# Frank Solutions for Class 9 Maths Chapter 14 Constructions of Triangles

Frank Solutions for Class 9 Maths Chapter 14 Constructions of Triangles consist of step-by-step answers to clear students’ doubts instantly. These solutions are curated by the subject experts at BYJU’S with years of experience and are easy to understand. Students who practise these solutions on a regular basis can gain expertise in solving tricky problems quickly. Moreover, it will assist them in working on the important questions necessary from an exam point of view.

Chapter 14 provides a clear idea of constructing triangles with step-by-step explanations, which will help students learn how to construct a triangle when the lengths of all three sides are given. Expert faculty prepared these solutions to help students score good marks in the final exams. They can download Frank Solutions for Class 9 Maths Chapter 14 Constructions of Triangles PDF free from the link given below and practise offline as well.

## Access Frank Solutions for Class 9 Maths Chapter 14 Construction of Triangles

1. Construct a triangle using the given data:

(i) AB = 6.5 cm, BC = 8.4 cm and AC = 7.2 cm

(ii) PQ = 4.8 cm, QR = 6.3 cm and PR = 5.5 cm

(iii) DE = 6.5 cm, EF = 5.8 cm and DF = 4.2 cm

Solution:

(i) AB = 6.5 cm, BC = 8.4 cm and AC = 7.2 cm

Steps of Construction:

1. Draw AB of length 6.5 cm

2. Taking A as the centre and radius 7.2 cm, draw an arc

3. Taking B as the centre and radius 8.4 cm, draw another arc to cut the first arc at point C

4. Now, join AC and BC

Hence,

Triangle ABC is the required triangle

(ii) PQ = 4.8 cm, QR = 6.3 cm and PR = 5.5 cm

Steps of Construction:

1. Draw a line segment PQ of length 4.8 cm

2. Taking P as the centre and radius 5.5 cm, draw an arc

3. Taking Q as the centre and radius 6.3 cm, draw another arc to cut the first arc at point R

4. Now, join PR and QR

Hence,

Triangle PQR is the required triangle

(iii) DE = 6.5 cm, EF = 5.8 cm and DF = 4.2 cm

Steps of Construction:

1. Draw a line segment DE of length 6.5 cm

2. Taking D as the centre and radius 4.2 cm, draw an arc

3. Taking E as the centre and radius 5.8 cm, draw another arc to cut the first arc at point F

4. Now, join DF and EF

Hence,

Triangle DEF is the required triangle

2. Construct a triangle using the given data:

(i) BC = 6 cm, AC = 5.0 cm and âˆ C = 600

(ii) XY = 5.2 cm, XZ = 6.5 cm and âˆ X = 750

(iii) PQ = 6.2 cm, QR = 9.0 cm and âˆ Q = 300

Solution:

(i) BC = 6 cm, AC = 5.0 cm and âˆ C = 600

Steps of Construction:

1. Draw a line segment BC of length 6 cm

2. Taking C as the centre, draw an arc to cut BC at point P

3. Taking P as the centre and the same radius, cut the arc at point Q

4. Draw a ray CX passing a point Q. CX makes an angle of 600 with BC

5. Taking C as the centre and radius 5 cm, cut an arc on CX and mark the point as A

6. Now, join AB

Therefore,

ABC is the required triangle

(ii) XY = 5.2 cm, XZ = 6.5 cm and âˆ X = 750

Steps of Construction:

1. Draw a line segment XY of length 5.2 cm

2. Taking X as the centre, draw an arc meeting XY at point L

3. Taking L as the centre and same radius, cut the arc at point M and then from M, with the same radius, cut the arc at point N

4. Taking M and N as centre bisect âˆ MXN thus formed to draw a ray XP.

5. Again, bisect the âˆ MXP. Let XR be the bisector. XR makes an angle of 750 with XY

6. Taking X as the centre and radius 6.5 cm, cut an arc on XR and mark the point as Z

7. Now, join YZ

Therefore,

XYZ is the required triangle

(iii) PQ = 6.2 cm, QR = 9.0 cm and âˆ Q = 300

Steps of Construction:

1. Draw a line segment PQ of length 6.2 cm

2. Taking Q as the centre, draw an arc meeting PQ at point M

3. Taking M as the centre and the same radius, cut the arc at point N

4. Join QN

5. Now, bisect âˆ NQP. Let QY be the bisector. QY makes an angle of 300 with PQ

6. Taking Q as the centre and radius 9 cm, cut an arc on QY and mark the point as R

7. Join PR

Therefore,

PQR is the required triangle

3. Construct a triangle using the given data:

(i) BC = 6.0 cm, âˆ B = 600 and âˆ C = 450

(ii) PQ = 6.2 cm, âˆ P = 1050 and âˆ Q = 450

(iii) DE = 5 cm, âˆ D = 750 and âˆ E = 600

Solution:

(i) BC = 6.0 cm, âˆ B = 600 and âˆ C = 450

Steps of Construction:

1. Draw a line segment BC of length 6 cm.

2. Taking B as the centre, draw an arc meeting BC at point M

3. Taking M as the centre and the same radius, cut the arc at point N

4. Produce BN to BX.

5. BX makes an angle of 600 with BC

6. Taking C as the centre, draw an arc meeting BC at point P

7. Taking P as the centre and same radius, cut the arc at Q and taking Q as the centre and same radius, cut the arc at R

8. Taking Q and R as the centre, cut arcs and draw CY perpendicular to BC

9. Bisect âˆ YCB. Let CZ be the bisector. Here, CZ makes an angle of 450 with BC

10. Mark the point as A, where BX and CZ meet each other

Hence, ABC is the required triangle

(ii) PQ = 6.2 cm, âˆ P = 1050 and âˆ Q = 450

Steps of Construction:

1. Draw a line segment PQ of length 6.2 cm

2. Taking P as the centre, draw an arc meeting PQ at point A

3. Taking A as the centre and same radius, cut the arc at point B and taking BQ as the centre and same radius, cut the arc at point C

4. Now, taking B and C as the centre, cut arcs and draw PM perpendicular to PQ

5. Let PX be the bisector to bisect âˆ MPC. PX makes an angle of 1050 with PQ

6. Taking Q as the centre, draw an arc meeting PQ at S

7. Taking S as the centre and same radius, cut the arc at T and taking T as the centre and same radius, cut the arc at U

8. Now, taking T and U as the centre, cut the arcs and draw QN perpendicular to PQ

9. Let QY be the bisector to bisect âˆ NQP. QY makes an angle of 450 with PQ

10. Mark the point as R, where PX and QY meet each other

Hence,

PQR is the required triangle

(iii) DE = 5 cm, âˆ D = 750 and âˆ E = 600

Steps of Construction:

1. Draw a line segment DE of length 5 cm

2. Taking D as the centre, draw an arc cutting DE at point P

3. Taking P as the centre and same radius, cut the arc at Q and then taking Q as the centre with the same radius, cut the arc at point R

4. Taking Q and R as the centre, bisect âˆ RDQ thus formed to draw a ray XD

5. Let DY be the bisector to bisect the âˆ XDQ. DY makes an angle of 750 with DE

6. Taking E as the centre, draw an arc meeting DE at point S

7. Taking S as the centre and the same radius, cut the arc at point T

8. Produce ET to EZ.

9. EZ makes an angle of 600 with DE

10. Mark the point as F, where DY and EZ meet each other

Therefore,

DEF is the required triangle

4. Construct a right-angled triangle in which:

(a) Side AB = 4.5 cm and hypotenuse AC = 7 cm

(b) Side DE = 6 cm and âˆ E = 300, âˆ D = 900

(c) QP = QR and hypotenuse PR = 7 cm

Solution:

(a) Side AB = 4.5 cm and hypotenuse AC = 7 cm

Steps of Construction:

1. Draw a line segment AB of length 4.5 cm

2. At B, construct a ray BP such as âˆ ABP = 900

3. Taking A as the centre and radius 7 cm, draw an arc to cut BP at point C

4. Now, join AC

Hence,

ABC is the required triangle

(b) Side DE = 6 cm and âˆ E = 300, âˆ D = 900

Steps of Construction:

1. Draw a line segment DE of length 6 cm

2. At D, construct a ray DP such that âˆ PDE = 900

3. Taking E as the centre, draw âˆ DEM of angle 300

4. Ray DP and ray EM intersect at point F

Therefore,

DEF is the required triangle

(c) QP = QR and hypotenuse PR = 7 cm

Steps of Construction:

1. Draw a line segment PR of length 7 cm

2. Draw a ray PT such that âˆ RPT = 450 and a ray RS such that âˆ PRS = 450

3. Here, ray RS and ray PT meets at point Q

Therefore,

PQR is the required triangle

Now,

In â–³PQR,

QP = QR â€¦â€¦ (given)

âˆ QPR = âˆ QRP â€¦â€¦ (angles opposite to two equal sides are equal)

Here, hypotenuse PR = 7cm,

So,

âˆ PQR = 900

Hence,

âˆ QPR + âˆ QRP = 900

âˆ QPR = âˆ QRP = 450

5. Construct an isosceles triangle in which:

(a) AB = AC, BC = 6 cm and âˆ B = 750

(b) XY = XZ, YZ = 5.5 cm and âˆ X = 600

(c) PQ = QR, PR = 4.5 cm and âˆ R = 600

Solution:

(a) AB = AC, BC = 6 cm and âˆ B = 750

In â–³ABC,

AB = AC â€¦â€¦ (given)

âˆ ACB = âˆ ABC = 750

Steps of Construction:

1. Draw a line segment BC of length 6 cm

2. Construct âˆ BCM = 750 and âˆ CBN = 750

3. Ray CM and ray BN meet at point A

Therefore,

ABC is a required triangle

(b) XY = XZ, YZ = 5.5 cm and âˆ X = 600

In â–³XYZ,

XY = XZ â€¦.. (given)

âˆ XZY = âˆ XYZ â€¦.. (1)

âˆ X = 600 â€¦â€¦ (given)

Now,

âˆ X + âˆ Y + âˆ Z = 1800

600 + âˆ Y + âˆ Y = 1800 â€¦â€¦â€¦ [from (1)]

We get,

2âˆ Y = 1800 – 600

2âˆ Y = 1200

We get,

âˆ Y = 600 = âˆ Z

Steps of Construction:

1. Draw a line segment YZ of length 5.5 cm

2. Construct an âˆ YZP = 600 and âˆ ZYQ = 600

3. Ray ZP and YQ meet at a point X

Therefore,

XYZ is the required triangle

(c) PQ = QR, PR = 4.5 cm and âˆ R = 600

In â–³PQR,

PQ = QR â€¦â€¦â€¦. (given)

âˆ PRQ = âˆ RPQ = 600

Steps of Construction:

1. Draw a line segment PR of length 4.5 cm

2. Construct âˆ PRU = 600 and âˆ RPV = 600

3. Ray RU and PV meet at a point Q

Therefore,

PQR is the required triangle

6. Construct an isosceles triangle using the given data:

(i) Altitude RM = 5 cm and vertex âˆ R = 1200

(ii) Altitude AD = 4 cm and vertex âˆ A = 900

(iii) Altitude XT = 6.8 cm and vertex âˆ X = 300

Solution:

(i) Altitude RM = 5 cm and vertex âˆ R = 1200

Steps of Construction:

1. Draw a line segment SU of any length

2. Take a point M on SU

3. Through the point M on SU, draw NM perpendicular to SU

4. Taking M as the centre and radius 5 cm, draw an arc to cut NM at point R

5. Construct âˆ MRP = âˆ MRQ = (1 / 2) Ã— 1200 = 600

(a)Taking R as the centre, draw an arc cutting RM at point L

(b)Taking L as the centre and the same radius, cut the arc at points X and Y

(c) Now, join RX and RY and produce them to T and V, respectively. RT and RV make an angle of measure 600 with RM

(d) Mark the points as P and Q where RT and RV meet SU

Hence, PQR is the required triangle

(ii) Altitude AD = 4 cm and vertex âˆ A = 900

Steps of Construction:

1. Draw a line segment SU of any length

2. Take a point D on SU

3. Through point D on SU, draw ND perpendicular to SU

4. Taking D as the centre and radius 4 cm, draw an arc to cut ND at point A

5. Construct âˆ DAB = âˆ DAC = (1 / 2) Ã— 900 = 450

(a) Taking A as the centre, draw an arc cutting AD at point L

(b)Taking L as the centre and the same radius, cut an arc at points X and Y

(c) Using X and Y, draw PQ perpendicular to AD

(d)Bisect âˆ PAD and âˆ QAD. Let AT and AV be the bisectors. AT and AV make an angle of 450 with AD

(e) Mark the points as B and C where AT and AV meet SU

Hence,

ABC is the required triangle

(iii) Altitude XT = 6.8 cm and vertex âˆ X = 300

Steps of Construction:

1. Draw a line segment SU of any length

2. Take a point T on SU

3. Through the point T on SU, draw NT perpendicular to SU

4. Taking T as the centre and radius 6.8 cm, draw an arc to cut NT at point X

5. Construct âˆ TXY = âˆ TXZ = (1 / 2) Ã— 300 = 150

(a) Taking X as the centre, draw an arc cutting XT at point L

(b) Taking L as the centre and the same radius, cut the arc at points P and Q

(c) Join PX and QX

(d) Bisect âˆ PXT and âˆ QXT. Let XA and XB be the bisectors

(e) Again, bisect âˆ AXT and âˆ BXT. Let XR and XV be the bisectors. XR and XV make an angle of 150 with XT

(f) Mark the points as Y and Z where XR and XV meet SU

Hence, XYZ is the required triangle

7. Construct an isosceles right-angled triangle whose hypotenuse is of length 6 cm.

Solution:

Let â–³UVW be the isosceles right-angled triangle,

Right angled at point U

Hypotenuse VW = 6 cm

UV = UW

âˆ UWV = âˆ UVW

âˆ U = 900

âˆ UWV + âˆ UVW = 900

2 âˆ UWV = 900

We get,

âˆ UWV = âˆ UVW = 450

Steps of Construction:

1. Draw a line segment of length 6 cm

2. Construct âˆ WVY = 450 and âˆ VWX = 450

3. Ray VY and ray WX meet at point U

Hence,

UVW is the required triangle

8. Construct an equilateral triangle using the data:

(i) Altitude AD = 5 cm

(ii) Altitude PM = 3.6 cm

(iii) Altitude OM = 5.8 cm

Solution:

(i) Altitude AD = 5 cm

Steps of Construction:

1. Draw a line segment PQ of any length

2. Through a point D on PQ, draw AD perpendicular to PQ such that AD = 5 cm

3. Through A, draw AB and AC making angles equal to 300 with AD and meeting PQ at points B and C, respectively

Hence,

ABC is the required triangle

(ii) Altitude PM = 3.6 cm

Steps of Construction:

1. Draw a line segment ST of any length

2. Through a point M on ST, draw PM perpendicular to ST such that PM is of length 3.6 cm

3. Through P, draw PQ and PR, making angles equal to 300 with PM and meeting ST at points Q and R, respectively

Hence,

PQR is the required triangle

(iii) Altitude OM = 5.8 cm

Steps of Construction:

1. Draw a line segment ST of any length

2. Through a point M on ST, draw PM perpendicular to ST such that OM is of length 5.8 cm

3. Through O, draw OQ and OR, making angles equal to 300 with OM and meeting ST at Q and R, respectively

Hence,

OQR is the required triangle

9. Construct a triangle using the following data:

(i) XY + YZ = 5.6 cm, XZ = 4.5 cm and âˆ X = 450

(ii) PQ + PR = 10.6 cm, QR = 4.8 cm and âˆ R = 450

(iii) DE + DF = 10.3 cm, EF = 6.4 cm and âˆ E = 750

Solution:

(i) XY + YZ = 5.6 cm, XZ = 4.5 cm and âˆ X = 450

Steps of Construction:

1. Draw a line segment XZ of length 4.5 cm

2. Taking X as the centre, construct âˆ SXZ = 450

3. Cut XT on XS such that XT = 5.6 cm

4. Join TZ

5. Draw a perpendicular bisector of TZ, which cuts XT at point Y

6. Join YZ

Hence,

XYZ is the required triangle

(ii) PQ + PR = 10.6 cm, QR = 4.8 cm and âˆ R = 450

Steps of Construction:

1. Draw a line segment QR of length 4.8 cm

2. Taking Q as the centre, construct âˆ SQR = 450

3. Cut QT on QS such that QT = 10.6 cm

4. Join TR

5. Draw a perpendicular bisector of TR, which cuts QT at point P

6. Now, join PR

Hence,

PQR is the required triangle

(iii) DE + DF = 10.3 cm, EF = 6.4 cm and âˆ E = 750

Steps of Construction:

1. Draw a line segment EF of length 6.4 cm

2. Taking E as centre, construct âˆ SEF = 750

3. Cut ET on ES such that ET = 10.3 cm

4. Join TF

5. Draw a perpendicular bisector of TF, which cuts ET at point D

6. Join DF

Hence,

DEF is the required triangle

10. Construct a triangle using the given data:

(i) PQ â€“ PR = 1.5 cm, QR = 6.0 cm and âˆ Q = 750

(ii) AB â€“ AC = 1.2 cm, BC = 6.0 cm and âˆ B = 600

(iii) XY â€“ XZ = 1.5 cm, YZ = 3.4 cm and âˆ X = 450

Solution:

(i) PQ â€“ PR = 1.5 cm, QR = 6.0 and âˆ Q = 450

Steps of Construction:

1. Draw a line segment QR of length 6 cm

2. Taking Q as the centre, draw âˆ TQR = 450

3. From Q, cut an arc of measure 1.5 cm on QT and name it as S

4. Now, join S and R

5. Draw a perpendicular bisector of SR, which cuts QT at point P

6. Join PR

Therefore, PQR is the required triangle

(ii) AB â€“ AC = 1.2 cm, BC = 6.0 cm and âˆ B = 600

Steps of Construction:

1. Draw a line segment BC of length = 6 cm

2. Taking B as the centre, draw âˆ TBC = 600

3. From B, cut an arc of measure 1.2 cm on BT and name it as point S

4. Now, join S and C

5. Draw a perpendicular bisector of SC which cuts BT at point A

6. Join AC

Therefore, ABC is the required triangle

(iii) XY â€“ XZ = 1.5 cm, YZ = 3.4 cm and âˆ X = 450

Steps of Construction:

1. Draw a line segment YZ of length 3.4 cm

2. Taking Y as the centre, draw âˆ TYZ = 450

3. From Y, cut an arc of measure 1.5 cm on YT and name it as S

4. Now, join S and Z

5. Draw a perpendicular bisector of SZ which cuts YT at point X

6. Join XZ

Therefore, XYZ is the required triangle

11. Construct a triangle using the given data:

(i) Perimeter of the triangle is 6.4 cm, and the base angles are 600 and 450

(ii) Perimeter of the triangle is 9 cm, and the base angles are 600 and 450

(iii) Perimeter of the triangle is 10.6 cm, and the base angles are 600 and 900

Solution:

(i) Perimeter of the triangle is 6.4 cm, and the base angles are 600 and 450

Steps of Construction:

1. Draw DE of measure 6.4 cm

2. Draw DP and EQ such that âˆ PDE = 450 and âˆ QED = 600

3. Draw AD and AE, the bisectors of angles PDE and QED, respectively, intersecting each other at point A

4. Draw perpendicular bisectors of AD and AE, intersecting DE at B and C, respectively

5. Now, join AB and AC

Therefore, ABC is the required triangle

(ii) Perimeter of the triangle is 9 cm, and the base angles are 600 and 450

Steps of Construction:

1. Draw DE of measure 9 cm

2. Draw DP and EQ, such that âˆ PDE = 450 and âˆ QED = 600

3. Draw AD and AE, the bisectors of angles PDE and QED, respectively, intersecting each other at point A

4. Draw perpendicular bisectors of AD and AE, intersecting DE at B and C, respectively

5. Now, join AB and AC

Therefore, ABC is the required triangle

(iii) Perimeter of the triangle is 10.6 cm, and the base angles are 600 and 900

Steps of Construction:

1. Draw DE of measure 10.6 cm

2. Draw DP and EQ such that âˆ PDE = 900 and âˆ QED = 600

3. Draw AD and AE, the bisectors of angles PDE and QED, respectively, intersecting each other at point A

4. Draw perpendicular bisectors of AD and AE, intersecting DE at B and C, respectively

5. Now, join AB and AC

Therefore, ABC is the required triangle

12. Construct a â–³XYZ with YZ = 7.5 cm, âˆ Y = 600 and âˆ Z = 450. Draw the bisectors of âˆ Y and âˆ Z. If these bisectors meet at O, measure angle YOZ.

Solution:

Steps of Construction:

1. Draw a line segment YZ of length 7.5 cm

2. Taking Y as the centre, draw an arc cutting YZ at point L

3. Taking L as the centre and the same radius, cut the arc at M

4. Join Y and M. Produce YM to S. Now, YS makes an angle of 600 with YZ

5. With Z as the centre, draw an arc cutting YZ at point P

6. Taking P as the centre and same radius, cut the arc at Q, and then taking Q as the centre and same radius, cut the arc at R. Now, using Q and R, draw UZ perpendicular to YZ

7. Bisect âˆ UZY. Let TZ be the bisector. TZ makes an angle of 450 with YZ

8. Bisect âˆ SYZ and âˆ TZY

9. Consider the point as O, where the bisectors of âˆ SYZ and âˆ TZY meet

10. On measuring, we get âˆ YOZ = 127.50

13. Construct a â–³RST with side ST = 5.4 cm, âˆ RST = 600 and the perpendicular from R on ST = 3.0 cm.

Solution:

Steps of Construction:

1. Draw a line segment ST of length 5.4 cm

2. Taking S as the centre, draw XS such that âˆ XST = 600

3. Draw a straight line PQ parallel to ST at a distance of 3 cm from ST

4. PQ meets XS at point R

5. Now, join RT

Therefore, âˆ RST is the required triangle with angle 600

14. Construct a â–³PQR with âˆ Q = 600, âˆ R = 450, and the perpendicular from P to QR be 3.5 cm. Measure PQ.

Solution:

Steps of Construction:

1. Draw a line segment ST of any length.

2. From any point Y on ST, draw XY perpendicular to ST.

3. Taking Y as the centre and radius 3.5 cm, mark a point P on XY.

4. Taking P as the centre, draw an arc cutting XY at point L.

5. With L as the centre and same radius, cut the arc at O and M. With M as the centre and same radius, cut the arc at N

6. Draw PZ perpendicular to XY using M and N.

7. Bisect the angles OPY and ZPY making 300 and 450 angles with PY, respectively.

(In triangle PQY, âˆ PQY = 600, âˆ QYP = 900, hence âˆ QPY = 300 and in triangle PYR, âˆ YRP = 450, âˆ RYP = 900, hence âˆ YPR = 450)

8. Now, join PQ and PR

9. PQR is the required triangle

10. On measuring, PQ = 4.1 cm.

15. Construct a â–³ABC, right angled at B, with a perimeter of 10 cm and one acute angle of 600.

Solution:

Steps of Construction:

1. Draw DE of length 10 cm

2. Draw DP and EQ such that âˆ PDE = 900 and âˆ QED = 600.

3. Draw AD and AE, the bisectors of angles PDE and QED, respectively, intersecting each other at point A.

4. Draw perpendicular bisectors AD and AE, intersecting DE at points B and C, respectively.

5. Now, join AB and AC.

Hence, ABC is the required triangle.