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Std 9 Mathematics Part 2 · Chapter 4

4. Constructions of Triangles

Solutions to Practice Sets 4.1 to 4.3 and Problem Set 4, with analysis, steps of construction and fair figures. Maharashtra State Board (Balbharati), Std 9 Mathematics Part 2 (Geometry).

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Practice Set 4.1

1Construct ΔPQR, in which QR = 4.2 cm, m∠Q = 40° and PQ + PR = 8.5 cm.

Analysis (rough figure):

Take seg QR = 4.2 cm and draw ray QT making an angle of 40° with QR at Q. Point P lies on ray QT.

Mark point D on ray QT such that QD = QP + PR = 8.5 cm. Since QP + PD = QD = QP + PR, we get PD = PR.

So point P is equidistant from D and R; it lies on the perpendicular bisector of seg DR. Hence P is the point where ray QT meets that perpendicular bisector.

Steps of construction:

  1. Draw seg QR of length 4.2 cm.
  2. Draw ray QT such that ∠TQR = 40°.
  3. Mark point D on ray QT such that QD = 8.5 cm.
  4. Draw seg DR.
  5. Construct the perpendicular bisector of seg DR.
  6. Name the point where the perpendicular bisector meets ray QT as P.
  7. Draw seg PR. ΔPQR is the required triangle.
PQRDT40°4.2 cm

Fair figure (reduced in size). Dashed lines are construction lines.

Check: in an accurate construction, QP ≈ 5.2 cm and PR ≈ 3.3 cm, and 5.2 + 3.3 = 8.5 cm.

2Construct ΔXYZ, in which YZ = 6 cm, XY + XZ = 9 cm, ∠XYZ = 50°.

Analysis (rough figure):

Take seg YZ = 6 cm and draw ray YT making an angle of 50° with YZ at Y. Point X lies on ray YT.

Mark point D on ray YT such that YD = YX + XZ = 9 cm. Since YX + XD = YD = YX + XZ, we get XD = XZ.

So point X is equidistant from D and Z; it lies on the perpendicular bisector of seg DZ. Hence X is the point where ray YT meets that perpendicular bisector.

Steps of construction:

  1. Draw seg YZ of length 6 cm.
  2. Draw ray YT such that ∠TYZ = 50°.
  3. Mark point D on ray YT such that YD = 9 cm.
  4. Draw seg DZ.
  5. Construct the perpendicular bisector of seg DZ.
  6. Name the point where the perpendicular bisector meets ray YT as X.
  7. Draw seg XZ. ΔXYZ is the required triangle.
XYZDT50°6 cm

Fair figure (reduced in size). Dashed lines are construction lines.

Check: in an accurate construction, YX ≈ 4.4 cm and XZ ≈ 4.6 cm, and 4.4 + 4.6 = 9 cm.

3Construct ΔABC, in which BC = 6.2 cm, ∠ACB = 50°, AB + AC = 9.8 cm.

Here the given angle is at C, so the ray is drawn at C and the sum is marked along it.

Analysis (rough figure):

Take seg BC = 6.2 cm and draw ray CP making an angle of 50° with CB at C. Point A lies on ray CP.

Mark point D on ray CP such that CD = CA + AB = 9.8 cm. Since CA + AD = CD = CA + AB, we get AD = AB.

So point A is equidistant from D and B; it lies on the perpendicular bisector of seg DB. Hence A is the point where ray CP meets that perpendicular bisector.

Steps of construction:

  1. Draw seg BC of length 6.2 cm.
  2. Draw ray CP such that ∠PCB = 50°.
  3. Mark point D on ray CP such that CD = 9.8 cm.
  4. Draw seg DB.
  5. Construct the perpendicular bisector of seg DB.
  6. Name the point where the perpendicular bisector meets ray CP as A.
  7. Draw seg AB. ΔABC is the required triangle.
ACBDP50°6.2 cm

Fair figure (reduced in size). Dashed lines are construction lines.

Check: in an accurate construction, CA ≈ 5.0 cm and AB ≈ 4.8 cm, and 5.0 + 4.8 = 9.8 cm.

4Construct ΔABC, in which BC = 3.2 cm, ∠ACB = 45° and perimeter of ΔABC is 10 cm.

AB + AC = perimeter − BC = 10 − 3.2 = 6.8 cm

Analysis (rough figure):

Take seg BC = 3.2 cm and draw ray CP making an angle of 45° with CB at C. Point A lies on ray CP.

Mark point D on ray CP such that CD = CA + AB = 6.8 cm. Since CA + AD = CD = CA + AB, we get AD = AB.

So point A is equidistant from D and B; it lies on the perpendicular bisector of seg DB. Hence A is the point where ray CP meets that perpendicular bisector.

Steps of construction:

  1. Draw seg BC of length 3.2 cm.
  2. Draw ray CP such that ∠PCB = 45°.
  3. Mark point D on ray CP such that CD = 6.8 cm.
  4. Draw seg DB.
  5. Construct the perpendicular bisector of seg DB.
  6. Name the point where the perpendicular bisector meets ray CP as A.
  7. Draw seg AB. ΔABC is the required triangle.
ACBDP45°3.2 cm

Fair figure (reduced in size). Dashed lines are construction lines.

Check: in an accurate construction, CA ≈ 4.0 cm and AB ≈ 2.8 cm, and 4.0 + 2.8 = 6.8 cm.

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Practice Set 4.2

1Construct ΔXYZ, such that YZ = 7.4 cm, ∠XYZ = 45° and XY − XZ = 2.7 cm.

Analysis (rough figure):

YX − XZ = 2.7 cm, so YX > XZ. Draw seg YZ = 7.4 cm and ray YL making an angle of 45° at Y; point X lies on ray YL.

Take point D on ray YL such that YD = 2.7 cm. Then Y − D − X and DX = YX − YD = YX − 2.7 = XZ.

So X is equidistant from D and Z, and lies on the perpendicular bisector of seg DZ.

Steps of construction:

  1. Draw seg YZ of length 7.4 cm.
  2. Draw ray YL such that ∠LYZ = 45°.
  3. Take point D on ray YL such that YD = 2.7 cm.
  4. Draw seg DZ and construct its perpendicular bisector.
  5. Name the point where the perpendicular bisector meets ray YL as X.
  6. Draw seg XZ. ΔXYZ is the required triangle.
XYZDL45°7.4 cm

Fair figure (reduced in size). Dashed lines are construction lines.

Check: in an accurate construction, YX ≈ 9.4 cm and XZ ≈ 6.7 cm, and 9.4 − 6.7 = 2.7 cm.

2Construct ΔPQR, such that QR = 6.5 cm, ∠PQR = 40° and PQ − PR = 2.5 cm.

Analysis (rough figure):

QP − PR = 2.5 cm, so QP > PR. Draw seg QR = 6.5 cm and ray QL making an angle of 40° at Q; point P lies on ray QL.

Take point D on ray QL such that QD = 2.5 cm. Then Q − D − P and DP = QP − QD = QP − 2.5 = PR.

So P is equidistant from D and R, and lies on the perpendicular bisector of seg DR.

Steps of construction:

  1. Draw seg QR of length 6.5 cm.
  2. Draw ray QL such that ∠LQR = 40°.
  3. Take point D on ray QL such that QD = 2.5 cm.
  4. Draw seg DR and construct its perpendicular bisector.
  5. Name the point where the perpendicular bisector meets ray QL as P.
  6. Draw seg PR. ΔPQR is the required triangle.
PQRDL40°6.5 cm

Fair figure (reduced in size). Dashed lines are construction lines.

Check: in an accurate construction, QP ≈ 7.3 cm and PR ≈ 4.8 cm, and 7.3 − 4.8 = 2.5 cm.

3Construct ΔABC, such that BC = 6 cm, ∠ABC = 100° and AC − AB = 2.5 cm.

Analysis (rough figure):

AC − AB = 2.5 cm, so AC > AB. Draw seg BC = 6 cm and ray BT making an angle of 100° at B; point A lies on ray BT.

Take point D on the opposite ray BS such that BD = 2.5 cm. Then AD = AB + BD = AB + 2.5 = AC.

So A is equidistant from D and C, and lies on the perpendicular bisector of seg DC.

Steps of construction:

  1. Draw seg BC of length 6 cm.
  2. Draw ray BT such that ∠TBC = 100°.
  3. Take point D on the opposite ray BS of ray BT such that BD = 2.5 cm.
  4. Draw seg DC and construct its perpendicular bisector.
  5. Name the point where the perpendicular bisector meets ray BT as A.
  6. Draw seg AC. ΔABC is the required triangle.
ABCDTS100°6 cm

Fair figure (reduced in size). Dashed lines are construction lines.

Check: in an accurate construction, AB ≈ 10.2 cm and AC ≈ 12.7 cm, and 12.7 − 10.2 = 2.5 cm.

Practice Set 4.3

1Construct ΔPQR, in which ∠Q = 70°, ∠R = 80° and PQ + QR + PR = 9.5 cm.

Analysis (rough figure):

Take points S and T on line QR such that SQ = PQ and RT = PR. Then ST = PQ + QR + RP = 9.5 cm.

In ΔSQP, QS = QP, so ∠QSP = ∠QPS. Their sum equals the exterior angle ∠PQR = 70° (theorem of remote interior angles), so each is 35°.

Similarly ∠RTP = 80° ÷ 2 = 40°. So ΔSPT can be drawn first.

Q lies on the perpendicular bisector of seg PS and R lies on the perpendicular bisector of seg PT.

Steps of construction:

  1. Draw seg ST of length 9.5 cm.
  2. Draw a ray at S making an angle of 35° with ST.
  3. Draw a ray at T making an angle of 40° with TS.
  4. Name the point of intersection of the two rays as P.
  5. Construct the perpendicular bisectors of seg PS and seg PT. Name the points where they meet seg ST as Q and R respectively.
  6. Draw seg PQ and seg PR. ΔPQR is the required triangle.
PQRST35°40°

Fair figure (reduced in size); ST = 9.5 cm. ΔPQR is shaded, with ∠Q = 70° and ∠R = 80°.

Check: in an accurate construction, PQ ≈ 3.9 cm, QR ≈ 2.0 cm, PR ≈ 3.7 cm, and the total is 9.5 cm.

2Construct ΔXYZ, in which ∠Y = 58°, ∠X = 46° and perimeter of triangle is 10.5 cm.

The given angles are at X and Y, so side XY is taken as the base.

Analysis (rough figure):

Take points P and Q on line XY such that PX = ZX and YQ = ZY. Then PQ = ZX + XY + YZ = 10.5 cm.

In ΔPXZ, XP = XZ, so ∠XPZ = ∠XZP. Their sum equals the exterior angle ∠ZXY = 46° (theorem of remote interior angles), so each is 23°.

Similarly ∠YQZ = 58° ÷ 2 = 29°. So ΔPZQ can be drawn first.

X lies on the perpendicular bisector of seg ZP and Y lies on the perpendicular bisector of seg ZQ.

Steps of construction:

  1. Draw seg PQ of length 10.5 cm.
  2. Draw a ray at P making an angle of 23° with PQ.
  3. Draw a ray at Q making an angle of 29° with QP.
  4. Name the point of intersection of the two rays as Z.
  5. Construct the perpendicular bisectors of seg ZP and seg ZQ. Name the points where they meet seg PQ as X and Y respectively.
  6. Draw seg ZX and seg ZY. ΔZXY is the required triangle.
ZXYPQ23°29°

Fair figure (reduced in size); PQ = 10.5 cm. ΔZXY is shaded, with ∠X = 46° and ∠Y = 58°.

Check: in an accurate construction, ZX ≈ 3.5 cm, XY ≈ 4.0 cm, ZY ≈ 3.0 cm, and the total is 10.5 cm.

3Construct ΔLMN, in which ∠M = 60°, ∠N = 80° and LM + MN + NL = 11 cm.

Analysis (rough figure):

Take points P and Q on line MN such that PM = LM and NQ = LN. Then PQ = LM + MN + NL = 11 cm.

In ΔPML, MP = ML, so ∠MPL = ∠MLP. Their sum equals the exterior angle ∠LMN = 60° (theorem of remote interior angles), so each is 30°.

Similarly ∠NQL = 80° ÷ 2 = 40°. So ΔPLQ can be drawn first.

M lies on the perpendicular bisector of seg LP and N lies on the perpendicular bisector of seg LQ.

Steps of construction:

  1. Draw seg PQ of length 11 cm.
  2. Draw a ray at P making an angle of 30° with PQ.
  3. Draw a ray at Q making an angle of 40° with QP.
  4. Name the point of intersection of the two rays as L.
  5. Construct the perpendicular bisectors of seg LP and seg LQ. Name the points where they meet seg PQ as M and N respectively.
  6. Draw seg LM and seg LN. ΔLMN is the required triangle.
LMNPQ30°40°

Fair figure (reduced in size); PQ = 11 cm. ΔLMN is shaded, with ∠M = 60° and ∠N = 80°.

Check: in an accurate construction, LM ≈ 4.3 cm, MN ≈ 2.8 cm, LN ≈ 3.8 cm, and the total is 11 cm.

Problem Set 4

1Construct ΔXYZ, such that XY + XZ = 10.3 cm, YZ = 4.9 cm, ∠XYZ = 45°.

Analysis (rough figure):

Take seg YZ = 4.9 cm and draw ray YT making an angle of 45° with YZ at Y. Point X lies on ray YT.

Mark point D on ray YT such that YD = YX + XZ = 10.3 cm. Since YX + XD = YD = YX + XZ, we get XD = XZ.

So point X is equidistant from D and Z; it lies on the perpendicular bisector of seg DZ. Hence X is the point where ray YT meets that perpendicular bisector.

Steps of construction:

  1. Draw seg YZ of length 4.9 cm.
  2. Draw ray YT such that ∠TYZ = 45°.
  3. Mark point D on ray YT such that YD = 10.3 cm.
  4. Draw seg DZ.
  5. Construct the perpendicular bisector of seg DZ.
  6. Name the point where the perpendicular bisector meets ray YT as X.
  7. Draw seg XZ. ΔXYZ is the required triangle.
XYZDT45°4.9 cm

Fair figure (reduced in size). Dashed lines are construction lines.

Check: in an accurate construction, YX ≈ 6.0 cm and XZ ≈ 4.3 cm, and 6.0 + 4.3 = 10.3 cm.

2Construct ΔABC, in which ∠B = 70°, ∠C = 60°, AB + BC + AC = 11.2 cm.

Analysis (rough figure):

Take points P and Q on line BC such that PB = AB and CQ = AC. Then PQ = AB + BC + CA = 11.2 cm.

In ΔPBA, BP = BA, so ∠BPA = ∠BAP. Their sum equals the exterior angle ∠ABC = 70° (theorem of remote interior angles), so each is 35°.

Similarly ∠CQA = 60° ÷ 2 = 30°. So ΔPAQ can be drawn first.

B lies on the perpendicular bisector of seg AP and C lies on the perpendicular bisector of seg AQ.

Steps of construction:

  1. Draw seg PQ of length 11.2 cm.
  2. Draw a ray at P making an angle of 35° with PQ.
  3. Draw a ray at Q making an angle of 30° with QP.
  4. Name the point of intersection of the two rays as A.
  5. Construct the perpendicular bisectors of seg AP and seg AQ. Name the points where they meet seg PQ as B and C respectively.
  6. Draw seg AB and seg AC. ΔABC is the required triangle.
ABCPQ35°30°

Fair figure (reduced in size); PQ = 11.2 cm. ΔABC is shaded, with ∠B = 70° and ∠C = 60°.

Check: in an accurate construction, AB ≈ 3.8 cm, BC ≈ 3.3 cm, AC ≈ 4.1 cm, and the total is 11.2 cm.

3The perimeter of a triangle is 14.4 cm and the ratio of lengths of its sides is 2 : 3 : 4. Construct the triangle.

Finding the sides: Let the sides be 2k, 3k and 4k.

2k + 3k + 4k = 14.4 ∴ 9k = 14.4 ∴ k = 1.6

Sides: 2k = 3.2 cm, 3k = 4.8 cm, 4k = 6.4 cm

Now construct ΔABC with BC = 6.4 cm, AC = 4.8 cm and AB = 3.2 cm (three sides given).

Steps of construction:

  1. Draw seg BC of length 6.4 cm.
  2. With B as centre and radius 3.2 cm, draw an arc on one side of BC.
  3. With C as centre and radius 4.8 cm, draw an arc cutting the first arc. Name the point of intersection A.
  4. Draw seg AB and seg AC. ΔABC is the required triangle.
ABC6.4 cm3.2 cm4.8 cm

Fair figure (reduced in size)

Sides: 3.2 cm, 4.8 cm, 6.4 cm; check: 3.2 + 4.8 + 6.4 = 14.4 cm.

4Construct ΔPQR, in which PQ − PR = 2.4 cm, QR = 6.4 cm and ∠PQR = 55°.

Analysis (rough figure):

QP − PR = 2.4 cm, so QP > PR. Draw seg QR = 6.4 cm and ray QL making an angle of 55° at Q; point P lies on ray QL.

Take point D on ray QL such that QD = 2.4 cm. Then Q − D − P and DP = QP − QD = QP − 2.4 = PR.

So P is equidistant from D and R, and lies on the perpendicular bisector of seg DR.

Steps of construction:

  1. Draw seg QR of length 6.4 cm.
  2. Draw ray QL such that ∠LQR = 55°.
  3. Take point D on ray QL such that QD = 2.4 cm.
  4. Draw seg DR and construct its perpendicular bisector.
  5. Name the point where the perpendicular bisector meets ray QL as P.
  6. Draw seg PR. ΔPQR is the required triangle.
PQRDL55°6.4 cm

Fair figure (reduced in size). Dashed lines are construction lines.

Check: in an accurate construction, QP ≈ 13.8 cm and PR ≈ 11.4 cm, and 13.8 − 11.4 = 2.4 cm.

This triangle is long and narrow (PQ is about 13.8 cm), so start QR near the bottom-left of the page to leave room for the ray.

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