middle line of the bigger angle, you get the smaller angle.

Exam Tip: Double-check that 108° divides evenly to 54° - this relationship must be clear before you begin the bisection.

Question 4. Using protractor, draw a right angle. Bisect it to get an angle of measure 45°.
Answer: A right angle measures 90°. Draw ray OA using a ruler. Using a protractor, construct ∠AOB = 90° at O. With O as center and suitable radius, draw an arc meeting sides OA and OB at points P and Q respectively. With center P and radius more than half of PQ, draw an arc. Using the same radius and Q as center, draw another arc intersecting the first arc at R. Join O and R, extending it to X. The angle ∠AOX now measures 45°, which is half of 90°. You can verify: ∠AOB = 90°, ∠AOX = 45°, ∠XOB = 45°.
In simple words: Dividing a right angle in half gives you two 45° angles. The line that splits it equally is called the bisector.

Exam Tip: Always verify both resulting angles with a protractor to confirm each measures 45° and they sum to 90°.

Question 5. Draw a linear pair of angles. Bisect each of the two angles. Verify that the two bisecting rays are perpendicular to each other.
Answer: A linear pair consists of two adjacent supplementary angles. Draw line AB and mark point O on it. Draw any ray OC from O, creating angles ∠AOC and ∠COB that together form 180°. Bisect ∠AOC by drawing ray OX with compass and ruler. Bisect ∠COB by drawing ray OY similarly. Now examine the angle ∠XOY formed between the bisecting rays. The sum ∠XOY = (1/2)∠AOC + (1/2)∠COB = (1/2)(∠AOC + ∠COB) = (1/2) × 180° = 90°. Since ∠XOY = 90°, the bisecting rays OX and OY are perpendicular to each other.
In simple words: When you split two angles that add to 180° equally, the two splitting lines always meet at a right angle.

Exam Tip: Use a protractor or set square to verify the 90° angle between the bisectors - this confirms the perpendicular relationship.

Question 6. Draw a pair of vertically opposite angles. Bisect each of the two angles. Verify that the bisecting rays are in the same line.
Answer: Draw two intersecting lines AB and CD at point O. This creates two pairs of vertically opposite angles. Vertically opposite angles are equal. Let us take ∠AOC and ∠BOD. Bisect ∠AOC by drawing the bisecting ray OX using compass and ruler. Bisect ∠BOD by drawing the bisecting ray OY similarly. Calculate: ∠XOA + ∠AOD + ∠DOY = (1/2)∠AOC + ∠AOD + (1/2)∠BOD. Since ∠AOC = ∠BOD (vertically opposite), we have ∠XOA + ∠AOD + ∠DOY = (1/2)∠AOD + ∠AOD + (1/2)∠AOD = ∠AOD + ∠AOD = 180°. This shows that points X, O, and Y are collinear, meaning OX and OY lie on the same straight line.
In simple words: When you bisect two opposite angles formed by intersecting lines, the two bisectors always form one straight line.

Exam Tip: Use a ruler to verify that the two bisecting rays align perfectly - they should form a single continuous line through O.

Question 7. Using ruler and compass only, draw a right angle.
Answer: Draw ray OA using a ruler. With center O and suitable radius, draw arc PQ intersecting ray OA at point P. With center P and radius exceeding half of PQ, draw an arc. Using the same radius and center Q, draw another arc cutting the first arc at point R. With centers C and R (where C is opposite P on arc PQ), draw two arcs of radius greater than half of CR that intersect at S. Join O and S, extending to B. The angle ∠AOB is the required right angle measuring 90°.
In simple words: Using equal compass distances and intersecting arcs, you can find a point that forms a perfect 90° angle with the original ray.

Exam Tip: Ensure the second pair of arcs intersects cleanly - this guarantees the perpendicularity of the final angle.

Question 8. Using ruler and compass only, draw an angle of measure 135°.
Answer: Draw line AB with point O on it. Using compass and ruler, construct a perpendicular to AB at O (as in Question 7), creating a 90° angle. Now bisect one of the 90° angles on line AB. Using compass with suitable radius and center O, draw arc PQ intersecting line AB at P and the perpendicular at Q. With centers P and Q and radius exceeding half of PQ, draw two arcs meeting at R. Join O and R. Angle ∠AOR = 45°. Therefore, angle ∠BOR = 90° + 45° = 135°, which is the required angle.
In simple words: Start with a right angle of 90°. Add half of another right angle (which is 45°) to get 135°.

Exam Tip: Verify your construction by measuring with a protractor - the angle must be exactly 135°, not approximately.

Question 9. Using a protractor, draw an angle of measure 72°. With this angle as given, draw angles of measure 36° and 54°.
Answer: Draw ray OA and use a protractor to construct ∠AOB = 72° at O. To create the 36° angle: Since 36° = 72° ÷ 2, bisect ∠AOB. With O as center and suitable radius, draw arc cutting OA and OB at P and Q respectively. With centers P and Q and radius more than half of PQ, draw two arcs intersecting at R. Join O and R extending to X. The angle ∠AOX = 36°. To create the 54° angle: Bisect the angle ∠XOB (which is 36°) to obtain 18°. Then ∠AOY = ∠AOX + ∠XOY = 36° + 18° = 54°, which is the required angle. Alternatively, note that 72° ÷ 2 = 36° and 36° + 18° = 54°.
In simple words: Start with 72°. Bisect it to get 36°. Then add half of 36° (which is 18°) to the first 36° to reach 54°.

Exam Tip: Mark and label all bisection points clearly - this prevents confusion when combining angles to create the final required angles.

Exercise 19.6

Question 1. Construct an angle of 60° with the help of compasses and bisect it by paper folding.
Answer: Draw ray OA using a ruler. With center O and suitable radius, mark arc cutting ray OA at P. Using the same radius and center P, mark another arc crossing the first arc at Q. Join O and Q, extending to B. Angle ∠AOB = 60° (formed by two equal radii). Now cut out the sector OPQ as a paper piece. Fold the paper so that line segments OP and OQ coincide. The crease formed at the fold line passes through O and marks the bisector of ∠AOB. The angle formed at O after folding equals 30°, confirming that the bisector divides the 60° angle into two equal 30° parts.
In simple words: Make a 60° angle with compass. Then fold the paper to bring the two sides together - the fold line is your bisector.

Exam Tip: Fold precisely so that both rays align exactly - an imprecise fold will not create an accurate bisector of the original angle.

Question 2. Construct the following angles with the help of ruler and compasses only.
(I) 30°
(II) 90°
Answer:
(I) 30°: Draw ray OA. With center O and suitable radius, draw arc PQ meeting ray OA at P. With the same radius and center P, draw another arc meeting the first arc at Q. Join O and Q extending to B; ∠AOB = 60°. Now bisect ∠AOB: With center O and suitable radius, draw arc meeting OA and OB at M and N. With centers M and N and radius more than half of MN, draw two arcs intersecting at R. Join O and R extending to C. The angle ∠AOC = 30°.
(II) 90°: Draw ray OA. Using the method in Question 7 of Exercise 19.5, construct a perpendicular at O using ruler and compass. With center O and suitable radius, draw arc PQ on ray OA at P. With center P and radius exceeding half of PQ, draw an arc. Using the same radius and center Q, draw another arc meeting the first at R. Join O and R, extending to B. Angle ∠AOB = 90°.
In simple words: For 30°, first make 60° by repeating an equal radius, then cut it in half. For 90°, use intersecting arcs positioned symmetrically to create a perpendicular.

Exam Tip: Double-check both angles with a protractor after construction to ensure accuracy - measurement errors compound when creating derived angles.

Question 1. (iii) 45°
Answer: To build an angle of 45°, first construct a 90° angle, then bisect it. Begin by drawing ray OA. Using a suitable radius with center at O, draw an arc that intersects OA at point P. Using the same radius and center at P, draw another arc cutting the first arc at point Q. Taking P and Q as centers with a radius greater than half of PQ, draw two arcs that intersect at point R. Draw QR and extend it to reach point B. The angle AOB formed is the required 45° angle.
In simple words: Make a right angle first, then cut it in half. That gives you 45 degrees.

Exam Tip: Always verify that your bisecting arcs (from P and Q) intersect clearly; if they do not, increase the radius slightly and redraw them.

Question 1. (iv) 135°
Answer: Start by drawing a horizontal line AB with point O at its center. Using a suitable radius and center at O, draw an arc that cuts AB at points P and Q. Using the same radius and center at P, draw an arc intersecting the first arc at point R. With Q and R as centers and a radius exceeding half of QR, draw two arcs meeting at point S. Draw OS and extend it to form ray OD. This creates a 90° angle BOD. Now bisect angle BOD by using Q and R as centers with radius greater than half of QR, drawing two arcs that cross at point T. Draw OT and extend it to ray OE. Angle BOE is the required 135° angle.
In simple words: Make a 90° angle, then add half of another 90° (which is 45°) to it. That gives 135 degrees.

Exam Tip: When extending rays beyond the arc intersections, use a straightedge to ensure accuracy and mark the final angle clearly.

Question 1. (v) 150°
Answer: Draw a line segment AB and position O at its midpoint. Using a convenient radius with center at O, create an arc intersecting the line at points P and Q. Using the same radius and center at Q, draw an arc cutting the first arc at point R. With P and R as centers and radius exceeding half of PR, draw two arcs that meet at point S. Draw OS and extend it to point C. Angle BOS = 150°. To verify: angle AOS = 30°, so angle BOS = 180° − 30° = 150°.
In simple words: Build the 30° angle on one side, then measure 150° as the angle on the other side along the straight line.

Exam Tip: Always check that supplementary angles sum to 180° when both are formed on a straight line.

Question 1. (vi) 105°
Answer: Draw ray OA and create a 90° angle by constructing rays OB and OC such that angle AOB = 90°. Next, bisect the 90° angle to get a 45° ray OD, making angle AOD = 45°. Then bisect angle DOB to produce ray OE with angle DOE = 22.5°. Therefore, angle AOE = 45° + 22.5° = 67.5°. Finally, angle AOF = 90° + 15° = 105° can be built by adding a 15° angle (obtained by bisecting 30°) to the 90° angle. Alternatively, construct 90° and 60° angles, then bisect the difference to get 105°.
In simple words: Build a 90° angle and a 60° angle, then split the remaining space between them to make 105 degrees.

Exam Tip: When combining multiple angle constructions, draw each intermediate ray lightly in pencil first, then darken the final answer rays only.

Question 3. Construct a rectangle whose adjacent sides are 8 cm and 3 cm.
Answer: Start by drawing a line segment AB measuring 8 cm. At point A, use a compass and ruler to establish angle BAX = 90°; similarly, at point B, construct angle ABY = 90°. Using compass and ruler, mark point D on ray AX so that AD = 3 cm. On ray BY, mark point C such that BC = 3 cm. Finally, draw the line segment CD to complete the figure. Rectangle ABCD is formed with sides 8 cm and 3 cm as required.
In simple words: Make one long side 8 cm, then draw two perpendicular sides of 3 cm each from its ends, and join the endpoints with a fourth side.

Exam Tip: Verify that all four angles are 90° by checking that opposite sides are equal and parallel.