Selina Concise Solutions for ICSE Class 10 Mathematics Chapter 19 Constructions Circles

Exercise 19

Question 1. Draw a circle of radius 3 cm. Mark a point P at a distance of 5 cm from the centre of the circle drawn. Draw two tangents PA and PB to the given circle and measure the length of each tangent.
Answer:
Steps of Construction:
1. Draw a circle with centre O and radius 3 cm.
2. From O, take a point P such that OP = 5 cm
3. Draw a bisector of OP which intersects OP at M.
4. With centre M, and radius OM, draw a circle which intersects the given circle at A and B.
5. Join AP and BP.
AP and BP are the required tangents.
On measuring AP = BP = 4 cm
In simple words: We draw a big circle around the midpoint of OP. This circle cuts our original circle at two points. Lines from P to these points are tangents.

[Diagram: Shows a circle with center O and radius 3 cm, point P at 5 cm distance, construction circle with center M, and two tangent lines PA and PB of length 4 cm each.]

📝 Teacher's Note: Show students that tangent lines from an outside point to a circle are always equal in length. Use a string and circle to demonstrate this fact.

🎯 Exam Tip: Always measure both tangents and write that they are equal. This proves your construction is correct. Show all construction steps clearly.

Question 2. Draw a circle of diameter of 9 cm. Mark a point at a distance of 7.5 cm from the centre of the circle. Draw tangents to the given circle from this exterior point. Measure the length of each tangent.
Answer:
1. Draw a circle of diameter 9 cm, taking O as the centre.
2. Mark a point P outside the circle, such that PO = 7.5 cm.
3. Taking OP as the diameter, draw a circle such that it cuts the earlier circle at A and B.
4. Join PA and PB.
Thus, PA and PB are required tangents. PA = PB = 6 cm
In simple words: We make a new circle with OP as diameter. Where this new circle cuts the original circle, we get our tangent points.

[Diagram: Shows construction with circle of diameter 9 cm, point P at 7.5 cm from center, auxiliary circle with OP as diameter, and tangents PA and PB of length 6 cm each.]

📝 Teacher's Note: Explain that diameter is twice the radius. So radius here is 4.5 cm. The construction uses the property that angle in semicircle is 90 degrees.

🎯 Exam Tip: Remember diameter = 2 × radius. Write radius = 4.5 cm clearly. Show that both tangents are equal length when measured.

Question 3. Draw a circle of radius 5 cm. Draw two tangents to this circle so that the angle between the tangents is 45°.
Answer:
Steps of Construction:
1. Draw a circle with centre O and radius BC = 5 cm
2. Draw arcs making an angle of 180° - 45° = 135° at O such that ∠AOB = 135°
3. At A and B, draw two rays making an angle of 90° at each point which meet each other at point P, outside the circle.
4. AP and BP are the required tangents which make an angle of 45° with each other at P.
In simple words: We find points A and B on the circle. Then we draw perpendicular lines at these points. These perpendicular lines meet at P and form our tangents.

[Diagram: Shows circle with center O, points A and B such that angle AOB = 135°, perpendicular rays at A and B meeting at P, forming 45° angle between tangents.]

📝 Teacher's Note: Teach the rule: if tangents make angle x, then the angle at center is 180° - x. This is a key property students must remember.

🎯 Exam Tip: Write the formula: Angle at center = 180° - angle between tangents. This shows you know the theory. Always draw perpendiculars at the tangent points.

Question 4. Draw a circle of radius 4.5 cm. Draw two tangents to this circle so that the angle between the tangents is 60°.
Answer:
Steps of Construction:
1. Draw a circle with centre O and radius BC = 4.5 cm
2. Draw arcs making an angle of 180° - 60° = 120° at O such that ∠AOB = 120°
3. At A and B, draw two rays making an angle of 90° at each point which meet each other at point P, outside the circle.
4. AP and BP are the required tangents which make an angle of 60° with each other at P.
In simple words: Same method as Question 3, but now we use 120° at the center because tangents make 60°.

[Diagram: Shows similar construction to Question 3 but with angle AOB = 120° and tangents making 60° angle at P.]

📝 Teacher's Note: This question reinforces the same concept as Question 3. Make students practice the formula with different angles.

🎯 Exam Tip: Calculate center angle first: 180° - 60° = 120°. Write this calculation clearly. Then follow the same steps as before.

Question 5. Using ruler and compasses only, draw an equilateral triangle of side 4.5 cm and draw its circumscribed circle. Measure the radius of the circle.
Answer:
Steps of construction:
1. Draw a line segment BC = 4.5 cm
2. With centers B and C, draw two arcs of radius 4.5 cm which intersect each other at A.
3. Join AC and AB.
4. Draw perpendicular bisectors of AC and BC intersecting each other at O.
5. With centre O, and radius OA or OB or OC draw a circle which will pass through A, B and C.
This is the required circumcircle of triangle ABC.
On measuring the radius OA = 2.6 cm
In simple words: We make an equilateral triangle first. Then we find the center by drawing perpendicular bisectors. The circumcircle passes through all three vertices.

[Diagram: Shows equilateral triangle ABC with circumscribed circle having center O and radius 2.6 cm.]

📝 Teacher's Note: Explain that circumcircle means the circle that goes through all three corners of the triangle. The center is where perpendicular bisectors meet.

🎯 Exam Tip: Draw perpendicular bisectors of any two sides - they meet at the circumcenter. Measure the radius carefully and write the value with units.

Question 6. Using ruler and compasses only. (i) Construct triangle ABC, having given BC = 7 cm, AB - AC = 1 cm and ∠ABC = 45°. (ii) Inscribe a circle in the △ABC constructed in (i) above. Measure its radius.
Answer:
Steps of Construction:
i) Construction of triangle:
• Draw a line segment BC = 7 cm
• At B, draw a ray BX making an angle of 45° and cut off BE = AB - AC = 1 cm
• Join EC and draw the perpendicular bisector of EC intersecting BX at A.
• Join AC.
△ABC is the required triangle.

ii) Construction of in circle:
• Draw angle bisectors of ∠ABC and ∠ACB intersecting each other at O.
• From O, draw perpendiculars OL to BC.
• O as centre and OL as radius draw circle which touches the sides of the △ABC.
This is the required in-circle of △ABC.
On measuring, radius OL = 1.8 cm
In simple words: We use a special method to construct the triangle when we know one side, one angle, and difference of other two sides. Then we draw the incircle using angle bisectors.

[Diagram: Shows triangle ABC constructed using the given conditions, with inscribed circle having center O and radius 1.8 cm.]

📝 Teacher's Note: This is a complex construction. Break it into steps. Show that incenter is where angle bisectors meet. The incircle touches all three sides.

🎯 Exam Tip: For incircle, always draw angle bisectors of any two angles. Where they meet is the incenter. Draw perpendicular from incenter to any side for the radius.

Question 7. Using ruler and compasses only, draw an equilateral triangle of side 5 cm. Draw its inscribed circle. Measure the radius of the circle.
Answer:
Steps of Construction:
1. Draw a line segment BC = 5 cm
2. With centers B and C, draw two arcs of 5 cm radius each which intersect each other at A.
3. Join AB and AC.
4. Draw angle bisectors of ∠B and ∠C intersecting each other at O.
5. From O, draw OL ⊥ BC.
6. Now with centre O and radius OL, draw a circle which will touch the sides of △ABC
On measuring, OL = 1.4 cm
In simple words: Make an equilateral triangle first. Find the incenter using angle bisectors. Draw the incircle that touches all three sides inside the triangle.

[Diagram: Shows equilateral triangle ABC with side 5 cm and inscribed circle with center O and radius 1.4 cm.]

📝 Teacher's Note: Compare this with Question 5. One circle goes through vertices (circumcircle), other touches sides (incircle). Make this difference very clear.

🎯 Exam Tip: For inscribed circle, use angle bisectors to find center. For circumscribed circle, use perpendicular bisectors. Don't mix these up in exams.

Question 8. Using ruler and compasses only,
(i) Construct a triangle ABC with the following data:
Base AB = 6 cm, BC = 6.2 cm and \( \angle CAB = 60° \)
(ii) In the same diagram, draw a circle which passes through the points A, B and C and mark its centre as O.
(iii) Draw a perpendicular from O to AB which meets AB in D.
(iv) Prove that AD = BD
Answer:
Steps of Construction:
1. Draw a line segment AB = 6 cm
2. At A, draw a ray making an angle of 60º with BC.
3. With B as centre and radius = 6.2 cm draw an arc which intersects AX ray at C.
4. Join BC.
△ABC is the required triangle.
5. Draw the perpendicular bisectors of AB and AC intersecting each other at O.
6. With centre O, and radius as OA or OB or OC, draw a circle which will pass through A, B and C.
7. From O, draw OD ⊥ AB.

Proof: In right △OAD and △OBD
OA = OB (radii of same circle)
Side OD = OD (common)
∴ △OAD ≅ △OBD (RHS)
⇒ AD = BD (CPCT)
 

[Diagram: This diagram shows a triangle ABC with a circumcircle. O is the circumcentre, and D is the foot of perpendicular from O to side AB.]

In simple words: We make a triangle first. Then we find the centre of a circle that goes through all three corners. When we drop a line straight down from this centre to the base, it cuts the base exactly in half.

📝 Teacher's Note: Show students that the circumcentre is equally far from all three vertices. When you drop a perpendicular from equal distances, it always lands in the middle. Use a simple example with three dots on paper.

🎯 Exam Tip: Always write "radii of same circle" and "RHS congruence" clearly. Then write CPCT to show AD = BD. These are the key words examiners look for.

Question 9. Using ruler and compasses only construct a triangle ABC in which BC = 4 cm, \( \angle ACB = 45° \) and perpendicular from A on BC is 2.5 cm. Draw a circle circumscribing the triangle ABC.
Answer:
Steps of Construction:
1. Draw a line segment BC = 4 cm.
2. At C, draw a perpendicular line CX and from it, cut off CE = 2.5 cm.
3. From E, draw another perpendicular line EY.
4. From C, draw a ray making an angle of 45º with CB, which intersects EY at A.
5. Join AB.
△ABC is the required triangle.
6. Draw perpendicular bisectors of sides AB and BC intersecting each other at O.
7. With centre O, and radius OB, draw a circle which will pass through A, B and C.
Measuring the radius OB = OC = OA = 2 cm
 

[Diagram: This diagram shows triangle ABC with the perpendicular from A to BC being 2.5 cm, and the circumcircle drawn around it.]

In simple words: We use the height of the triangle (2.5 cm) to find where point A should be. Then we draw a circle that touches all three corners of the triangle.

📝 Teacher's Note: Explain that the perpendicular distance helps us locate point A precisely. The height is like dropping a stone straight down from A to the base BC.

🎯 Exam Tip: Draw the perpendicular first to locate A correctly. Then find the circumcentre by drawing perpendicular bisectors of any two sides. Write all measurements clearly.

Question 10. Perpendicular bisectors of the sides AB and AC of a triangle ABC meet at O.
(i) What do you call the point O?
(ii) What is the relation between the distances OA, OB and OC?
(iii) Does the perpendicular bisector of BC pass through O?
Answer:
1. O is called the circumcentre of circumcircle of △ABC.
2. OA, OB and OC are the radii of the circumcircle.
3. Yes, the perpendicular bisector of BC will pass through O.
 

[Diagram: This diagram shows triangle ABC with perpendicular bisectors of all three sides meeting at point O, which is the circumcentre.]

In simple words: Point O is the centre of a circle that goes through all three corners of the triangle. It is equally far from all three corners. All three perpendicular bisectors meet at this point.

📝 Teacher's Note: Use three students standing in a triangle. Ask where you would stand to be equally far from all three. That spot is like the circumcentre. All students can relate to this.

🎯 Exam Tip: Write "circumcentre" and "radii are equal" clearly. Always mention that all three perpendicular bisectors meet at the same point O.

Question 11. The bisectors of angles A and B of a scalene triangle ABC meet at O.
i) What is the point O called?
ii) OR and OQ are drawn perpendiculars to AB and CA respectively. What is the relation between OR and OQ?
iii) What is the relation between angle ACO and angle BCO?
Answer:
1. O is called the in centre of the in circle of △ABC.
2. OR and OQ are the radii of the in circle and OR = OQ.
3. OC is the bisector of angle C
∴ \( \angle ACO = \angle BCO \)
 

[Diagram: This diagram shows triangle ABC with angle bisectors meeting at point O (incentre), and perpendiculars OR and OQ drawn to the sides.]

In simple words: Point O is the centre of a circle that fits inside the triangle and touches all three sides. This point is equally far from all three sides of the triangle.

📝 Teacher's Note: Show students how a coin placed inside a triangle touches all sides when rolled to the right spot. That spot is like the incentre. It helps them visualize easily.

🎯 Exam Tip: Write "incentre" and "equal perpendicular distances to sides" clearly. Remember that angle bisectors create equal angles on both sides.

Question 12.
i) Using ruler and compasses only, construct a triangle ABC in which AB = 8 cm, BC = 6 cm and CA = 5 cm.
ii) Find its in centre and mark it I.
iii) With I as centre, draw a circle which will cut off 2 cm chords from each side of the triangle.
Answer:
Steps of Construction:
1. Draw a line segment BC = 6 cm.
2. With centre B and radius 8 cm draw an arc.
3. With centre C and radius 5 cm draw another arc which intersects the first arc at A.
4. Join AB and AC.
△ABC is the required triangle.
5. Draw the angle bisectors of \( \angle B \) and \( \angle A \) intersecting each other at I. Then I is the in centre of the triangle ABC
6. Through I, draw ID ⊥ AB
7. Now from D, cut off \( DP = DQ = \frac{2}{2} = 1 \text{ cm} \)
8. With centre I, and radius IP or IQ, draw a circle which will intersect each side of triangle ABC cutting chords of 2 cm each.
 

[Diagram: This diagram shows triangle ABC with incentre I and a circle that cuts 2 cm chords from each side of the triangle.]

In simple words: First we make the triangle using the three given side lengths. Then we find the incentre. Finally we draw a special circle that cuts small 2 cm pieces from each side.

📝 Teacher's Note: Explain that the chord length is the straight line distance between two points where the circle cuts the side. Students often confuse this with the arc length.

🎯 Exam Tip: First construct the triangle, then find incentre using angle bisectors. The radius for cutting 2 cm chords needs careful calculation from the perpendicular distance.

Question 13. Construct an equilateral triangle ABC with side 6 cm. Draw a circle circumscribing the triangle ABC.
Answer:
Steps of Construction:
1. Draw a line segment BC = 6 cm
2. With centers B and C, draw two arcs of radius 6 cm which intersect each other at A.
3. Join AC and AB.
4. Draw perpendicular bisectors of AC, AB and BC intersecting each other at O.
5. With centre O, and radius OA or OB or OC draw a circle which will pass through A, B and C.
This is the required circumcircle of triangle ABC.
 

[Diagram: This diagram shows an equilateral triangle ABC with all sides 6 cm and the circumcircle drawn around it with centre O.]

In simple words: An equilateral triangle has all sides equal. We make it using compass arcs. Then we draw a circle that goes through all three corners of this triangle.

📝 Teacher's Note: Show students that in an equilateral triangle, the circumcentre is at the exact middle. It's like balancing the triangle on a pin at that point.

🎯 Exam Tip: Use compass to make equal arcs for the equilateral triangle. Then draw perpendicular bisectors to find circumcentre O. The circle passes through all three vertices.

Question 14. Construct a circle, inscribing an equilateral triangle with side 5.6 cm.
Answer:
Steps of Construction:
1. Draw a line segment BC = 5.6 cm
2. With centers B and C, draw two arcs of 5.6 cm radius each which intersect each other at A.
3. Join AB and AC.
4. Draw angle bisectors of \( \angle B \) and \( \angle C \) intersecting each other at O.
5. From O, draw OL ⊥ BC.
6. Now with centre O and radius OL, draw a circle which will touch the sides of △ABC
This is the required circle.
 

[Diagram: This diagram shows an equilateral triangle ABC with an inscribed circle (incircle) that touches all three sides of the triangle.]

In simple words: We make an equilateral triangle first. Then we draw a circle inside it that just touches all three sides. This circle fits perfectly inside the triangle.

📝 Teacher's Note: Explain that an inscribed circle touches the sides from inside, while a circumscribed circle goes through the vertices. Students often mix these up.

🎯 Exam Tip: Use angle bisectors to find the incentre for inscribed circle. The radius is the perpendicular distance from incentre to any side.

Question 15. Draw a circle circumscribing a regular hexagon of side 5 cm.
Answer:

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In simple words: A regular hexagon has 6 equal sides. We need to draw a circle that goes through all 6 corners of this hexagon.

📝 Teacher's Note: For a regular hexagon, the circumradius equals the side length. This is a special property that makes construction easier.

🎯 Exam Tip: Remember that in a regular hexagon, the distance from centre to any vertex equals the side length. Use this to draw the circumcircle.

Question 16. Draw an inscribing circle of a regular hexagon of side 5.8 cm.
Answer:
Steps of Construction:
1. Draw a line segment AB = 5.8 cm
2. At A and B, draw rays making an angle of 120° each and cut off AF = BC = 5.8 cm
3. Again F and C, draw rays making an angle of 120° each and cut off FE = CD = 5.8 cm.
4. Join DE. Then ABCDEF is the regular hexagon.
5. Draw the bisectors of ∠A and ∠B intersecting each other at O.
6. From O, draw OL ⊥ AB
7. With centre O and radius OL, draw a circle which touches the sides of the hexagon.
This is the required in circle of the hexagon.
[Diagram: This diagram shows a regular hexagon ABCDEF with side 5.8 cm and angle bisectors meeting at center O, with an inscribed circle touching all sides of the hexagon.]
In simple words: We make a six-sided shape first. Then we find the center by drawing angle bisectors. The inscribed circle goes inside and touches all sides of the hexagon.

📝 Teacher's Note: Show students that an inscribed circle touches all sides from inside. It is different from circumscribed circle which goes around the shape from outside.

🎯 Exam Tip: Always draw angle bisectors to find the center. Write "inscribed circle" clearly. Show that the circle touches all sides of the hexagon.

Question 17. Construct a regular hexagon of side 4 cm. Construct a circle circumscribing the hexagon.
Answer:
Steps of Construction:
1. Draw a circle of radius 4 cm with centre O
2. Since the interior angle of regular hexagon is 60°, draw radii OA and OB such that ∠AOB = 60°
3. Cut off arcs BC, CD, EF and each equal to arc AB on given circle
4. Join AB, BC, CD, DE, EF, FA to get required regular hexagon ABCDEF in a given circle.
The circle is the required circumcircle, circumscribing the hexagon.
[Diagram: This diagram shows a circle with radius 4 cm containing a regular hexagon ABCDEF, where all vertices of the hexagon touch the circle.]
In simple words: We draw a circle first. Then we divide it into 6 equal parts using 60° angles. We join these points to make a hexagon. The circle goes around the hexagon from outside.

📝 Teacher's Note: Explain that in a regular hexagon, each central angle is 60°. This is because 360° ÷ 6 = 60°. Students often forget this basic division.

🎯 Exam Tip: Start with the circle first, not the hexagon. Use 60° angles to mark points on the circle. Write "circumscribing circle" clearly.

Question 18. Draw a circle of radius 3.5 cm. Mark a point P outside the circle at a distance of 6 cm from the centre. Construct two tangents from P to the given circle. Measure and write down the length of one tangent.
Answer:
Steps of Construction:
1. Draw a line segment OP = 6 cm
2. With centre O and radius 3.5 cm, draw a circle
3. Draw the midpoint of OP
4. With centre M and diameter OP, draw a circle which intersect the circle at T and S
5. Join PT and PS.
PT and PS are the required tangents. On measuring the length of PT = PS = 4.8 cm
[Diagram: This diagram shows a circle with center O and radius 3.5 cm, point P outside at distance 6 cm, and two tangent lines PT and PS touching the circle at points T and S.]
In simple words: We draw a circle and mark a point outside it. Then we use a special method with another circle to find where the tangent lines touch. Both tangent lines have the same length.

📝 Teacher's Note: Remind students that tangent lines from an external point are always equal in length. This is a very important property they must remember.

🎯 Exam Tip: Always measure and write the length of tangents. Both tangents will be equal. Write the construction steps clearly in order.

Question 19. Construct a triangle ABC in which base BC = 5.5 cm, AB = 6 cm and m ∠ABC =120°.
(i) Construct a circle circumscribing the triangle ABC.
(ii) Draw a cyclic quadrilateral ABCD so that D is equidistant from B and C.
Answer:
Steps of Construction:
(i)
a. Draw a line BC = 5.4 cm.
b. Draw AB = 6 cm, such that m ∠ABC = 120°.
c. Construct the perpendicular bisectors of AB and BC, such that they intersect at O.
d. Draw a circle with O as the radius.
(ii)
(e) Extend the perpendicular bisector of BC, such that it intersects the circle at D.
(f) Join BD and CD.
(g) Here BD = DC.
[Diagram: This diagram shows triangle ABC with circumscribed circle, and point D on the circle such that ABCD forms a cyclic quadrilateral with BD = DC.]
In simple words: We make a triangle first. Then we find the center by drawing perpendicular bisectors. The circumscribed circle goes around the triangle. Point D is placed so it is equal distance from B and C.

📝 Teacher's Note: Show students that perpendicular bisectors meet at the circumcenter. A cyclic quadrilateral has all four vertices on the same circle.

🎯 Exam Tip: Draw perpendicular bisectors carefully to find the circumcenter. Make sure D is on the perpendicular bisector of BC to be equidistant from B and C.

Question 20. Using a ruler and compasses only:
(i) Construct a triangle ABC with the following data: AB = 3.5 cm, BC = 6 cm and ∠ABC = 120°.
(ii) In the same diagram, draw a circle with BC as diameter. Find a point P on the circumference of the circle which is equidistant from AB and BC.
(iii) Measure ∠BCP.
Answer:
Steps of constructions:
1. Draw a line segment BC = 6 cm.
At B, draw a ray BX making an angle of 120° with BC.
With B as centre and radius 3.5 cm, cut-off AB = 3.5 cm.
Join AC
Thus, ABC is the required triangle.
2. Draw perpendicular bisector MN of BC which cuts BC at point o.
With O as centre and radius = OB, draw a circle.
Draw angle bisector of ∠ABC which meets the circle at point P.
Thus, point P is equidistant from AB and BC
3. On measuring, ∠BCP = 30°
[Diagram: This diagram shows triangle ABC with angle 120° at B, a circle with BC as diameter, and point P on the circle where the angle bisector of ∠ABC meets the circle.]
In simple words: We make triangle ABC first. Then we draw a circle using BC as diameter. Point P is found using the angle bisector - this makes P equal distance from both sides AB and BC.

📝 Teacher's Note: Explain that angle bisector makes equal angles with both sides. Any point on angle bisector is equidistant from the two sides of the angle.

🎯 Exam Tip: Use angle bisector to find point equidistant from two sides. Measure ∠BCP carefully with protractor. Write all measurements clearly.

Question 21. Construct a ∆ABC with BC = 6.5 cm, AB = 5.5 cm, AC = 5 cm. Construct the incircle of the triangle. Measure and record the radius of the incircle.
Answer:
Steps of construction:
1. Draw BC = 6.5 cm.
2. With B as centre, draw an arc of radius 5.5 cm.
3. With C as centre, draw an arc of radius 5 cm.
Let this arc meets the previous arc at A.
4. Join AB and AC to get ∆ABC.
5. Draw the bisectors of ∠ABC and ∠ACB.
Let these bisectors meet each other at O.
6. Draw ON ⊥ BC.
7. With O as centre and radius ON, draw a incircle that touches all the sides of ∆ABC
8. By measurement, radius ON = 1.5 cm
[Diagram: This diagram shows triangle ABC with all three sides given, angle bisectors meeting at incenter O, and an inscribed circle touching all three sides of the triangle.]
In simple words: We make triangle ABC using the three given sides. Then we draw angle bisectors to find the center. The incircle goes inside and touches all three sides of the triangle.

📝 Teacher's Note: Remind students that incenter is where angle bisectors meet. The incircle always touches all three sides of the triangle from inside.

🎯 Exam Tip: Draw angle bisectors of any two angles to find incenter. Measure the radius by drawing perpendicular from center to any side. Write the measured value clearly.

Question 22. Construct a triangle ABC with AB = 5.5 cm, AC = 6 cm and ∠BAC = 105°. Hence:
(i) Construct the locus of points equidistant from BA and BC.
(ii) Construct the locus of points equidistant from B and C.
(iii) Mark the point which satisfies the above two loci as P. Measure and write the length of PC.
Answer:
Steps of construction:
1. Draw AB = 5.5 cm
2. Construct ∠BAR = 105°
3. With centre A and radius 6 cm, cut off arc on AR at C.
4. Join BC. ABC is the required triangle.
5. Draw angle bisector BD of ∠ABC, which is the locus of points equidistant from BA and BC.
6. Draw perpendicular bisector EF of BC, which is the locus of points equidistant from B and C.
7. BD and EF intersect each other at point P.
Thus, P satisfies the above two loci.
By measurement, PC = 4.8 cm
In simple words: We drew a triangle first. Then we found two special lines. The angle bisector divides the angle into two equal parts. The perpendicular bisector cuts the side BC into two equal parts. Where these lines meet is our answer point P.

📝 Teacher's Note: Show students that angle bisector means equal distance from two sides. Perpendicular bisector means equal distance from two points. Where they meet gives both conditions together.

🎯 Exam Tip: Always write "locus of points equidistant" clearly. Show the intersection point P clearly in your diagram. Measure PC carefully with ruler.

Question 23. Construct a regular hexagon of side 5 cm. Hence construct all its lines of symmetry and name them. (2016)
Answer:
Steps of construction:
1. Draw AF measuring 5 cm using a ruler.
2. With A as the centre and radius equal to AF, draw an arc above AF.
3. With F as the centre, and same radius cut the previous arc at Z
4. With Z as the centre, and same radius draw a circle passing through A and F.
5. With A as the centre and same radius, draw an arc to cut the circle above AF at B.
6. With B as the centre and same radius, draw an arc to cut the circle at C.
7. Repeat this process to get remaining vertices of the hexagon at D and E.
8. Join consecutive arcs on the circle to form the hexagon.
9. Draw the perpendicular bisectors of AF, FE and DE.
10. Extend the bisectors of AF, FE and DE to meet CD, BC and AB at X, L and O respectively.
11. Join AD, CF and EB.
These are the 6 lines of symmetry of the regular hexagon.
In simple words: A regular hexagon has 6 equal sides. It has 6 lines of symmetry - 3 lines joining opposite corners and 3 lines through middle of opposite sides. Each line divides the hexagon into two equal mirror parts.

📝 Teacher's Note: Use a paper hexagon and fold it along each line of symmetry. Students can see how one half fits exactly on the other half. This makes symmetry very clear.

🎯 Exam Tip: A regular hexagon always has exactly 6 lines of symmetry. Draw all 6 lines clearly and label them. Count them to make sure you have all 6.

Question 24. Draw a line AB = 5 cm. Mark a point C on AB such that AC = 3 cm. Using a ruler and a compass only, construct:
(i) A circle of radius 2.5 cm, passing through A and C.
(ii) Construct two tangents to the circle from the external point B. Measure and record the length of the tangents. (2016)
Answer:
Steps for construction:
1. Draw AB = 5 cm using a ruler.
2. With A as the centre cut an arc of 3 cm on AB to obtain C.
3. With A as the centre and radius 2.5 cm, draw an arc above AB.
4. With same radius, and C as the centre draw an arc to cut the previous arc and mark the intersection as O.
5. With O as the centre and radius 2.5 cm, draw a circle so that points A and C lie on the circle formed.
6. Join OB.
7. Draw the perpendicular bisector of OB to obtain the mid-point of OB, M.
8. With the M as the centre and radius equal to OM, draw a circle to cut the previous circle at points P and Q.
9. Join PB and QB. PB and QB are the required tangents to the given circle from exterior point B.
QB = PB = 3 cm
That is, length of each tangent is 3 cm.
In simple words: A tangent is a line that touches a circle at exactly one point. From any outside point, we can draw exactly two tangents to a circle. Both tangents have the same length.

📝 Teacher's Note: Show students that tangent means "touching". Like a wheel touching the road at one point. From outside the circle, both tangent lengths are always equal.

🎯 Exam Tip: Always measure both tangent lengths to check they are equal. Write "tangent length = 3 cm" clearly. Show the construction steps neatly.

Solution 25.
Answer:
Steps of construction:
1. Draw a line AB = 7 cm
2. Taking P as centre and same radius, draw an arc of a circle which intersects AB at M.
3. Taking M as centre and with the same radius as before drawn an arc intersecting previously drawn arc, at point N.
4. Draw the ray AX passing through N, then ∠XAB = 60°
5. Taking A as centre and radius equal to 5 cm, draw an arc cutting AX at C.
6. Join BC
7. The required triangle ABC is obtained.
8. Draw angle bisector of ∠CAB and ∠ABC
9. Mark their intersection as O
10. With O as center, draw a circle with radius OD
In simple words: We first made a 60° angle using compass. Then we made a triangle. The incircle is the biggest circle that fits inside the triangle. It touches all three sides.

📝 Teacher's Note: Tell students that incircle always touches all three sides of the triangle. The center is where angle bisectors meet. This point is equally distant from all three sides.

🎯 Exam Tip: First construct the triangle completely. Then find the incenter by drawing angle bisectors. The incircle radius is perpendicular distance from center to any side.

Solution 26.
Answer:
Steps for construction:
1. Draw BC = 6.8 cm.
2. Mark point D where BD = DC = 3.4 cm which is mid-point of BC.
3. Mark a point A which is intersection of arcs AD = 4.4 cm and AB = 5 cm from a point D and B respectively.
4. Join AB, AD and AC. ABC is the required triangle.
5. Draw bisectors of angle B and angle C which are ray BX and CY where I is the in centre of a circle.
6. Draw in circle of a triangle ABC.
In simple words: We used the given measurements to find where point A should be. Then we drew the incircle - the circle that fits inside the triangle and touches all three sides.

📝 Teacher's Note: Show students how we use two measurements (AD = 4.4 cm and AB = 5 cm) to find the exact position of point A. Where two arcs cross gives us the point.

🎯 Exam Tip: Always mark the midpoint D first. Use two arcs to locate point A exactly. Draw the triangle first, then construct the incircle using angle bisectors.

Solution 27.
Steps for construction:

1. Draw concentric circles of radius 4 cm and 6 cm with centre of O.
2. Take point P on the outer circle.
3. Join OP.
4. Draw perpendicular bisectors of OP where M is the midpoint of OP.
5. Take a distance of a point O from the point M and mark arcs from M on the inner circle it cuts at point A and B respectively.
6. Join PA and PB.

We observe that PA and PB are tangents from outer circle to inner circle are equal of a length 4.5 cm each.
In simple words: We draw two circles inside each other. Then we find a point P on the big circle. From P, we draw two lines that just touch the small circle. These touching lines are called tangents.

[Diagram: This diagram shows two concentric circles with center O. Point P is on the outer circle. Two tangent lines PA and PB are drawn from P to touch the inner circle at points A and B. Both tangents have equal length of 4.5 cm.]

📝 Teacher's Note: Show students that tangents from a point to a circle are always equal in length. Use a coin and draw tangents to help them see this rule.

🎯 Exam Tip: Always write that tangents from an external point to a circle are equal. This is the key property examiners want to see.

Solution 28.
Steps for construction:

1. Draw BC = 7.2 cm.
2. Draw an angle ABC = 90° using compass.
3. Draw BD perpendicular to AC using compass.
4. Join BD.
5. Draw perpendicular bisectors of AB and BC which intersect at I, where I is the circumcentre of a circle.
6. Draw circumcircle using circumcentre I. we get radius of a circle is 4.7 cm.

In simple words: We make a right triangle first. Then we find the center point that is equal distance from all three corners. From this center, we draw a circle that passes through all three corners of the triangle.

[Diagram: This diagram shows a right triangle ABC with BC = 7.2 cm and angle ABC = 90°. Point I is the circumcenter where perpendicular bisectors meet. A circle is drawn with center I passing through all three vertices A, B, and C, with radius 4.7 cm.]

📝 Teacher's Note: Tell students that the circumcenter of a right triangle is always at the middle of the longest side (hypotenuse). This makes construction easier.

🎯 Exam Tip: Write "circumcenter" and "circumcircle" correctly. Show that perpendicular bisectors meet at the circumcenter. Measure and write the radius value clearly.