Class 10 Notes

Class 10 Mathematics Notes Chapter 11 (Constructions) – Mathematics Book

Philoid

Philoid

— 7 min read

Mathematics
Alright class, let's get straight into Chapter 11, 'Constructions'. Now, pay close attention. While the NCERT has rationalized the syllabus for the 2023-24 board exams onwards, removing this chapter, concepts from it might still be relevant for certain government exams that follow older patterns or test fundamental geometric skills. So, understanding these basic constructions is beneficial.

This chapter primarily deals with using a ruler (straightedge) and compasses to draw geometric figures accurately. We'll focus on three main types of constructions that were part of the original syllabus.

Key Concepts in Chapter 11: Constructions (Based on Original NCERT Syllabus)

1. Division of a Line Segment in a Given Ratio (Internally)

  • Objective: To divide a given line segment AB into two parts such that the ratio of their lengths is m:n, where m and n are positive integers.
  • Mathematical Principle: Based on the Basic Proportionality Theorem (Thales Theorem). If a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, the other two sides are divided in the same ratio.
  • Steps of Construction (Example: Divide AB in ratio 3:2):
    1. Draw the line segment AB of the given length.
    2. Draw any ray AX, making an acute angle (∠BAX) with AB.
    3. Locate m + n = 3 + 2 = 5 points (A₁, A₂, A₃, A₄, A₅) on AX such that AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅. (This is done by setting the compass to a convenient radius and marking arcs of equal length starting from A).
    4. Join the last point A₅ to B (i.e., join BA₅).
    5. Through the point A<0xE2><0x82><0x98> (where m=3), draw a line parallel to A₅B. This is done by constructing an angle at A₃ equal to ∠AA₅B. Let this line intersect AB at point C.
    6. Then, C is the point dividing AB internally in the ratio 3:2. So, AC : CB = 3 : 2.
  • Justification: Since A₃C is parallel to A₅B (by construction), in ΔABA₅, by the Basic Proportionality Theorem, we have AA₃ / A₃A₅ = AC / CB. By construction, AA₃ consists of 3 equal parts and A₃A₅ consists of 2 equal parts. Therefore, AA₃ / A₃A₅ = 3 / 2. Hence, AC / CB = 3 / 2.
  • Exam Tip: The minimum number of points to be marked on the ray AX is m + n.

2. Construction of a Triangle Similar to a Given Triangle as per a Given Scale Factor

  • Objective: To construct a triangle whose sides are in a given ratio (scale factor) with the corresponding sides of a given triangle.
  • Scale Factor (k): The ratio of the sides of the new triangle to the corresponding sides of the given triangle.
    • If k < 1, the new triangle is smaller than the given triangle.
    • If k > 1, the new triangle is larger than the given triangle.
  • Mathematical Principle: Based on Similarity Criteria (AAA, SSS, SAS) and the Basic Proportionality Theorem.
  • Steps of Construction (Example: Construct ΔA'BC' similar to ΔABC with scale factor 3/4):
    1. Construct the given triangle ΔABC with the given dimensions.
    2. Draw any ray BX making an acute angle with BC on the side opposite to vertex A.
    3. Locate 4 points (the greater of 3 and 4 in the ratio 3/4) B₁, B₂, B₃, B₄ on BX such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄.
    4. Join B₄ (the denominator point) to C (B₄C).
    5. Through B₃ (the numerator point), draw a line parallel to B₄C, intersecting BC at C'. (Construct ∠BB₃C' = ∠BB₄C).
    6. Through C', draw a line parallel to AC, intersecting AB at A'. (Construct ∠BC'A' = ∠BCA).
    7. Then, ΔA'BC' is the required triangle similar to ΔABC with sides 3/4 of the corresponding sides of ΔABC.
  • Steps for Scale Factor > 1 (Example: Scale factor 5/3):
    1. Construct ΔABC.
    2. Draw ray BX. Locate 5 points (greater of 5 and 3) B₁, B₂, B₃, B₄, B₅.
    3. Join B₃ (denominator point) to C.
    4. Through B₅ (numerator point), draw a line parallel to B₃C, intersecting the extended line segment BC at C'.
    5. Through C', draw a line parallel to AC, intersecting the extended line segment BA at A'.
    6. Then ΔA'BC' is the required larger triangle.
  • Justification: Based on BPT and AA similarity criterion. For k=3/4 example: BC'/BC = BB₃/BB₄ = 3/4. Since C'A' || CA, ΔA'BC' ~ ΔABC (AA similarity), so A'B/AB = BC'/BC = A'C'/AC = 3/4.
  • Exam Tip: Always join the point corresponding to the denominator of the scale factor to the vertex of the base. Draw the parallel line from the point corresponding to the numerator. Extend sides if the scale factor is greater than 1.

3. Construction of Tangents to a Circle from a Point Outside it

  • Objective: To draw the pair of tangents from a given external point P to a given circle with center O and radius r.
  • Mathematical Principle:
    • The tangent at any point of a circle is perpendicular to the radius through the point of contact.
    • The lengths of tangents drawn from an external point to a circle are equal.
    • The angle in a semicircle is a right angle (90°).
  • Steps of Construction:
    1. Draw the circle with center O and given radius r.
    2. Mark the external point P at the given distance from O.
    3. Join OP.
    4. Construct the perpendicular bisector of the line segment OP. Let M be the midpoint of OP.
    5. Taking M as the center and MO (or MP) as the radius, draw a second circle (or arcs).
    6. This second circle will intersect the original circle at two points, say T₁ and T₂.
    7. Join PT₁ and PT₂.
    8. PT₁ and PT₂ are the required tangents to the circle from point P.
  • Justification: Join OT₁ and OT₂ (radii of the original circle). Since OP is the diameter of the second circle (centered at M), the angle ∠OT₁P is the angle in a semicircle, hence ∠OT₁P = 90°. Since OT₁ is the radius and PT₁ is a line meeting the radius at 90° at the point of contact T₁, PT₁ must be a tangent. Similarly, ∠OT₂P = 90°, so PT₂ is also a tangent. Also, by Pythagoras in ΔOT₁P and ΔOT₂P, or by RHS congruence, PT₁ = PT₂.
  • Exam Tip: Remember to find the midpoint of the line joining the center and the external point by constructing the perpendicular bisector.

General Tips for Construction:

  • Use a well-sharpened pencil, a ruler with clear markings, and accurate compasses.
  • Draw thin, clear lines.
  • Do not erase the construction arcs, as they show the method used.
  • Label all points clearly.
  • Briefly write the steps of construction if asked, but focus on accuracy in drawing.
  • Understand the justification behind each construction.

Multiple Choice Questions (MCQs)

Here are 10 MCQs based on the concepts from Chapter 11 'Constructions':

  1. To divide a line segment AB in the ratio 5:7, first, a ray AX is drawn such that ∠BAX is an acute angle and then points A₁, A₂, A₃, ... are located at equal distances on the ray AX. The minimum number of these points required is:
    (a) 5
    (b) 7
    (c) 10
    (d) 12

  2. To divide a line segment AB in the ratio m:n (m, n are positive integers), draw a ray AX so that ∠BAX is acute. Then mark points on AX at equal distances. The point B is joined to which point on AX?
    (a) A<0xE2><0x82><0x98>
    (b) A<0xE2><0x82><0x99>
    (c) A<0xE2><0x82><0x98>₊<0xE2><0x82><0x99>
    (d) The last point marked

  3. The construction of dividing a line segment internally in a given ratio is based on which theorem?
    (a) Pythagoras Theorem
    (b) Basic Proportionality Theorem
    (c) Midpoint Theorem
    (d) Alternate Segment Theorem

  4. To construct a triangle similar to a given ΔABC with its sides 4/7 of the corresponding sides of ΔABC, first draw a ray BX such that ∠CBX is acute and X lies on the opposite side of A w.r.t BC. Then locate points B₁, B₂, B₃,... B₇ on BX at equal distances. The next step is to join:
    (a) B₄ to C
    (b) B₇ to C
    (c) B₄ to A
    (d) B₇ to A

  5. In the construction described in Q4, a line through which point is drawn parallel to B₇C?
    (a) B₁
    (b) B₃
    (c) B₄
    (d) B₅

  6. To draw a pair of tangents to a circle which are inclined to each other at an angle of 60°, it is required to draw tangents at the endpoints of those two radii of the circle, the angle between which is:
    (a) 120°
    (b) 60°
    (c) 90°
    (d) 150°
    (Hint: Consider the quadrilateral formed by the center, the external point, and the points of contact.)

  7. To draw a pair of tangents to a circle of radius 5 cm from a point P which is 13 cm away from the center O, the first step is to find the midpoint M of OP. The radius of the circle drawn with center M will be:
    (a) 5 cm
    (b) 13 cm
    (c) 6.5 cm
    (d) 8 cm

  8. The construction of tangents from an external point to a circle relies on the property that:
    (a) The angle in a semicircle is 90°.
    (b) All chords are equidistant from the center.
    (c) The angle subtended by an arc at the center is double the angle at the circumference.
    (d) Opposite angles of a cyclic quadrilateral are supplementary.

  9. If you construct a ΔA'BC' similar to a given ΔABC with a scale factor of 5/3, then ΔA'BC' will be:
    (a) Smaller than ΔABC
    (b) Larger than ΔABC
    (c) Congruent to ΔABC
    (d) Right-angled

  10. When constructing tangents from an external point P to a circle with center O, the line segment OP is used as:
    (a) A radius of the intersecting circle
    (b) A tangent to the intersecting circle
    (c) A chord of the intersecting circle
    (d) The diameter of the intersecting circle

Answer Key for MCQs:

  1. (d) 12 (Since 5 + 7 = 12)
  2. (c) A<0xE2><0x82><0x98>₊<0xE2><0x82><0x99> (The last point, which corresponds to m+n)
  3. (b) Basic Proportionality Theorem
  4. (b) B₇ to C (Join the point corresponding to the denominator)
  5. (c) B₄ (Draw parallel line from the point corresponding to the numerator)
  6. (a) 120° (Angle between radii + angle between tangents = 180°)
  7. (c) 6.5 cm (Radius is MO = OP/2 = 13/2 = 6.5 cm)
  8. (a) The angle in a semicircle is 90°. (This ensures the radius is perpendicular to the tangent at the point of contact).
  9. (b) Larger than ΔABC (Since the scale factor 5/3 > 1)
  10. (d) The diameter of the intersecting circle (The circle whose center is the midpoint M of OP)

Remember to practice these constructions using a ruler and compass to gain proficiency. Even if the chapter isn't in the current board syllabus, the geometric principles involved are fundamental. Good luck with your preparation!

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