ML Aggarwal Class 6 Maths Solutions Chapter 13 Practical Geometry

Question 1. Construct a circle of radius:
(i) 2 cm
(ii) 3.5 cm
Answer: To draw a circle with radius 2 cm, first mark a point O as the centre on your paper. Open the compass and place its pointer at the zero mark of a ruler, then adjust it so the pencil end touches the 2 cm mark. Keep this opening fixed. Place the compass pointer at O and slowly rotate the pencil end around to create the circle.
For the second circle with radius 3.5 cm, follow the same steps: mark point O as the centre, set the compass width to 3.5 cm using the ruler, position the pointer at O, and rotate the pencil end to draw the circle.
In simple words: Measure your distance with the compass using a ruler. Put the sharp point at the centre and spin the pencil end around to draw the circle.

Exam Tip: Always use a ruler to set the compass width accurately - this ensures your circle has the exact radius needed. Hold the compass firmly at the knob to prevent slipping while drawing.

Question 2. With the same centre O, draw two circles of radii 2.6 cm and 4.1 cm.
Answer: Begin by marking a single point O on your paper, which will serve as the common centre for both circles. Use your compass and ruler to set the opening to 2.6 cm. With the pointer at O, rotate the pencil end slowly to create the first circle. Next, adjust the compass width to 4.1 cm without moving point O. Again, place the compass pointer at the same point O and rotate the pencil end to draw the second circle. Both circles now share the same centre but have different radii, making them concentric circles.
In simple words: Draw two circles at the same centre point. The inner circle has a smaller radius and the outer circle has a larger radius.

Exam Tip: Keep point O fixed for both circles - this is the key to drawing concentric circles correctly. Carefully reset your compass width between drawing each circle.

Question 3. Draw any circle and mark points A, B and C such that
(i) A is on the circle.
(ii) B is in the interior of the circle.
(iii) C is in the exterior of the circle.
Answer: First, select a point O on your paper to serve as the centre. Draw a circle using any suitable radius by setting your compass and rotating it around O. Once the circle is complete, mark point A directly on the curved line itself - this ensures A lies on the circle. Next, mark point B somewhere inside the circle, between the centre and the circle's edge - this places B in the interior region. Finally, mark point C outside the circle, away from the curved line - this puts C in the exterior region. You have now shown all three positions relative to a circle.
In simple words: Point A sits exactly on the circle's line. Point B is inside the circle. Point C is outside the circle.

Exam Tip: Make sure your points are clearly marked and easy to see. The difference between the three regions - on, inside, and outside - should be visually obvious in your diagram.

Question 4. Draw a circle and any two of its (non-perpendicular) diameters. If you join the ends of these diameters, what is the figure obtained? What figure is obtained if the diameters are perpendicular to each other?
Answer: Start by marking a point O and drawing a circle of any suitable size. Next, draw two diameters AC and BD that do not meet at right angles - they should cross at O but slant at an angle other than 90 degrees. Join the four endpoints A, B, C, and D in order around the circle. The shape you get is a rectangle. Now suppose the two diameters were perpendicular, meeting at exactly 90 degrees. If you were to draw perpendicular diameters and connect their four endpoints, the resulting shape would be a square instead.
In simple words: When two diameters are not perpendicular, connecting their ends makes a rectangle. When they are perpendicular, connecting their ends makes a square.

Exam Tip: Remember that both a rectangle and a square have four right angles and opposite sides equal. The key difference is that a square also has all four sides equal length.

Question 5. Let A, B be the centres of two circles of equal radii; draw them so that each one of them passes through the centre of the other. Let them intersect at C and D. Examine whether \( \overline{AB} \) and \( \overline{CD} \) are at right angles.
Answer: Draw a line segment AB of any suitable length. This segment will be the distance between the two circle centres. Now, with A as the centre and radius equal to the length AB, draw your first circle. Using the same radius (AB), draw a second circle with B as its centre. These two circles will intersect at two points; call these points C and D. Draw the line segment CD connecting these intersection points. When you measure the angle between line segments AB and CD where they meet, you will find they form a right angle of 90 degrees. This shows that AB and CD are perpendicular to each other.
In simple words: The line connecting the two circle centres and the line connecting the two intersection points cross each other at a perfect right angle.

Exam Tip: This is a key geometric property - when two equal circles pass through each other's centres, the line joining the centres is always perpendicular to the line joining the intersection points. Use a protractor to verify the 90-degree angle.

Question 6. Construct a line segment of length 6.3 cm using ruler and compass.
Answer: Draw a line l and mark a point A on it. Set your compass by placing its pointer end at the zero mark of a ruler and opening it so the pencil end reaches the 6.3 cm mark on the ruler - this gives your compass the exact width needed. Without changing this opening, place the compass pointer at point A and draw an arc that crosses line l at another point, which you can label B. The distance from A to B is now exactly 6.3 cm.
In simple words: Use the ruler to set the compass to 6.3 cm, then draw an arc from point A to mark point B. The segment AB is your required length.

Exam Tip: Be precise when setting the compass width using the ruler - even small errors will make your segment the wrong length. Do not move the compass opening after setting it.

Question 7. Construct \( \overline{AB} \) of length 8.3 cm. From this cut off \( \overline{AC} \) of length 5.6 cm. Measure the length of \( \overline{BC} \).
Answer: Draw a line l and mark a point A on it. Set your compass to 8.3 cm using a ruler, place the pointer at A, and draw an arc crossing line l at point B. Now you have a segment AB of length 8.3 cm. Next, adjust your compass to 5.6 cm. With the pointer still at A, draw another arc that cuts segment AB at point C. This creates AC of length 5.6 cm. Now measure the remaining segment BC with your ruler. Since B is 8.3 cm from A and C is 5.6 cm from A in the same direction, the distance BC equals 8.3 cm minus 5.6 cm, which gives 2.7 cm.
In simple words: Create a 8.3 cm segment, then mark a point 5.6 cm from one end. The leftover piece measures 2.7 cm.

Exam Tip: Always subtract the smaller measurement from the larger one to find the remaining segment: BC = AB - AC.

Question 8. Draw any line segment \( \overline{PQ} \). Without measuring \( \overline{PQ} \), construct a copy of \( \overline{PQ} \).
Answer: Begin by drawing any line segment PQ of unknown length. Set your compass by placing its pointer at P and opening it until the pencil end meets Q exactly - this compass opening now holds the length of PQ. Draw another line l and select a point R on it. Without changing your compass opening, place the pointer at R and draw an arc cutting line l at point S. The segment RS now has exactly the same length as PQ, making it a perfect copy.
In simple words: Use the compass to copy the length from PQ, then transfer it to a new line to create an identical segment RS.

Exam Tip: The compass "remembers" the length - do not adjust it between measuring PQ and drawing the arc at R. This technique is useful whenever you need to copy or transfer distances.

Question 9. Given some line segment \( \overline{AB} \), whose length you do not know, construct \( \overline{PQ} \) such that the length of \( \overline{PQ} \) is twice that of \( \overline{AB} \).
Answer: Draw a line segment AB of any length. Now draw another line l and mark a point P on it. Set your compass by placing its pointer at A and adjusting it so the pencil end touches B - this compass setting equals the length AB. Without changing this opening, place the compass pointer at P and draw an arc crossing line l at point R, making PR equal to AB. Keep the compass at the same width and place the pointer now at R. Draw another arc crossing line l at point Q, making RQ also equal to AB. Since you have placed two equal segments end-to-end, the total distance PQ equals PR plus RQ, which is AB plus AB, or twice AB.
In simple words: Take the compass width from AB. Mark off this length twice in a row on line l. The total distance is double the original.

Exam Tip: This method works by marking equal segments consecutively. For three times a length, mark three consecutive segments; for half a length, divide by bisecting.

Question 10. Take a line segment of length 10 cm. From \( \overline{PQ} \), cut off \( \overline{PA} \) of length 4.3 cm and \( \overline{BQ} \) of length 2.5 cm. Measure the length of segment \( \overline{AB} \).
Answer: Construct a line segment PQ of length 10 cm using ruler and compass. From point P, set your compass to 4.3 cm, place the pointer at P, and draw an arc crossing PQ at point A. This marks PA as 4.3 cm. From point Q, set your compass to 2.5 cm, place the pointer at Q, and draw an arc crossing PQ at point B. This marks BQ as 2.5 cm. Now measure the remaining segment AB. Since the total length PQ is 10 cm and we have removed 4.3 cm from P's end and 2.5 cm from Q's end, the length of AB equals 10 minus 4.3 minus 2.5, which is 3.2 cm.
In simple words: Start with a 10 cm line. Remove 4.3 cm from one end and 2.5 cm from the other end. What remains measures 3.2 cm.

Exam Tip: When removing segments from both ends, subtract both amounts from the total: AB = PQ - PA - BQ.

Question 11. Given two line segments \( \overline{AB} \) and \( \overline{CD} \) of lengths 7.5 cm and 4.6 cm respectively. Construct line segments.
(i) \( \overline{PQ} \) of length equal to the sum of the lengths of \( \overline{AB} \) and \( \overline{CD} \)
(ii) \( \overline{XY} \) of length equal to the difference of the lengths of \( \overline{AB} \) and \( \overline{CD} \)
Verify these lengths by measurements.
Answer:
(i) Construction of \( \overline{PQ} \) = \( \overline{AB} \) + \( \overline{CD} \):
Draw a line l and mark point P on it. Set your compass to the length of AB (7.5 cm) and place the pointer at P, drawing an arc to mark point R on line l, making PR equal to 7.5 cm. Now adjust your compass to the length of CD (4.6 cm). With the pointer at R, draw another arc crossing line l at point Q, making RQ equal to 4.6 cm. The total length PQ is now 7.5 plus 4.6, which equals 12.1 cm. Measure PQ with a ruler to verify this result.
(ii) Construction of \( \overline{XY} \) = \( \overline{AB} \) - \( \overline{CD} \):
Draw a line m and mark point Z on it. Set your compass to the length of AB (7.5 cm), place the pointer at Z, and draw an arc crossing line m at point Y, making ZY equal to 7.5 cm. Now adjust your compass to the length of CD (4.6 cm). With the pointer at Y, draw an arc going backwards along the line to mark point X, making YX equal to 4.6 cm. The segment XY now measures the difference: 7.5 minus 4.6, which equals 3.1 cm. Measure XY with a ruler to verify.
In simple words: To add two lengths, mark them one after the other on a line. To find the difference, mark the larger length first, then measure backwards from the end by the smaller length.

Exam Tip: Always verify your construction by measuring with a ruler - this confirms your compass work was accurate. For addition, segments are placed end-to-end. For subtraction, one segment is taken from inside the other.

Exercise 13.2

Question 1. Draw a line segment of length 5.6 cm. Draw a perpendicular to it from a point A outside by using ruler and compass.
Answer: Steps:
1. Draw a line segment PQ of length 5.6 cm.
2. Mark a point A outside the line segment PQ.
3. With A as centre and any suitable radius, draw an arc to cut PQ or the line containing PQ at points C and D.
4. With C and D as centres, draw two arcs of equal radius (greater than 1/2 CD) cutting each other at Q' on the other side of PQ.
5. Join A and Q'. Let AQ' intersect PQ at N.
Hence, AN is the required perpendicular from point A to the line segment PQ.
In simple words: Use arcs drawn from two points on the line to find where a perpendicular from an outside point should go. This perpendicular meets the line at a right angle.

Exam Tip: Ensure the radius in step 4 is clearly greater than half the distance CD - this guarantees the arcs will intersect properly on both sides.

Question 2. Draw a line segment AB of length 6.2 cm. Draw a perpendicular to it at a point M on AB by using ruler and compass.
Answer: Steps:
1. Draw a line segment AB of length 6.2 cm.
2. Mark any point M on AB.
3. With M as centre and any suitable radius, draw an arc to cut AB at points C and D.
4. With C and D as centres, draw two arcs of equal radius (greater than 1/2 CD) cutting each other at point Q.
5. Draw a line passing through points M and Q.
Hence, MQ is the required perpendicular to AB at the point M.
In simple words: Mark a point on the line, then use compass arcs from equal distances on either side to find the perpendicular direction at that point.

Exam Tip: The point M can be anywhere on AB - the method works identically whether M is at the midpoint or closer to either end.

Question 3. Draw a line l and take a point P on it. Through P, draw a line segment PQ perpendicular to l. Now draw a perpendicular to PQ at Q (use ruler and compass).
Answer: Steps:
1. Draw a line l and mark a point P on it.
2. With P as centre and any suitable radius, draw an arc to cut the line l at points C and D.
3. With C and D as centres, draw two arcs of equal radius (greater than 1/2 CD) cutting each other at point Q.
4. Join P and Q. Then PQ is the required line segment perpendicular to l at point P.
5. Produce PQ beyond Q to form a line PQ'.
6. With Q as centre and any suitable radius, draw an arc to cut the line PQ' at points E and F.
7. With E and F as centres, draw two arcs of equal radius (greater than 1/2 EF) on both sides cutting each other at points R and S.
8. Draw a line passing through points S, Q and R.
Hence, RS is the required perpendicular to PQ at point Q.
In simple words: First, construct a perpendicular from a point on a line. Then, from a point on that perpendicular, construct another perpendicular using the same method.

Exam Tip: This demonstrates that perpendiculars constructed using the same method on a prior perpendicular create lines parallel to the original line.

Question 4. Draw a line segment AB of length 6.4 cm and construct its axis of symmetry (use ruler and compass).
Answer: The axis of symmetry of a line segment is its perpendicular bisector.
Steps:
1. Draw a line segment AB of length 6.4 cm.
2. With A as centre and any suitable radius (greater than 1/2 AB), draw arcs on each side of AB.
3. With B as centre and the same radius, draw arcs on each side of AB to cut the previous arcs at C and D.
4. Draw line CD.
Hence, CD is the required axis of symmetry (perpendicular bisector) of line segment AB.
In simple words: The axis of symmetry is a line that divides a shape into two identical halves. For a line segment, this line cuts it in half and stands at a right angle to it.

Exam Tip: Label the points where the arcs meet clearly - these points lie on the perpendicular bisector, and joining them gives the axis of symmetry.

Question 5. Draw the perpendicular bisector of XY whose length is 8.3 cm. (i) Take any point P on the bisector drawn. Examine whether PX = PY. (ii) If M is mid-point of XY, what can you say about the lengths MX and MY?
Answer: Steps:
1. Draw a line segment XY of length 8.3 cm.
2. With X as centre and any suitable radius (greater than 1/2 XY), draw arcs on each side of XY.
3. With Y as centre and the same radius, draw arcs on each side of XY to cut the previous arcs at C and D.
4. Draw line CD. This line CD is the required perpendicular bisector of XY.
5. Mark any point P on the bisector CD.
6. Let CD meet XY at point M.
(i) On measuring PX and PY with the help of a ruler or divider, we find that PX = PY. Hence, yes, PX = PY.
(ii) Since M lies on the perpendicular bisector of XY and M is the mid-point of XY, therefore, MX = MY. Hence, the lengths MX and MY are equal.
In simple words: Any point on the perpendicular bisector of a line segment is the same distance from both endpoints. The midpoint divides the segment into two equal parts.

Exam Tip: This property - that all points on a perpendicular bisector are equidistant from the endpoints - is a key theorem used in many geometry problems and constructions.

Question 6. Draw a line segment of length 8.8 cm. Using ruler and compass, divide it into four equal parts. Verify by actual measurement.
Answer: Steps:
1. Draw a line segment AB of length 8.8 cm.
2. Construct the perpendicular bisector of AB. Let it meet AB at the point M. Then M is the mid-point of AB, so AM = MB = 4.4 cm.
3. Construct the perpendicular bisector of AM. Let it meet AM at the point N. Then N is the mid-point of AM, so AN = NM = 2.2 cm.
4. Construct the perpendicular bisector of MB. Let it meet MB at the point O. Then O is the mid-point of MB, so MO = OB = 2.2 cm.
Thus, the line segment AB is divided into four equal parts AN, NM, MO and OB. On measuring with a ruler, we find AN = NM = MO = OB = 2.2 cm. Hence, the line segment has been divided into four equal parts, each of length 2.2 cm.
In simple words: By finding the midpoint, then finding the midpoints of each half, you create four segments of equal size. Each segment is one quarter of the original length.

Exam Tip: Always verify your construction by measuring each part - this confirms the method worked and demonstrates understanding of the relationship between the segments.

Question 7. With PQ of length 5.6 cm as diameter, draw a circle.
Answer: Steps:
1. Draw a line segment PQ of length 5.6 cm.
2. Construct the perpendicular bisector of PQ. Let it meet PQ at point M. Then M is the mid-point of PQ.
3. With M as centre and radius equal to MP (= 1/2 PQ = 2.8 cm), draw a circle.
Hence, the required circle with PQ as diameter is drawn.
In simple words: The centre of a circle with a given diameter lies at the midpoint of that diameter. The radius is half the diameter length.

Exam Tip: Remember that the radius equals exactly half the diameter - so if the diameter is 5.6 cm, the radius must be 2.8 cm for the construction to be accurate.

Question 8. Draw a circle with centre C and radius 4.2 cm. Draw any chord AB. Construct the perpendicular bisector of AB and examine if it passes through C.
Answer: Steps:
1. Mark a point C on the sheet of paper.
2. With C as centre and radius 4.2 cm, draw a circle.
3. Mark any two points A and B on the circle and join them. AB is a chord of the circle.
4. With A as centre and any suitable radius (greater than 1/2 AB), draw arcs on each side of AB.
5. With B as centre and the same radius, draw arcs on each side of AB to cut the previous arcs at P and Q.
6. Draw a line passing through points P and Q.
On examining, we find that the perpendicular bisector PQ of chord AB passes through the centre C of the circle. Hence, yes, the perpendicular bisector of AB passes through C.
In simple words: The perpendicular bisector of any chord in a circle always goes through the centre of that circle.

Exam Tip: This is an important property of circles - use it to locate the centre of a circle if you know a chord, or to verify that a point is the true centre.

Question 9. Draw a circle of radius 3.5 cm. Draw any two of its (non-parallel) chords. Construct the perpendicular bisectors of these chords. Where do they meet?
Answer: Steps:
1. Mark a point O on the sheet of paper as the centre of the circle.
2. With O as centre and radius 3.5 cm, draw a circle.
3. Draw any two non-parallel chords AB and CD of the circle.
4. Construct the perpendicular bisector of chord AB. For this, take A and B as centres and equal radius (greater than 1/2 AB), draw arcs on both sides of AB, and join the points of intersection of the arcs.
5. Similarly, construct the perpendicular bisector of chord CD.
On examining, we find that the perpendicular bisectors of the two chords meet at the centre O of the circle.
In simple words: When you draw the perpendicular bisectors of any two chords in a circle, they always cross at the centre point of that circle.

Exam Tip: This property offers a practical method to find the centre of a circle - draw any two chords, construct their perpendicular bisectors, and their intersection point is the centre.

Exercise 13.3

Question 1. Draw an angle of 80° and make a copy of it using ruler and compass.
Answer: Follow these steps to draw and copy an 80° angle:

1. Start by drawing a ray OA.

2. Place the protractor's centre at O and align its zero line with OA. Mark a point at 80°.

3. Remove the protractor and connect O with the marked point, extending it to B. This gives ∠AOB = 80°.

4. To copy this angle, draw any ray PQ.

5. Using O as the centre and any suitable radius, draw an arc that crosses OA at C and OB at D.

6. Using P as the centre and the same radius, draw an arc that meets PQ at R.

7. Measure the distance CD with a compass.

8. With R as the centre and radius equal to CD, draw an arc to meet the previous arc at S.

9. Join PS and extend it to form ray PT. Then ∠QPT = ∠AOB = 80°.

Thus, ∠QPT is a copy of ∠AOB.
In simple words: Draw the first angle using a protractor. Then use a compass to copy the arc distances from the original angle onto a new angle, creating an exact copy.

Exam Tip: Ensure the compass radius remains unchanged from step 5 to step 6 - this is crucial for the copy to be accurate. Always verify that both angles measure 80°.

Question 2. Draw an angle of measure 127° and construct its bisector.
Answer: Follow these steps to draw a 127° angle and bisect it:

1. Start by drawing a ray OA.

2. Position the protractor's centre at O and its zero line along OA. Mark a point at 127°.

3. Remove the protractor and join O with the marked point, extending it to B. This creates ∠AOB = 127°.

4. Using O as the centre and any suitable radius, draw an arc that intersects OA at C and OB at D.

5. Using C as the centre and any suitable radius (greater than \( \frac{1}{2}CD \)), draw an arc. Also, using D as the centre with the same radius, draw another arc to meet the previous arc at E.

6. Join OE and extend it to form a ray.

Ray OE is the required bisector of ∠AOB, and ∠AOE = ∠EOB = 63.5°.
In simple words: Draw the angle with a protractor. Then use two equal compass arcs from points C and D to find point E, which lies on the angle's bisector.

Exam Tip: The radius used in step 5 must be more than half of CD to ensure the arcs intersect properly. Check that both resulting angles are equal at 63.5°.

Question 3. Draw ∠POQ = 64°. Also draw its line of symmetry.
Answer: The line of symmetry of an angle is its angle bisector. Follow these steps:

1. Draw a ray OP.

2. Place the protractor's centre at O and align its zero line with OP. Mark a point at 64°.

3. Remove the protractor and join O with the marked point, extending it to Q. This creates ∠POQ = 64°.

4. Using O as the centre and any suitable radius, draw an arc that meets OP at C and OQ at D.

5. Using C as the centre and any suitable radius (greater than \( \frac{1}{2}CD \)), draw an arc. Also, using D as the centre with the same radius, draw another arc to meet the previous arc at E.

6. Join OE and extend it to form a ray.

Ray OE is the line of symmetry of ∠POQ, and ∠POE = ∠EOQ = 32°.
In simple words: Draw the angle using a protractor. Then find the bisector by creating two equal arcs from the points where the first arc meets the angle's sides.

Exam Tip: The line of symmetry divides the angle into two perfectly equal parts. Always verify that both angles are exactly 32°.

Question 4. Draw a right angle and construct its bisector.
Answer: Follow these steps to draw a right angle and find its bisector:

1. Draw any line l and select a point O on it.

2. Using O as the centre and a suitable radius, draw an arc that cuts line l at points A and B.

3. Using A and B as centres, draw two arcs with equal radius (greater than \( \frac{1}{2}AB \)) that meet each other at C. Join OC and extend it to form ray OC. Then ∠BOC = 90°.

4. To bisect ∠BOC, use O as the centre and any suitable radius to draw an arc meeting OB at P and OC at Q.

5. Using P and Q as centres with equal radii (greater than \( \frac{1}{2}PQ \)), draw two arcs that meet each other at D.

6. Join OD and extend it to form a ray.

Ray OD is the bisector of the right angle ∠BOC, and ∠BOD = ∠DOC = 45°.
In simple words: First, build a right angle using perpendicular arcs from points on a line. Then bisect this 90° angle using the same arc method to create two 45° angles.

Exam Tip: The right angle must be exactly 90° before bisecting. Check that the two resulting angles from bisection are each precisely 45°.

Question 5. Draw an angle of 152° and divide it into four equal parts.
Answer: Follow these steps to draw a 152° angle and divide it into four equal parts:

1. Draw a ray OA.

2. Place the protractor's centre at O and its zero line along OA. Mark a point at 152°.

3. Remove the protractor and join O with the marked point, extending it to B. This creates ∠AOB = 152°.

4. Using O as the centre and any suitable radius, draw an arc that meets OA at C and OB at D.

5. Using C and D as centres, draw arcs with equal radius (greater than \( \frac{1}{2}CD \)) that meet each other at E. Join OE and extend it. OE bisects ∠AOB, so ∠AOE = ∠EOB = 76°. Let OE meet the arc at point F.

6. To bisect ∠AOE: using C and F as centres, draw arcs with equal radius that meet each other at G. Join OG and extend it.

7. Similarly, bisect ∠EOB: using F and D as centres, draw arcs with equal radius that meet each other at H. Join OH and extend it.

OG bisects ∠AOE and OH bisects ∠EOB. Therefore, ∠AOG = ∠GOE = ∠EOH = ∠HOB = 38°. Also: ∠AOE = ∠AOG + ∠GOE = 76° and ∠EOB = ∠EOH + ∠HOB = 76°.

The angle of 152° is divided into four equal parts, each measuring 38°.
In simple words: Bisect the large angle into two 76° angles. Then bisect each of those into two 38° angles, giving you four equal parts.

Exam Tip: Dividing into four parts requires two rounds of bisection. Verify all intermediate angles: 76° for the first bisection and 38° for each final part.

Question 6. Draw an angle of measure 45° and bisect it.
Answer: Follow these steps to draw a 45° angle and find its bisector:

1. Draw any line l and select a point O on it.

2. Using O as the centre and a suitable radius, draw two arcs that cut line l at points A and B.

3. Using A and B as centres, draw two arcs with equal radius (greater than \( \frac{1}{2}AB \)) that meet each other at C. Join OC and extend it. Then ∠BOC = 90°.

4. Bisect ∠BOC by drawing two equal arcs from points B and F, meeting each other at point D. Join OD - this is the bisector. Then ∠BOD = 45°.

5. To bisect ∠BOD, use O as the centre and any suitable radius to draw an arc meeting OB at P and OD at Q.

6. Using P and Q as centres with equal radii (greater than \( \frac{1}{2}PQ \)), draw two arcs that meet each other at E.

7. Join OE and extend it to form a ray.

Ray OE is the bisector of ∠BOD = 45°, and ∠BOE = ∠EOD = 22.5°.
In simple words: First make a right angle, then bisect it to get 45°, then bisect the 45° angle to get two 22.5° angles.

Exam Tip: Ensure step 3 creates exactly 90°, step 4 reduces it to 45°, and step 7 creates the final 22.5° bisector. Always double-check your angle measurements.

Exercise 13.4

Question 1. Construct a square of side 4 cm, using ruler and protractor.
Answer: Follow these steps to construct a 4 cm square:

1. Use a ruler to draw a line segment AB = 4 cm.

2. At A, use a protractor to draw ∠XAB = 90°. From AX, measure and cut off AD = 4 cm.

3. At B, use a protractor to draw ∠YBA = 90°. From BY, measure and cut off BC = 4 cm.

4. Join CD.

ABCD is the required square with side 4 cm.
In simple words: Draw the bottom side with a ruler. Use a protractor to make two 90° angles at the ends. Measure 4 cm up from each end and join the top corners.

Exam Tip: All four sides must be exactly 4 cm and all angles must be exactly 90°. Use the protractor carefully to ensure right angles at both A and B.

Question 2. Construct a rectangle of sides 6 cm and 3 cm, using ruler and compass.
Answer: Follow these steps to construct a rectangle with sides 6 cm and 3 cm:

1. Use a ruler to draw a line segment AB = 6 cm.

2. At A, construct a perpendicular AX to AB using a ruler and compass. From AX, measure and cut off AD = 3 cm.

3. At B, construct a perpendicular BY to AB using a ruler and compass. From BY, measure and cut off BC = 3 cm.

4. Join CD.

ABCD is the required rectangle with sides 6 cm and 3 cm.
In simple words: Draw a 6 cm base line. Build perpendicular lines 3 cm tall at each end. Connect the tops to complete the rectangle.

Exam Tip: The opposite sides must be equal - AB and CD should both be 6 cm, while AD and BC should both be 3 cm. All angles must be 90°.

Question 3. Using a ruler and protractor, construct a rectangle in which one of the diagonals divides the opposite angles into 50° and 40°.
Answer: Follow these steps to construct the rectangle:

1. Use a ruler to draw a line segment AB of any suitable length.

2. At B, use a protractor to draw ∠ABY = 90°.

3. At A, use a protractor to draw ∠BAX = 90°.

4. At A, use a protractor to construct ∠BAC = 50° such that AC lies inside the right angle ABY.

5. At A, use a protractor to draw ∠CAX = 40° (so that ∠BAC = 50°). Ray AX is perpendicular to AB.

6. From AX, cut off AD = BC using a compass (so that AD = BC and ABCD becomes a rectangle).

7. Join CD.

ABCD is the required rectangle in which diagonal AC divides the opposite angles into 50° and 40°.
In simple words: Draw a base line and two perpendiculars at its ends. Then use a protractor to mark how the diagonal should split the corner angle into 50° and 40°. Complete the rectangle by matching the side lengths.

Exam Tip: The diagonal AC must split angle A into exactly 50° and 40° parts. Verify that opposite angles are equal and the figure has right angles at all corners.

Question 4. Construct a toy house tent as given in the adjoining figure.
Answer: Follow these steps to construct the toy house tent:

1. Use a ruler to draw a line segment AB = 5 cm.

2. At A, construct a perpendicular AX to AB. From AX, cut off AD = 4 cm.

3. At B, construct a perpendicular BY to AB. From BY, cut off BC = 4 cm.

4. Join DC. ABCD is now a rectangle with length 5 cm and breadth 4 cm.

5. Using D as the centre and radius = 4 cm, draw an arc.

6. Using C as the centre and radius = 4 cm, draw another arc that meets the previous arc at point P (the apex of the tent roof).

7. Join DP and CP to complete the tent shape.

The tent is now complete, consisting of the rectangular base ABCD with a triangular roof formed by point P above the line DC.
In simple words: Build a rectangle with a 5 cm base and 4 cm sides. Then use two equal compass arcs from the top corners to find the peak point of the roof, and draw the roof sides.

Exam Tip: The two arcs in steps 5 and 6 must have equal radius (4 cm) to ensure the roof is symmetric. The apex point P should be equidistant from both D and C.

Question 5. Construct a square of side 5 cm, with a small circle of radius 1 cm inside it, having the same centre as the square.
Answer: Start by drawing a line segment AB that measures 5 cm. At point A, use your ruler and compass to draw a perpendicular line AX to AB. Mark off a length of 5 cm on this line and label the endpoint D. In the same way, at point B, construct a perpendicular line BY to AB and mark off 5 cm to get point C. Now join C to D. The shape ABCD is your square with sides of 5 cm. To find the centre, draw both diagonals AC and BD - they meet at a point O, which is the centre of the square. Using O as the centre and setting your compass to a radius of 1 cm, draw the circle. The result is a square ABCD with a small circle of radius 1 cm inside, both sharing the same centre O.
In simple words: Make a square with 5 cm sides. Find the middle point where the two diagonals cross. Draw a tiny circle (1 cm across) from that middle point.

Exam Tip: The centre of a square is always found where its two diagonals intersect. Check that both the square's sides and the circle's radius match the given measurements before finalizing your answer.

Question 6. Can you construct a square in the centre of a rectangle as shown in the adjoining figure?
Answer: Yes, we are able to construct a square in the centre of a rectangle. Start by drawing a line segment AB measuring 10 cm with a ruler. At point A, draw a perpendicular AX to AB using ruler and compass, then cut off 4 cm along AX to mark point D. Repeat this at point B - draw a perpendicular BY and mark off 4 cm to get point C. Connect C and D. Now ABCD forms your rectangle with length 10 cm and width 4 cm. Next, find the midpoint M of side AB by drawing its perpendicular bisector - this gives you AM = MB = 5 cm. From M, measure 2 cm toward A and label it P, then measure 2 cm toward B and label it Q. This makes PQ = 4 cm. At P, draw a vertical line that meets CD at point S. At Q, draw another vertical line that meets CD at point R. The shape PQRS is a square with side 4 cm positioned at the centre of rectangle ABCD.
In simple words: Make a rectangle first. Find the middle of the bottom side. Mark two spots that are 2 cm away from the middle on each side. Draw vertical lines from these spots to the top - you get a square in the middle.

Exam Tip: Remember that the square's side length will always equal the rectangle's shorter dimension. Use the perpendicular bisector method to accurately locate the middle point of the longer side.

Objective Type Questions - Mental Maths

Question 1. Fill in the blanks:
(i) A ruler is used to draw line ..... and to measure their .....
(ii) A divider is used to compare .....
(iii) A compass is used to draw circles or arcs of .....
(iv) A protractor is used to draw and measure .....
(v) The set squares are two triangular pieces having angles of ........ and .....
(vi) To bisect a line segment of length 7 cm, the opening of the compass should be more than .....
(vii) The perpendicular bisector of a line segment is also its line of .....
Answer:
(i) A ruler is used to draw line segments and to measure their lengths.
(ii) A divider is used to compare lengths of line segments.
(iii) A compass is used to draw circles or arcs of circles.
(iv) A protractor is used to draw and measure angles.
(v) The set squares are two triangular pieces having angles of 30°, 60°, 90° and 45°, 45°, 90°.
(vi) To bisect a line segment of length 7 cm, the opening of the compass should be more than 3.5 cm.
(vii) The perpendicular bisector of a line segment is also its line of symmetry.
In simple words: A ruler draws and measures straight lines. A divider compares how long two lines are. A compass draws round shapes. A protractor measures angles - the corners. Set squares have fixed angles. For bisecting, open your compass wider than half the segment.

Exam Tip: Learn the specific angles of each set square (30-60-90 and 45-45-90) and the compass opening rule: it must be more than half the segment length for accurate bisection.

Question 2. State whether the following statements are true (T) or false (F):
(i) There is only one set square in a geometry box.
(ii) An angle can be copied with the help of a ruler and compass.
(iii) The perpendicular bisector of a line segment can be drawn by paper folding.
(iv) A perpendicular to a line from a given point not on it can be drawn by paper folding.
(v) A 45° - 45° - 90° set square and a protractor have the same number of line(s) of symmetry.
Answer:
(i) False. There are two set squares in a geometry box - one with angles 30°, 60°, 90° and the other with angles 45°, 45°, 90°.
(ii) True. An angle can be copied with the help of a ruler and compass.
(iii) True. The perpendicular bisector of a line segment can be drawn by paper folding.
(iv) True. A perpendicular to a line from a given point not on it can be drawn by paper folding.
(v) True. Both a 45° - 45° - 90° set square and a protractor have 1 line of symmetry each.
In simple words: A geometry box has two different set squares, not one. You can copy angles and draw perpendicular lines using basic tools or folding paper. Both the 45-45-90 triangle and the protractor have exactly one line where they match if folded.

Exam Tip: Remember that paper folding is a valid method for geometric constructions in exams. Always check symmetry by visualizing a fold line - it should divide the shape into two matching halves.

Question 3. A circle of any radius can be constructed with the help of a:
(1) ruler
(2) divider
(3) compass
(4) protractor
Answer: (3) compass
In simple words: A compass is the tool that draws circles. You open it to the radius you want and spin it around a centre point.

Exam Tip: The compass is the only geometric tool designed specifically for drawing circles and arcs of any size.

Question 4. The instrument in a geometry box having the shape of a semicircle is:
(1) ruler
(2) divider
(3) compass
(4) protractor
Answer: (4) protractor
In simple words: The protractor is shaped like a half-circle and is used for measuring angles.

Exam Tip: Protractors come in semicircular (180°) and full circular (360°) versions. The question refers to the standard semicircular one.

Question 5. The instrument used to measure an angle is:
(1) ruler
(2) protractor
(3) divider
(4) compass
Answer: (2) protractor
In simple words: Use a protractor whenever you need to find out how many degrees an angle is.

Exam Tip: Make sure the protractor's baseline aligns with one ray of the angle and the centre hole sits exactly at the angle's vertex.

Question 6. Which of the following angles cannot be constructed using ruler and compass?
(1) 15°
(2) 45°
(3) 75°
(4) 85°
Answer: (4) 85°
In simple words: With a ruler and compass, you can make 60° and 90° angles. By splitting these angles in half over and over, you can make 15°, 30°, 45°, 75° and many others. But 85° cannot be made this way.

Exam Tip: Any angle made from halving or adding/subtracting combinations of 60° and 90° can be constructed. 85° doesn't fit this pattern, making it impossible to construct with standard tools.

Question 7. The number of perpendiculars that can be drawn to a line from a point not on it is:
(1) 1
(2) 2
(3) 4
(4) infinitely many
Answer: (1) 1
In simple words: From any point away from a line, you can draw only one straight line that meets the original line at a right angle.

Exam Tip: This is a fundamental geometry principle - perpendicularity is unique. Only one perpendicular exists from an external point to a given line.

Question 8. The number of perpendicular bisectors that can be drawn of a given line segment is:
(1) 0
(2) 1
(3) 2
(4) infinitely many
Answer: (2) 1
In simple words: Every line segment has exactly one perpendicular bisector - a line that cuts it in half at a right angle.

Exam Tip: The perpendicular bisector of a line segment is unique and passes through the segment's midpoint at 90 degrees.

Question 9. The number of lines of symmetry in a picture of a divider is:
(1) 0
(2) 1
(3) 2
(4) 4
Answer: (2) 1
In simple words: A divider looks like a V-shape with two legs. If you draw a line down the middle between the two legs, each half mirrors the other. That is one line of symmetry.

Exam Tip: Symmetry lines must divide a shape so that both sides are identical mirror images. A divider has just one such line running through its centre.

Question 10. The number of lines of symmetry in a picture of a compass is:
(1) 0
(2) 1
(3) 2
(4) none of these
Answer: (1) 0
In simple words: A compass has one end with a pencil and one end with a sharp pointer. These two ends are different from each other, so no line can divide it into matching halves.

Exam Tip: For a line of symmetry to exist, folding the shape along that line must produce two identical halves. A compass's asymmetric ends prevent this.

Question 11. The number of lines of symmetry in a ruler is:
(1) 0
(2) 1
(3) 2
(4) 4
Answer: (3) 2
In simple words: A ruler is a rectangle. It has one line of symmetry running the long way and another running the short way. If you fold it either way down the middle, both halves match perfectly.

Exam Tip: Rectangular shapes typically have two lines of symmetry - one along the length and one along the width - unless they are squares, which have four.

Question 12. The number of lines of symmetry in a 30° - 60° - 90° set square is:
(1) 0
(2) 1
(3) 2
(4) 3
Answer: (1) 0
In simple words: A 30-60-90 triangle has all three sides of different lengths. Because no two sides match, no fold line can make two identical halves. So it has zero lines of symmetry.

Exam Tip: A scalene triangle (all sides different) has no lines of symmetry. An isosceles triangle has one, and an equilateral triangle has three.

Question 13. The number of lines of symmetry in a protractor is:
(1) 0
(2) 1
(3) 2
(4) more than 2
Answer: (2) 1
In simple words: A protractor is shaped like a half-circle. It has one line of symmetry - a vertical line that runs through the centre and the 90° marking, dividing it into two matching halves.

Exam Tip: Semicircular protractors have exactly one vertical line of symmetry at the 90° mark. This is where the two halves are perfectly balanced.

Question 14. Statement I: We can construct angles measuring 15°, 30° and 60° using a ruler and a compass. Statement II: We can draw angle bisectors using a ruler and a compass.
(1) Statement I is true but Statement II is false.
(2) Statement I is false but Statement II is true.
(3) Both Statement I and Statement II are true.
(4) Both Statement I and Statement II are false.
Answer: (3) Both Statement I and Statement II are true
In simple words: You can make a 60° angle with compass and ruler. Split it in half and you get 30°. Split that in half and you get 15°. You can also split any angle in half - that is called bisecting. Both statements are correct.

Exam Tip: Statement I is true because 60° is constructible and halving it twice gives 30° then 15°. Statement II is true because angle bisectors are a standard ruler-and-compass construction method.

Question 15. Statement I: The smallest length that can be measured with the ruler in your geometry box is 0.001 m. Statement II: 1 cm = 100 mm
(1) Statement I is true but Statement II is false.
(2) Statement I is false but Statement II is true.
(3) Both Statement I and Statement II are true.
(4) Both Statement I and Statement II are false.
Answer: (1) Statement I is true but Statement II is false
In simple words: A ruler in a geometry box has markings in millimetres. The tiniest measurement you can make with it is 1 mm, which equals 0.001 m. However, 1 cm is equal to 10 mm, not 100 mm.

Exam Tip: Always recall the standard metric conversions - 1 cm = 10 mm and 1 m = 1000 mm. These facts are essential for evaluating statements about measurement.

Question 16. Statement I: On the lower reading scale of a protractor, as we move from left to right, the value of the angle decreases. Statement II: A protractor has two reading scales for measuring an angle.
(1) Statement I is true but Statement II is false.
(2) Statement I is false but Statement II is true.
(3) Both Statement I and Statement II are true.
(4) Both Statement I and Statement II are false.
Answer: (3) Both Statement I and Statement II are true
In simple words: Look at a protractor's lower scale - 0 degrees is on the right side and 180 degrees is on the left side. So as you move from left to right, the numbers get smaller. A protractor also has an inner scale and an outer scale to measure angles in both directions.

Exam Tip: When using a protractor, always identify which scale you need before reading an angle. Both scales are present for flexibility in measurement.

Question 17. Statement I: A protractor is a semicircle. Statement II: The angle between the zero line and the central line of a protractor is 90°.
(1) Statement I is true but Statement II is false.
(2) Statement I is false but Statement II is true.
(3) Both Statement I and Statement II are true.
(4) Both Statement I and Statement II are false.
Answer: (3) Both Statement I and Statement II are true
In simple words: A protractor has the shape of half a circle, which is called a semicircle. The middle line of a protractor is perpendicular (at a right angle) to the zero line, making the angle between them exactly 90 degrees.

Exam Tip: Remember that a protractor's design is based on semicircular geometry. The perpendicular relationship between the central and zero lines is a fundamental feature of its structure.

Question 18. Statement I: A circle is a closed curve such that its every point is at a fixed distance from the centre. Statement II: When two protractors are placed along their straight edges, the resulting figure is a circle.
(1) Statement I is true but Statement II is false.
(2) Statement I is false but Statement II is true.
(3) Both Statement I and Statement II are true.
(4) Both Statement I and Statement II are false.
Answer: (3) Both Statement I and Statement II are true
In simple words: A circle is a closed shape where every point on it is the same distance away from a center point - that distance is called the radius. Since each protractor is shaped like half a circle, placing two protractors side by side along their flat edges combines the two halves into one complete circle.

Exam Tip: Understand that a circle's definition is based on the constant distance property (radius). Recognize that two semicircles always combine to form a complete circle.

Check Your Progress

Question 1. Draw a line segment \( \overline{AB} \) of length 5.4 cm. Construct a perpendicular at A by using a ruler and compass.
Answer:
Steps:
1. Draw a line segment \( \overline{AB} \) of length 5.4 cm.
2. Extend the line BA beyond A to any convenient point. This makes A a point on a complete line.
3. Using A as the center and a suitable radius, draw an arc that crosses the extended line BA at point C and the segment AB at point D.
4. With C and D as centers, draw two arcs with the same radius that meet each other at point P. The radius must be greater than half of \( CD \).
5. Join points A and P, then extend this line further.

Therefore, AP is the required perpendicular to \( \overline{AB} \) at point A.
In simple words: You extend the line, mark equal distances from A on both sides, then use those points to find where a perpendicular line should be. That perpendicular line meets the original line at a right angle.

Exam Tip: The key to this construction is ensuring the arc radius from C and D is large enough to intersect on both sides of the line. This guarantees an accurate perpendicular.

Question 2. Draw a line segment \( \overline{PQ} \) of length 6.8 cm. Draw a perpendicular to it from a point A outside PQ by using a ruler and compass.
Answer:
Steps:
1. Draw a line segment \( \overline{PQ} \) of length 6.8 cm.
2. Mark a point A that lies outside the line segment \( \overline{PQ} \).
3. Using A as the center and a suitable radius, draw an arc that cuts the line containing \( \overline{PQ} \) at two points C and D.
4. With C and D as centers, draw two arcs with the same radius (greater than half of \( CD \)) that meet each other at point Q' on the opposite side of \( \overline{PQ} \).
5. Connect A to Q'. The line AQ' intersects \( \overline{PQ} \) at point N.

Therefore, AN is the required perpendicular from point A to the line segment \( \overline{PQ} \).
In simple words: From a point outside a line, you find two spots where an arc crosses the line, then use those spots to locate the perpendicular point. The perpendicular always meets the line at a right angle.

Exam Tip: When constructing a perpendicular from an external point, the arc from that point must be large enough to intersect the line at two distinct points. This is essential for accuracy.

Question 3. Draw a line segment \( \overline{AB} \) of length 6.5 cm and construct its axis of symmetry.
Answer: The axis of symmetry of a line segment is the same as its perpendicular bisector - a line that cuts the segment in half and meets it at a right angle.

Steps:
1. Draw a line segment \( \overline{AB} \) of length 6.5 cm.
2. Using A as the center and a suitable radius (greater than half of \( AB \)), draw arcs on both sides of the segment.
3. Using B as the center and the same radius, draw arcs on both sides of the segment to intersect the previous arcs at points C and D.
4. Draw a line connecting C and D.

Therefore, CD is the required axis of symmetry (perpendicular bisector) of line segment \( \overline{AB} \).
In simple words: The axis of symmetry is a line that splits a segment into two equal halves and crosses it at a 90-degree angle. It is also called the perpendicular bisector.

Exam Tip: Make sure the compass radius is set to more than half the segment's length. This ensures the arcs from both endpoints will intersect above and below the segment clearly.

Question 4. Draw \( \angle AOB = 76° \) with the help of a protractor. Bisect this angle by using a ruler and compass. Measure the two parts by your protractor and see how accurate you are.
Answer:
Steps:
1. Draw a ray OA.
2. Position the protractor so its center sits at O and the zero line aligns with ray OA. Mark a point at the 76° mark.
3. Remove the protractor and join O to the marked point, extending it to form ray OB. This gives you \( \angle AOB = 76° \).
4. Using O as the center and a suitable radius, draw an arc that meets ray OA at C and ray OB at D.
5. Using C as the center and a suitable radius (greater than half of \( CD \)), draw an arc. Also draw an arc with D as the center using the same radius. Let these arcs meet at E.
6. Join O to E and extend it to form ray OE.

Then OE is the bisector of \( \angle AOB \).

On measuring the two parts with a protractor, we find: \( \angle AOE = 38° \) and \( \angle EOB = 38° \).

Therefore, ray OE divides \( \angle AOB \) into two equal parts of 38° each.
In simple words: Draw the angle with a protractor, then use compass arcs from both sides of the angle to find the point where the bisector should go. The bisector splits the angle into two equal smaller angles.

Exam Tip: When bisecting an angle with a compass, the arc radius from points C and D must be more than half of CD to ensure proper intersection. Always measure the bisected parts to verify accuracy.

Question 5. By using a ruler and compass, construct an angle of 135° and bisect it. Measure any one part by protractor and see how accurate you are.
Answer:
Steps:
1. Draw a straight line AOA'.
2. Build a perpendicular OC to line AOA' at O. This creates \( \angle AOC = 90° \).
3. Bisect \( \angle A'OC \) by drawing two equal arcs from center O that cut line OA' at D and OC at E. From D and E, draw two arcs of equal length (greater than half of \( ED \)) that meet at B. Join O to B. OB is the bisector of \( \angle A'OC \), so \( \angle COB = 45° \).
4. Therefore, \( \angle AOB = \angle AOC + \angle COB = 90° + 45° = 135° \).
5. Now bisect \( \angle AOB \). With O as the center, draw two equal arcs that cut OA at P and OB at Q.
6. With P and Q as centers and equal radii (greater than half of \( PQ \)), draw two arcs that meet at R.
7. Join O to R and extend it to form ray OR.

Then OR is the bisector of \( \angle AOB \).

On measuring one of the parts with a protractor, we find: \( \angle AOR = 67.5° \) (and \( \angle ROB = 67.5° \)).

Therefore, ray OR divides \( \angle AOB \) into two equal parts of 67.5° each.
In simple words: To make a 135° angle, start with a 90° perpendicular and add a 45° angle (which is half of 90°). Then divide the 135° angle in half to get two 67.5° angles using the bisection method.

Exam Tip: Building a 135° angle requires combining two simpler angles (90° and 45°). When bisecting, always ensure compass arcs have sufficient radius to intersect clearly. Measurement verification confirms your construction accuracy.