Constructions Notes

11

Constructions

Class 10 Mathematics • Complete Board Exam Notes

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Chapter Roadmap

1. Basics of Geometrical Construction 2. Division of a Line Segment 3. Division in a Given Ratio 4. Similar Triangles 5. Scale Factor 6. Constructing Similar Triangles 7. Tangents to a Circle 8. Why the Construction Works 9. Board-Exam Questions 10. Ten Detailed Solved Questions 11. Exam Strategy 12. Quick Revision

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Chapter Goal

Construction questions are not ordinary numerical questions. You must use a ruler and compass to create a required figure accurately. In the exam, write the construction steps clearly, keep the figure neat, and give a short justification wherever required.

1. Basics of Geometrical Construction

1.1 Essential Instruments

• Ruler: used to draw straight lines and measure lengths when permitted.
• Compass: used to draw arcs and transfer equal lengths.
• Pencil: use a sharp pencil for accurate construction.
• Protractor: may be useful for checking, but standard construction methods mainly depend on ruler and compass.

Construction: A geometrical construction is a precise method of creating a required figure using basic geometrical tools and known properties.

1.2 Basic Construction Ideas

• Draw a line segment of known length.
• Draw an arc with a fixed radius.
• Transfer a length from one position to another using a compass.
• Locate the intersection of two arcs or lines.
• Use parallel lines and proportionality to divide a segment.

Exam habit: Keep construction lines light and final required lines darker. Mark important points with capital letters.

Solved Example 1 — Instrument

Which instrument is mainly used to draw an arc with a fixed radius?

1. An arc must have all its points at a fixed distance from its centre.
2. A compass maintains a fixed opening and therefore a fixed radius.

Answer: Compass

Solved Example 2 — Transfer a Length

How can a length \(AB\) be copied to another ray?

1. Open the compass so that its two tips are at \(A\) and \(B\).
2. Without changing the compass opening, place one tip at the new starting point.
3. Draw an arc cutting the new ray.
4. The new segment has length equal to \(AB\).

Answer: Use the compass to transfer the length without changing its opening.

Solved Example 3 — Construction Accuracy

Why should the compass opening not be changed while transferring a length?

1. The compass opening represents the exact length being transferred.
2. Changing the opening would change the length.

Answer: To preserve the exact required length.

2. Division of a Line Segment

2.1 Dividing a Segment into Equal Parts

A line segment can be divided into any required number of equal parts using a standard parallel-line construction.

Core idea: Draw an auxiliary ray from one endpoint, mark equal intervals on that ray, join the last marked point to the other endpoint, and draw parallels through the remaining marked points.

2.2 Construction: Divide \(AB\) into \(n\) Equal Parts

1. Draw the given line segment \(AB\).
2. From \(A\), draw an auxiliary ray \(AX\) making a convenient angle with \(AB\).
3. Using the same compass opening, mark \(n\) equal points \(A_1,A_2,\ldots,A_n\) on \(AX\).
4. Join \(A_n\) to \(B\).
5. Through \(A_1,A_2,\ldots,A_{n-1}\), draw lines parallel to \(A_nB\).
6. The points where these parallels meet \(AB\) divide \(AB\) into \(n\) equal parts.

\[ \boxed{\text{Equal intervals on an auxiliary ray } \Rightarrow \text{ equal divisions on }AB} \]

Solved Example 1 — Divide into 2 Equal Parts

Construct the midpoint of \(AB\) using the equal-division method.

1. Draw \(AB\).
2. From \(A\), draw ray \(AX\).
3. Mark two equal intervals on \(AX\): \(A_1,A_2\).
4. Join \(A_2B\).
5. Through \(A_1\), draw a line parallel to \(A_2B\).
6. Let it meet \(AB\) at \(M\).

Answer: \(M\) divides \(AB\) into two equal parts.

Solved Example 2 — Divide into 3 Equal Parts

Describe the construction to divide \(AB\) into three equal parts.

1. Draw \(AB\) and an auxiliary ray \(AX\).
2. Mark three equal intervals \(A_1,A_2,A_3\) on \(AX\).
3. Join \(A_3B\).
4. Draw parallels to \(A_3B\) through \(A_1\) and \(A_2\).
5. The two intersections on \(AB\) give the required three equal parts.

Answer: \(AB\) is divided into three equal parts.

Solved Example 3 — Divide into 5 Equal Parts

What is the key construction step when dividing \(AB\) into five equal parts?

1. Mark five equal intervals on an auxiliary ray from \(A\).
2. Join the fifth point to \(B\).
3. Draw four parallels through the first four marked points.

Answer: The four parallels create five equal divisions on \(AB\).

3. Division in a Given Ratio

3.1 Internal Division

To divide a line segment \(AB\) in the ratio \(m:n\), construct \(m+n\) equal intervals on an auxiliary ray and use a parallel line.

\[ \boxed{AP:PB=m:n} \]

3.2 Construction Steps for Ratio \(m:n\)

1. Draw \(AB\).
2. Draw an auxiliary ray \(AX\).
3. Mark \(m+n\) equal intervals on \(AX\).
4. Let the last point be \(A_{m+n}\).
5. Join \(A_{m+n}\) to \(B\).
6. From the point \(A_m\), draw a parallel to \(A_{m+n}B\).
7. Let it meet \(AB\) at \(P\).
8. Then \(P\) divides \(AB\) internally in the ratio \(m:n\).

Solved Example 1 — Ratio \(1:2\)

Construct a point \(P\) dividing \(AB\) internally in the ratio \(1:2\).

1. Draw \(AB\).
2. Draw ray \(AX\).
3. Mark three equal intervals on \(AX\).
4. Join the third point to \(B\).
5. Through the first point, draw a parallel to this joining line.
6. Its intersection with \(AB\) is \(P\).

Answer: \(AP:PB=1:2\).

Solved Example 2 — Ratio \(2:3\)

Describe the construction for dividing \(AB\) in the ratio \(2:3\).

1. Mark \(2+3=5\) equal intervals on an auxiliary ray.
2. Join the fifth marked point to \(B\).
3. Through the second marked point, draw a parallel to the joining line.
4. The intersection with \(AB\) gives \(P\).

Answer: \(AP:PB=2:3\).

Solved Example 3 — Ratio \(3:4\)

How many equal intervals must be marked on the auxiliary ray to divide \(AB\) in the ratio \(3:4\)?

1. Add the ratio numbers: \[ 3+4=7. \]
2. Therefore mark seven equal intervals.

Answer: \(7\) equal intervals.

4. Similar Triangles

4.1 Meaning of Similarity

Two triangles are similar when their corresponding angles are equal and their corresponding sides are proportional.

\[ \boxed{\triangle ABC\sim\triangle PQR} \] \[ \boxed{\frac{AB}{PQ}=\frac{BC}{QR}=\frac{CA}{RP}} \]

4.2 AA Similarity

If two angles of one triangle are equal to two corresponding angles of another triangle, the triangles are similar by the AA criterion.

\[ \boxed{\text{Two corresponding equal angles}\Rightarrow\text{similar triangles}} \]

Solved Example 1 — Identify AA

If \(\angle A=\angle P\) and \(\angle B=\angle Q\), can \(\triangle ABC\) and \(\triangle PQR\) be declared similar?

1. Two corresponding angles are equal.
2. Therefore the AA similarity criterion applies.

Answer: Yes, \(\triangle ABC\sim\triangle PQR\) by AA.

Solved Example 2 — Find a Corresponding Side

If \(\triangle ABC\sim\triangle PQR\), \(AB=6\) cm, \(PQ=9\) cm and \(BC=8\) cm, find \(QR\).

1. Corresponding sides are proportional: \[ \frac{AB}{PQ}=\frac{BC}{QR}. \]
2. Substitute: \[ \frac69=\frac8{QR}. \]
3. Simplify: \[ \frac23=\frac8{QR}. \]
4. Cross multiply: \[ 2QR=24. \]
5. Therefore: \[ QR=12\text{ cm}. \]

Answer: \(12\) cm

Solved Example 3 — Scale Factor

If corresponding sides of two similar triangles are \(5\) cm and \(15\) cm, find the scale factor from the smaller triangle to the larger triangle.

1. Scale factor: \[ k=\frac{\text{larger corresponding side}}{\text{smaller corresponding side}}. \]
2. Therefore: \[ k=\frac{15}{5}=3. \]

Answer: Scale factor \(=3\)

5. Scale Factor

5.1 Enlarging or Reducing a Triangle

When a triangle is constructed similar to a given triangle, the scale factor determines whether the new triangle is larger or smaller.

Scale Factor	Result
\(k>1\)	Enlarged triangle
\(k=1\)	Same size
\(0<k<1\)	Reduced triangle

\[ \boxed{\text{New side}=k\times\text{corresponding original side}} \]

Solved Example 1 — Enlargement

A triangle is enlarged by scale factor \(2\). If a corresponding side is \(7\) cm, find the new side.

1. Use: \[ \text{New side}=k\times\text{old side}. \]
2. Substitute: \[ 2\times7=14. \]

Answer: \(14\) cm

Solved Example 2 — Reduction

A similar triangle has scale factor \(\frac23\) relative to the original. If an original side is \(15\) cm, find the corresponding new side.

1. Use: \[ \text{New side}=\frac23(15). \]
2. Therefore: \[ \text{New side}=10\text{ cm}. \]

Answer: \(10\) cm

Solved Example 3 — Find Scale Factor

A side changes from \(12\) cm to \(18\) cm. Find the scale factor.

1. Use: \[ k=\frac{18}{12}. \]
2. Simplify: \[ k=\frac32. \]

Answer: \(\frac32\)

6. Constructing Similar Triangles

6.1 When the New Triangle is Smaller

Suppose a triangle \(ABC\) is given and a similar triangle is required with scale factor \(\frac mn\), where \(m

Construction principle: First construct a ray from one vertex, mark \(n\) equal parts, join the \(n\)-th point to the appropriate vertex, and draw a parallel through the \(m\)-th point. The parallel creates the required proportional side.

6.2 Standard Construction Steps

1. Draw the given triangle \(ABC\).
2. From the vertex \(B\), draw an auxiliary ray \(BX\) making an acute angle with \(BC\).
3. Mark \(n\) equal points \(B_1,B_2,\ldots,B_n\) on \(BX\).
4. Join \(B_n\) to \(C\).
5. Through \(B_m\), draw a line parallel to \(B_nC\).
6. Let the parallel meet \(BC\) at \(C'\).
7. Through \(C'\), draw a line parallel to \(AC\) to meet \(AB\) at \(A'\).
8. Then \(\triangle A'BC'\) is similar to \(\triangle ABC\) in the required ratio.

Solved Example 1 — Ratio \(2:3\)

Describe the construction of a triangle similar to a given triangle with scale factor \(\frac23\).

1. Draw the given triangle \(ABC\).
2. From \(B\), draw ray \(BX\).
3. Mark three equal points on \(BX\).
4. Join the third point to \(C\).
5. Through the second point, draw a parallel to the joining line.
6. Use a second parallel to obtain the corresponding vertex on \(AB\).
7. The new triangle has corresponding sides in ratio \(2:3\).

Answer: Required similar triangle has scale factor \(\frac23\).

Solved Example 2 — Ratio \(3:5\)

How many equal intervals are needed on the auxiliary ray to construct a similar triangle in the ratio \(3:5\)?

1. The denominator gives the total number of intervals.
2. Therefore mark five equal intervals.
3. Use the third point to draw the required parallel.

Answer: \(5\) equal intervals, using the third point.

Solved Example 3 — Enlargement

For a scale factor \(\frac43\), should the construction use the third point or fourth point for the corresponding proportional step?

1. The ratio is \(4:3\).
2. The total number of equal intervals is based on the denominator \(3\) when using the inverse enlargement construction, or equivalently the construction can be arranged with the required larger triangle using the fourth and third marks.
3. The key requirement is that the final corresponding sides satisfy: \[ \frac{\text{new side}}{\text{old side}}=\frac43. \]

Answer: Construct the proportional division so the final scale factor is \(\frac43\).

7. Tangents to a Circle

7.1 Tangent from an External Point

From a point \(P\) outside a circle with centre \(O\), two tangents can be constructed to the circle.

\[ \boxed{\text{If }P\text{ is outside the circle, exactly two tangents can be drawn from }P.} \]

7.2 Construction of Tangents

1. Let \(P\) be the external point and \(O\) the centre of the circle.
2. Join \(OP\).
3. Find the midpoint \(M\) of \(OP\).
4. With \(M\) as centre and radius \(MO\), draw a circle.
5. This new circle intersects the given circle at \(T_1\) and \(T_2\).
6. Join \(PT_1\) and \(PT_2\).
7. These two lines are the required tangents.

Why this works: Since \(M\) is the midpoint of \(OP\), \(MO=MP\). For any intersection point \(T\), the angle \(OTP\) is \(90^\circ\). Therefore \(PT\) is perpendicular to the radius \(OT\), so \(PT\) is a tangent.

Solved Example 1 — Number of Tangents

How many tangents can be constructed from an external point \(P\)?

1. An external point lies outside the circle.
2. Two tangent positions are possible.

Answer: Two tangents.

Solved Example 2 — First Construction Step

What should be done first when constructing tangents from \(P\) to a circle with centre \(O\)?

1. Join the external point \(P\) to the centre \(O\).

Answer: Draw \(OP\).

Solved Example 3 — Why Midpoint is Used

Why is the midpoint of \(OP\) used in the tangent construction?

1. Let \(M\) be the midpoint of \(OP\).
2. Then \(MO=MP\).
3. A circle centred at \(M\) therefore creates a right-angle condition at its intersections with the original circle.
4. This makes the lines from \(P\) to the intersection points perpendicular to the corresponding radii.

Answer: It creates the \(90^\circ\) angle required for a tangent.

8. Why the Construction Works

8.1 Proof of Tangent Construction

Proof

Let the auxiliary circle with centre \(M\) meet the given circle at \(T\). Since \(M\) is the midpoint of \(OP\), \(MO=MP\). Also \(MT=MO\), because \(T\) lies on the auxiliary circle.

1. Therefore: \[ MO=MP=MT. \]
2. So \(O,P,T\) lie on a circle with centre \(M\).
3. Since \(OP\) is a diameter of this auxiliary circle, the angle subtended by diameter \(OP\) at \(T\) is a right angle.
4. Thus: \[ \angle OTP=90^\circ. \]
5. Since \(OT\) is a radius of the original circle and \(PT\perp OT\), \(PT\) is a tangent.

Hence the construction gives a tangent to the given circle.

8.2 Similar-Triangle Construction Logic

The parallel line creates equal corresponding angles. Therefore the smaller or larger triangle is similar to the original triangle by AA. Similarity then gives proportional corresponding sides, which produces the required scale factor.

Solved Example 1 — Similarity Reason

Why does drawing a parallel line help in constructing a similar triangle?

1. Parallel lines create equal corresponding angles.
2. Therefore two angles of the new triangle match the original triangle.
3. By AA, the triangles are similar.

Answer: The parallel creates the equal angles needed for AA similarity.

Solved Example 2 — Proportional Sides

If two triangles are similar with scale factor \(2:3\), what is the ratio of their corresponding sides?

1. Similarity means corresponding sides are proportional.
2. Therefore: \[ \frac{\text{new side}}{\text{old side}}=\frac23. \]

Answer: \(2:3\)

Solved Example 3 — Construction Verification

After completing a construction, what should be checked first?

1. Check that all required points are correctly labelled.
2. Check the proportional ratio or tangent condition.
3. Check that the construction lines and final lines are clearly distinguishable.

Answer: Verify the required geometric condition before finalising the figure.

9. Board-Exam Questions

Board Pattern 1

Question: Divide a line segment of length \(8\) cm internally in the ratio \(3:5\).

Construction: Draw the segment, construct an auxiliary ray, mark \(3+5=8\) equal intervals, join the eighth point to the other endpoint and draw a parallel through the third point.

Result: The required point divides the segment in ratio \(3:5\).

Board Pattern 2

Question: Construct a triangle similar to a given triangle with scale factor \(\frac34\).

Construction: Draw the given triangle, construct an auxiliary ray from a suitable vertex, mark four equal intervals, use the third point to create the required parallel, and complete the new triangle.

Result: New triangle is similar with scale factor \(\frac34\).

Board Pattern 3

Question: Construct two tangents to a circle from a point outside it.

Construction: Join the external point to the centre, bisect this segment, draw the auxiliary circle with the midpoint as centre, mark its intersections with the given circle, and join these intersection points to the external point.

Result: Two tangents are obtained.

Board Pattern 4

Question: Why are the lines obtained in the tangent construction actually tangents?

Answer: The construction gives a \(90^\circ\) angle between each joining line and the corresponding radius. A line perpendicular to the radius at the point on the circle is a tangent.

Key result: Radius \(\perp\) tangent.

10. Ten Detailed Solved Questions

Question 1

Divide a line segment \(AB\) into four equal parts. Write the construction steps.

1. Draw \(AB\).
2. Draw auxiliary ray \(AX\).
3. Mark four equal intervals on \(AX\).
4. Join the fourth point to \(B\).
5. Draw parallels through the first three marked points.
6. These parallels meet \(AB\) at three points.

Answer: \(AB\) is divided into four equal parts.

Question 2

Divide \(AB=10\) cm internally in the ratio \(2:3\).

1. Total ratio parts: \[ 2+3=5. \]
2. Construct five equal intervals on an auxiliary ray.
3. Use the second point to draw the required parallel.
4. The resulting point \(P\) satisfies: \[ AP:PB=2:3. \]
5. Since total length is \(10\) cm: \[ AP=\frac25(10)=4\text{ cm}, \] \[ PB=\frac35(10)=6\text{ cm}. \]

Answer: \(AP=4\) cm and \(PB=6\) cm.

Question 3

A line segment \(AB=12\) cm is divided in the ratio \(3:1\). Find the two parts.

1. Total parts: \[ 3+1=4. \]
2. One part: \[ \frac{12}{4}=3\text{ cm}. \]
3. First part: \[ 3(3)=9\text{ cm}. \]
4. Second part: \[ 1(3)=3\text{ cm}. \]

Answer: \(9\) cm and \(3\) cm.

Question 4

Two similar triangles have corresponding sides \(8\) cm and \(12\) cm. Find the scale factor from the first to the second.

1. Use: \[ k=\frac{\text{new side}}{\text{old side}}. \]
2. Substitute: \[ k=\frac{12}{8}=\frac32. \]

Answer: \(\frac32\)

Question 5

A similar triangle is required with scale factor \(\frac25\). How many equal points should be marked on the auxiliary ray?

1. The ratio is \(2:5\).
2. The total number of intervals is: \[ 5. \]
3. The second marked point is used for the proportional parallel.

Answer: Mark \(5\) equal intervals and use the \(2\)-nd point.

Question 6

Why are the triangles produced by the similar-triangle construction similar?

1. A constructed line is parallel to a side of the original triangle.
2. Therefore corresponding angles are equal.
3. One angle at the common vertex is also equal.
4. Hence two corresponding angles are equal.
5. By AA similarity, the triangles are similar.

Answer: By AA similarity criterion.

Question 7

Construct tangents from an external point \(P\) to a circle with centre \(O\).

1. Join \(OP\).
2. Find midpoint \(M\) of \(OP\).
3. With centre \(M\) and radius \(MO\), draw a circle.
4. Let it meet the given circle at \(T_1,T_2\).
5. Join \(PT_1\) and \(PT_2\).

Answer: \(PT_1\) and \(PT_2\) are the required tangents.

Question 8

Explain why \(\angle OT_1P=90^\circ\) in the tangent construction.

1. \(M\) is the midpoint of \(OP\).
2. Thus \(OP\) is a diameter of the auxiliary circle centred at \(M\).
3. The angle subtended by a diameter at any point on the circle is a right angle.
4. Therefore: \[ \angle OT_1P=90^\circ. \]

Answer: \(\angle OT_1P=90^\circ\).

Question 9

If a triangle is enlarged by scale factor \(3\), what happens to each corresponding side?

1. Scale factor \(=3\).
2. Every corresponding side is multiplied by \(3\).
3. For example, a \(4\) cm side becomes: \[ 3\times4=12\text{ cm}. \]

Answer: Every corresponding side becomes three times as long.

Question 10

A point \(P\) divides \(AB=20\) cm in the ratio \(3:2\). Find \(AP\) and \(PB\), and state how the construction can be performed.

1. Total parts: \[ 3+2=5. \]
2. One part: \[ \frac{20}{5}=4\text{ cm}. \]
3. Therefore: \[ AP=3(4)=12\text{ cm}, \] \[ PB=2(4)=8\text{ cm}. \]
4. For construction, mark five equal intervals on an auxiliary ray and use the third point to draw the required parallel.

Answer: \(AP=12\) cm, \(PB=8\) cm.

11. Exam Strategy

1. Read the ratio carefully.

For a division ratio \(m:n\), remember the total number of auxiliary intervals is \(m+n\).

2. Keep the compass opening fixed.

Whenever equal intervals or a length is being transferred, do not accidentally change the compass opening.

3. For similar triangles, identify the scale factor first.

If the required ratio is \(m:n\), plan the number of auxiliary divisions before drawing the parallels.

4. For tangent construction, remember the sequence:

\[ OP\rightarrow\text{midpoint of }OP\rightarrow\text{auxiliary circle}\rightarrow T_1,T_2\rightarrow PT_1,PT_2. \]

Common mistakes:

• Changing compass opening while marking equal intervals.
• Using the wrong marked point for the required ratio.
• Drawing an inaccurate parallel.
• Forgetting to join the midpoint construction correctly in tangent problems.
• Not writing the final conclusion.
• Making construction lines too dark and hiding the final figure.

12. Quick Revision Sheet

Concept	Formula / Key Fact
Equal Division	Mark equal intervals on an auxiliary ray and use parallels
Ratio \(m:n\)	\(m+n\) equal auxiliary intervals
Internal Division	\(AP:PB=m:n\)
Similarity	\(\triangle ABC\sim\triangle PQR\)
AA Criterion	Two corresponding angles equal
Scale Factor	\(k=\frac{\text{new side}}{\text{old side}}\)
Smaller Triangle	\(0<k<1\)
Larger Triangle	\(k>1\)
External Point	Two tangents can be drawn
Tangent Construction	Join \(OP\), bisect \(OP\), draw auxiliary circle, join intersection points to \(P\)
Tangent Condition	\(OT\perp PT\)
Compass	Transfers lengths and draws arcs
Parallel Lines	Create equal corresponding angles and proportional sides

30-Second Final Revision

• Division: auxiliary ray + equal intervals + parallel lines.
• Ratio \(m:n\): mark \(m+n\) equal intervals.
• Similar triangle: use parallel lines and AA similarity.
• Scale factor: new side ÷ corresponding old side.
• Tangents: join external point to centre, find midpoint, draw auxiliary circle.
• Final tangent check: radius is perpendicular to tangent.
• Exam presentation: figure → construction steps → result/justification.

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Class 10 Mathematics • Chapter 11: Constructions

Prepared in easy, exam-focused handwritten style by University Scope.

Practise every construction with a sharp pencil, accurate compass opening and clear labelling.