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Std 10 Mathematics Part 2 · Chapter 4

4. Geometric Constructions

Solutions to Practice Sets 4.1, 4.2 and Problem Set 4, with analysis, steps of construction and fair figures. Maharashtra State Board (Balbharati), Std 10 Mathematics Part 2 (Geometry).

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Practice Set 4.1

1Δ ABC ∼ Δ LMN. In Δ ABC, AB = 5.5 cm, BC = 6 cm, CA = 4.5 cm. Construct Δ ABC and Δ LMN such that BCMN = 54.

Analysis:

Δ ABC ∼ Δ LMN, so their corresponding sides are proportional.

ABLM = BCMN = CANL = 54

∴ 5.5LM = 6MN = 4.5NL = 54

∴ LM = 5.5 × 45 = 4.4 cm, MN = 6 × 45 = 4.8 cm, NL = 4.5 × 45 = 3.6 cm

∴ LM = 4.4 cm, MN = 4.8 cm, LN = 3.6 cm

Steps of construction:

  1. Draw seg BC = 6 cm. With B as centre and radius 5.5 cm draw an arc. With C as centre and radius 4.5 cm draw another arc cutting the first at A. Join AB and AC.
  2. Draw seg MN = 4.8 cm. With M as centre and radius 4.4 cm draw an arc. With N as centre and radius 3.6 cm draw another arc cutting the first at L. Join LM and LN.
6 cm5.5 cm4.5 cm4.8 cm4.4 cm3.6 cmABCLMN

Δ ABC and Δ LMN (green). Green lines show the required figure. Thin blue lines are construction lines.

2Δ PQR ∼ Δ LTR. In Δ PQR, PQ = 4.2 cm, QR = 5.4 cm, PR = 4.8 cm. Construct Δ PQR and Δ LTR, such that PQLT = 34.

Analysis:

Δ PQR ∼ Δ LTR and vertex R is common.

PQLT = QRTR = PRLR = 34

So the sides of Δ LTR are longer. T lies on ray RQ with RT = 43 × RQ, and L lies on ray RP.

RT = 43 × 5.4 = 7.2 cm, RL = 43 × 4.8 = 6.4 cm, LT = 43 × 4.2 = 5.6 cm.

Steps of construction:

  1. Construct Δ PQR with QR = 5.4 cm, PQ = 4.2 cm and PR = 4.8 cm.
  2. Draw a ray from R on the side of RQ opposite to P. Mark 4 equal parts on it with the compass (points 1, 2, 3, 4).
  3. Join point 3 to Q. Through point 4 draw a line parallel to 3Q. Let it meet ray RQ at T.
  4. Through T draw a line parallel to QP. Let it meet ray RP at L.
  5. Δ LTR is the required triangle.
12345.4 cm4.8 cm4.2 cmRPQLT

Δ PQR and Δ LTR (green). Green lines show the required figure. Thin blue lines are construction lines.

3Δ RST ∼ Δ XYZ. In Δ RST, RS = 4.5 cm, ∠ RST = 40°, ST = 5.7 cm. Construct Δ RST and Δ XYZ, such that RSXY = 35.

Analysis:

Δ RST ∼ Δ XYZ, so ∠ XYZ = ∠ RST = 40° and RSXY = STYZ = 35.

∴ 4.5XY = 5.7YZ = 35

∴ XY = 4.5 × 53 = 7.5 cm and YZ = 5.7 × 53 = 9.5 cm

∴ XY = 7.5 cm, YZ = 9.5 cm, ∠ XYZ = 40°

Steps of construction:

  1. Draw seg ST = 5.7 cm. At S draw ∠ TSR = 40° with SR = 4.5 cm. Join RT.
  2. Draw seg YZ = 9.5 cm. At Y draw ∠ ZYX = 40° with YX = 7.5 cm. Join XZ.
40°40°5.7 cm4.5 cm9.5 cm7.5 cmRSTXYZ

Δ RST and Δ XYZ (green)

4Δ AMT ∼ Δ AHE. In Δ AMT, AM = 6.3 cm, ∠ TAM = 50°, AT = 5.6 cm. AMAH = 75. Construct Δ AHE.

Analysis:

Δ AMT ∼ Δ AHE and vertex A is common.

AMAH = ATAE = 75

So the sides of Δ AHE are shorter. H lies on seg AM with AH = 57 × AM, and E lies on seg AT.

AH = 57 × 6.3 = 4.5 cm and AE = 57 × 5.6 = 4 cm.

Steps of construction:

  1. Construct Δ AMT with AM = 6.3 cm, ∠ TAM = 50° and AT = 5.6 cm.
  2. Draw a ray from A on the side of AM opposite to T. Mark 7 equal parts on it with the compass.
  3. Join point 7 to M. Through point 5 draw a line parallel to 7M. Let it meet seg AM at H.
  4. Through H draw a line parallel to MT. Let it meet seg AT at E.
  5. Δ AHE is the required triangle.
12345676.3 cm5.6 cm50°ATMEH

Δ AMT and Δ AHE (green). Green lines show the required figure. Thin blue lines are construction lines.

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Practice Set 4.2

1Construct a tangent to a circle with centre P and radius 3.2 cm at any point M on it.

Steps of construction:

  1. Draw a circle with centre P and radius 3.2 cm. Take any point M on it.
  2. Draw ray PM and extend it to X.
  3. Draw line l perpendicular to ray PX through point M.
  4. Line l is the required tangent at M.
3.2 cmPMXl

Line l ⊥ radius PM, so line l is the tangent at M.

2Draw a circle of radius 2.7 cm. Draw a tangent to the circle at any point on it.

Steps of construction:

  1. Draw a circle with centre O and radius 2.7 cm. Take any point P on it.
  2. Draw ray OP and extend it to X.
  3. Draw line l perpendicular to ray OX through point P.
  4. Line l is the required tangent at P.
2.7 cmOPXl

Line l is the tangent at P.

3Draw a circle of radius 3.6 cm. Draw a tangent to the circle at any point on it without using the centre.

Steps of construction:

  1. Draw a circle of radius 3.6 cm. Take any point C on it.
  2. Draw chord CB and an inscribed ∠ CAB.
  3. With A as centre and a convenient radius, draw an arc cutting AB at M and AC at N.
  4. With the same radius and C as centre, draw an arc cutting chord CB at R.
  5. With R as centre and radius equal to MN, draw an arc cutting the previous arc at D. Draw line CD.
  6. Line CD is the required tangent. (∠ BCD ≅ ∠ CAB, by the converse of the tangent-secant angle theorem.)
ABCMNRD

Line CD is the tangent at C.

4Draw a circle of radius 3.3 cm. Draw a chord PQ of length 6.6 cm. Draw tangents to the circle at points P and Q. Write your observation about the tangents.

Chord PQ = 6.6 cm = 2 × 3.3 cm, so chord PQ is a diameter. It passes through the centre O.

Steps of construction:

  1. Draw a circle with centre O and radius 3.3 cm. Draw diameter PQ = 6.6 cm.
  2. Draw line l ⊥ seg PQ at P and line m ⊥ seg PQ at Q.
  3. Lines l and m are the required tangents.
6.6 cmlmOPQ

Tangents at the ends of diameter PQ

Observation: The tangents at P and Q are parallel to each other.

Both are perpendicular to the same diameter PQ.

5Draw a circle with radius 3.4 cm. Draw a chord MN of length 5.7 cm in it. Construct tangents at point M and N to the circle.

Steps of construction:

  1. Draw a circle with centre O and radius 3.4 cm. Draw chord MN = 5.7 cm.
  2. Draw radii OM and ON.
  3. Draw a line perpendicular to OM at M and a line perpendicular to ON at N.
  4. These lines are the required tangents. They meet at T.
5.7 cmOMNT

Tangents at M and N meet at T.

Chord MN is not a diameter, so the tangents are not parallel. They meet at T, and TM = TN (tangent segment theorem).

6Draw a circle with centre P and radius 3.4 cm. Take point Q at a distance 5.5 cm from the centre. Construct tangents to the circle from point Q.

Steps of construction:

  1. Draw a circle with centre P and radius 3.4 cm. Take point Q such that PQ = 5.5 cm.
  2. Draw seg PQ. Draw its perpendicular bisector to get the midpoint M.
  3. With M as centre and radius MP, draw a circle. Let it cut the given circle at A and B.
  4. Draw lines QA and QB. These are the required tangents.
3.4 cm5.5 cmPQMAB

QA and QB are the tangents. ∠ PAQ = ∠ PBQ = 90° (angles in a semicircle).

On measuring, QA = QB ≈ 4.3 cm.

7Draw a circle with radius 4.1 cm. Construct tangents to the circle from a point at a distance 7.3 cm from the centre.

Steps of construction:

  1. Draw a circle with centre O and radius 4.1 cm. Take point P such that OP = 7.3 cm.
  2. Draw seg OP. Draw its perpendicular bisector to get the midpoint M.
  3. With M as centre and radius MO, draw a circle. Let it cut the given circle at A and B.
  4. Draw lines PA and PB. These are the required tangents.
4.1 cm7.3 cmOPMAB

PA and PB are the tangents.

On measuring, PA = PB ≈ 6 cm.

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Problem Set 4

1Select the correct alternative for each of the following questions.
  • (1) The number of tangents that can be drawn to a circle at a point on the circle is …… — (C) 1
    Only one line is perpendicular to the radius at that point.
  • (2) The maximum number of tangents that can be drawn to a circle from a point outside it is …… — (A) 2
    From an outside point, exactly two tangents can be drawn.
  • (3) If Δ ABC ∼ Δ PQR and ABPQ = 75, then …… — (A) Δ ABC is bigger.
    AB > PQ, so the sides of Δ ABC are longer.
2Draw a circle with centre O and radius 3.5 cm. Take point P at a distance 5.7 cm from the centre. Draw tangents to the circle from point P.

Steps of construction:

  1. Draw a circle with centre O and radius 3.5 cm. Take point P such that OP = 5.7 cm.
  2. Draw seg OP. Draw its perpendicular bisector to get the midpoint M.
  3. With M as centre and radius MO, draw a circle. Let it cut the given circle at A and B.
  4. Draw lines PA and PB. These are the required tangents.
3.5 cm5.7 cmOPMAB

PA and PB are the tangents.

On measuring, PA = PB ≈ 4.5 cm.

3Draw any circle. Take any point A on it and construct tangent at A without using the centre of the circle.

The figure below uses the letters of the textbook method: the given point is C and the tangent is line CD. For this question, name the given point A instead.

Steps of construction:

  1. Draw a circle of radius 3 cm. Take any point C on it.
  2. Draw chord CB and an inscribed ∠ CAB.
  3. With A as centre and a convenient radius, draw an arc cutting AB at M and AC at N.
  4. With the same radius and C as centre, draw an arc cutting chord CB at R.
  5. With R as centre and radius equal to MN, draw an arc cutting the previous arc at D. Draw line CD.
  6. Line CD is the required tangent. (∠ BCD ≅ ∠ CAB, by the converse of the tangent-secant angle theorem.)
ABCMNRD

Line CD is the tangent at the point on the circle.

4Draw a circle of diameter 6.4 cm. Take a point R at a distance equal to its diameter from the centre. Draw tangents from point R.

Radius = 6.42 = 3.2 cm and OR = 6.4 cm.

Steps of construction:

  1. Draw a circle with centre O and radius 3.2 cm. Take point R such that OR = 6.4 cm.
  2. Draw seg OR. Draw its perpendicular bisector to get the midpoint M.
  3. With M as centre and radius MO, draw a circle. Let it cut the given circle at A and B.
  4. Draw lines RA and RB. These are the required tangents.
3.2 cm6.4 cmORMAB

RA and RB are the tangents. Here the midpoint M of OR lies on the given circle.

On measuring, RA = RB ≈ 5.5 cm.

5Draw a circle with centre P. Draw an arc AB of 100° measure. Draw tangents to the circle at point A and point B.

Steps of construction:

  1. Draw a circle with centre P and any radius.
  2. Draw radius PA. At P draw ∠ APB = 100° with B on the circle. Then m(arc AB) = 100°.
  3. Draw a line perpendicular to PA at A and a line perpendicular to PB at B.
  4. These lines are the required tangents. They meet at T.
100°PABT

∠ APB = 100°, so m(arc AB) = 100°.

In □ PATB, ∠ A = ∠ B = 90°, so ∠ ATB = 360° − 90° − 90° − 100° = 80°.

6Draw a circle of radius 3.4 cm and centre E. Take a point F on the circle. Take another point A such that E − F − A and FA = 4.1 cm. Draw tangents to the circle from point A.

E − F − A, so EA = EF + FA = 3.4 + 4.1 = 7.5 cm.

Steps of construction:

  1. Draw a circle with centre E and radius 3.4 cm. Take point F on it.
  2. Draw ray EF and mark A on it such that FA = 4.1 cm.
  3. Draw the perpendicular bisector of seg EA to get its midpoint M.
  4. With M as centre and radius ME, draw a circle. Let it cut the given circle at C and D.
  5. Draw lines AC and AD. These are the required tangents.
3.4 cm4.1 cmEAMFCD

AC and AD are the tangents from A.

On measuring, AC = AD ≈ 6.7 cm.

7Δ ABC ∼ Δ LBN. In Δ ABC, AB = 5.1 cm, ∠ B = 40°, BC = 4.8 cm, ACLN = 47. Construct Δ ABC and Δ LBN.

Analysis:

Δ ABC ∼ Δ LBN and vertex B is common.

ABLB = BCBN = ACLN = 47

So the sides of Δ LBN are longer. N lies on ray BC with BN = 74 × BC, and L lies on ray BA.

BN = 74 × 4.8 = 8.4 cm and BL = 74 × 5.1 ≈ 8.9 cm.

Steps of construction:

  1. Construct Δ ABC with AB = 5.1 cm, ∠ B = 40° and BC = 4.8 cm.
  2. Draw a ray from B on the side of BC opposite to A. Mark 7 equal parts on it with the compass.
  3. Join point 4 to C. Through point 7 draw a line parallel to 4C. Let it meet ray BC at N.
  4. Through N draw a line parallel to CA. Let it meet ray BA at L.
  5. Δ LBN is the required triangle.
12345674.8 cm5.1 cm40°BACLN

Δ ABC and Δ LBN (green). Green lines show the required figure. Thin blue lines are construction lines.

8Construct Δ PYQ such that, PY = 6.3 cm, YQ = 7.2 cm, PQ = 5.8 cm. If YZYQ = 65, then construct Δ XYZ similar to Δ PYQ.

Analysis:

Δ XYZ ∼ Δ PYQ and vertex Y is common.

YZYQ = XYPY = XZPQ = 65

So the sides of Δ XYZ are longer. Z lies on ray YQ with YZ = 65 × YQ, and X lies on ray YP.

YZ = 65 × 7.2 = 8.64 cm, XY = 65 × 6.3 = 7.56 cm, XZ = 65 × 5.8 = 6.96 cm.

Steps of construction:

  1. Construct Δ PYQ with YQ = 7.2 cm, PY = 6.3 cm and PQ = 5.8 cm.
  2. Draw a ray from Y on the side of YQ opposite to P. Mark 6 equal parts on it with the compass.
  3. Join point 5 to Q. Through point 6 draw a line parallel to 5Q. Let it meet ray YQ at Z.
  4. Through Z draw a line parallel to QP. Let it meet ray YP at X.
  5. Δ XYZ is the required triangle.
1234567.2 cm6.3 cm5.8 cmYPQXZ

Δ PYQ and Δ XYZ (green). Green lines show the required figure. Thin blue lines are construction lines.

The lengths 8.64 cm and 7.56 cm cannot be measured exactly with a scale. That is why the equal-parts method is used here.

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