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

Frank Solutions for Class 9 Maths Chapter 14 Constructions of Triangles consists of step by step answers to clear students’ doubts instantly. These solutions curated by teachers with years of experience are easy to understand. Students who practise these solutions regularly, gain expertise in solving tricky problems within a short duration of time. The solutions assist students to work on the important questions, which are necessary from an exam point of view. Here, students can make use of Frank Solutions for Class 9 Maths Chapter 14 Constructions of Triangles PDF free, from the links given below

Chapter 14 provides an idea of constructing triangles with step by step explanation. Students study how to construct a triangle when the lengths of all three sides are given. Expert faculty prepare these solutions to help students in the exams.

## 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 centre and radius 7.2 cm, draw an arc

3. Taking B as 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 centre and radius 5.5 cm, draw an arc

3. Taking Q as 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 centre and radius 4.2 cm, draw an arc

3. Taking E as 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 centre, draw an arc to cut BC at point P

3. Taking P as 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 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 centre, draw an arc meeting XY at point L

3. Taking L as centre and same radius, cut the arc at point M and then from M, with 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 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 centre, draw an arc meeting PQ at point M

3. Taking M as centre and 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 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 centre, draw an arc meeting BC at point M

3. Taking M as centre and 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 centre, draw an arc meeting BC at point P

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

8. Taking Q and R as 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 centre, draw an arc meeting PQ at point A

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

4. Now, taking B and C as 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 centre, draw an arc meeting PQ at S

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

8. Now, taking T and U as 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 centre, draw an arc cutting DE at point P

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

4. Taking Q and R as 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 centre, draw an arc meeting DE at point S

7. Taking S as centre and 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 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 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 meets at a 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 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 centre, draw an arc cutting RM at point L

(b)Taking L as centre and same radius cut the arc at point 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 the point D on SU draw ND perpendicular to SU

4. Taking D as 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 centre, draw an arc cutting AD at point L

(b)Taking L as centre and 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 are 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 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 centre, draw an arc cutting XT at point L

(b) Taking L as centre and same radius, cut the arc at point 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 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 centre, construct âˆ SQR = 450

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

4. Join TR

5. Draw 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 perpendicular bisector of TF which cut 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 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 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 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 triangle is 6.4 cm, and the base angles are 600 and 450

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

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

Solution:

(i) Perimeter of 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 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 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 centre, draw an arc cutting YZ at point L

3. Taking L as centre and 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 centre, draw an arc cutting YZ at point P

6. Taking P as centre and same radius, cut the arc at Q, and then taking Q as 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 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 centre and radius 3.5 cm, mark a point P on XY.

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

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