# Frank Solutions for Chapter 14 Construction of Triangles Class 9 Mathematics ICSE

### Exercise 14.1

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

(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 C = 60°

(ii) XY = 5.2 cm, XZ = 6.5 cm and X = 75°

(iii) PQ = 6.2 cm, QR = 9.0 cm and Q = 30°

(i) BC = 6 cm, AC = 5.0 cm and ∠C = 60°

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 60° 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 ∠X = 75°

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 ∠MXN thus formed to draw a ray XP.
5. Again bisect the ∠MXP. Let XR be the bisector. XR makes an angle of 75° 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 ∠Q = 30°

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 ∠NQP. Let QY be the bisector. QY makes an angle of 30° 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, B = 60° and C = 45°

(ii) PQ = 6.2 cm, P = 105° and Q = 45°

(iii) DE = 5 cm, D = 75° and E = 60°

(i) BC = 6.0 cm, ∠B = 60° and ∠C = 45°

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 60° 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 ∠YCB. Let CZ be the bisector. Here, CZ makes an angle of 45° with BC
10. Mark the point as A, where BX and CZ meet each other
11. Hence, ABC is the required triangle

(ii) PQ = 6.2 cm, ∠P = 105° and ∠Q = 45°

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 ∠MPC. PX makes an angle of 105° 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 ∠NQP. QY makes an angle of 450with PQ
10. Mark the point as R, where PX and QY meet each other

Hence,

PQR is the required triangle

(iii) DE = 5 cm, ∠D = 75° and ∠E = 60°

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 ∠RDQ thus formed to draw a ray XD
5. Let DY be the bisector, to bisect the ∠XDQ. DY makes an angle of 750with 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 60° 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 E = 30°, D = 90°

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

(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 ∠ABP = 90°
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 ∠E = 30°, ∠D = 90°

Steps of Construction:

1. Draw a line segment DE of length 6 cm
2. At D, construct a ray DP such that ∠PDE = 90°
3. Taking E as centre, draw ∠DEM of angle 30°
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 ∠RPT = 45° and a ray RS such that ∠PRS = 45°
3. Here, ray RS and ray PT meets at point Q

Therefore,

PQR is the required triangle.

Now,

In △PQR,

QP = QR …(given)

∠QPR = ∠QRP …(angles opposite to two equal sides are equal)

Here, hypotenuse PR = 7cm,

So,

∠PQR = 90°

Hence,

∠QPR + ∠QRP = 90°

∠QPR = ∠QRP = 45°

5. Construct an isosceles triangle in which:

(a) AB = AC, BC = 6 cm and B = 75°

(b) XY = XZ, YZ = 5.5 cm and X = 60°

(c) PQ = QR, PR = 4.5 cm and R = 60°

(a) AB = AC, BC = 6 cm and ∠B = 750

In △ABC,

AB = AC …(given)

∠ACB = ∠ABC = 75°

Steps of Construction:

1. Draw a line segment BC of length 6 cm
2. Construct ∠BCM = 75° and ∠CBN = 75°
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 ∠X = 60°

In △XYZ,

XY = XZ …(given)

∠XZY = ∠XYZ …(1)

∠X = 60° …(given)

Now,

∠X + ∠Y + ∠Z = 180°

⇒ 600 + ∠Y + ∠Y = 180° …[from (1)]

We get,

2∠Y = 180° – 60°

⇒ 2∠Y = 120°

We get,

∠Y = 60° = ∠Z

Steps of Construction:

1. Draw a line segment YZ of length 5.5 cm
2. Construct an ∠YZP = 60° and ∠ZYQ = 60°
3. Ray ZP and YQ meet at a point X

Therefore,

XYZ is the required triangle

(c) PQ = QR, PR = 4.5 cm and ∠R = 60°

In △PQR,

PQ = QR …(given)

∠PRQ = ∠RPQ = 60°

Steps of Construction:

1. Draw a line segment PR of length 4.5 cm
2. Construct ∠PRU = 60° and ∠RPV = 60°
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 R = 120°

(ii) Altitude AD = 4 cm and vertex A = 90°

(iii) Altitude XT = 6.8 cm and vertex X = 30°

(i) Altitude RM = 5 cm and vertex ∠R = 120°

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 ∠MRP = ∠MRQ = (1/2) × 120° = 60°

(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 60° 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 ∠A = 90°

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 ∠DAB = ∠DAC = (1 / 2) × 90° = 45°

(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 ∠PAD and ∠QAD. Let AT and AV are the bisectors. AT and AV make an angle of 45° 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 ∠X = 30°

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 ∠TXY = ∠TXZ = (1 / 2) × 30° = 15°

(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 ∠PXT and ∠QXT. Let XA and XB be the bisectors

(e) Again bisect ∠AXT and ∠BXT. Let XR and XV be the bisectors. XR and XV make an angle of 15° 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.

Let △UVW be the isosceles right-angled triangle,

Right angled at point U

Hypotenuse VW = 6 cm

UV = UW

⇒ ∠UWV = ∠UVW

⇒ ∠U = 90°

∠UWV + ∠UVW = 90°

⇒ 2 ∠UWV = 90°

We get,

∠UWV = ∠UVW = 45°

Steps of Construction:

1. Draw a line segment of length 6 cm
2. Construct ∠WVY = 45° and ∠VWX = 45°
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

(i) Altitude AD = 5 cm

Steps of Construction:

Draw a line segment PQ of any length

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

Through A, draw AB and AC making angles equal to 30° 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 30° 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 30° 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 X = 45°

(ii) PQ + PR = 10.6 cm, QR = 4.8 cm and R = 45°

(iii) DE + DF = 10.3 cm, EF = 6.4 cm and E = 75°

(i) XY + YZ = 5.6 cm, XZ = 4.5 cm and ∠X = 45°

Steps of Construction:

1. Draw a line segment XZ of length 4.5 cm
2. Taking X as centre, construct ∠SXZ = 45°
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 ∠R = 45°

Steps of Construction:

1. Draw a line segment QR of length 4.8 cm
2. Taking Q as centre, construct ∠SQR = 45°
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 ∠E = 75°

Steps of Construction:

1. Draw a line segment EF of length 6.4 cm
2. Taking E as centre, construct ∠SEF = 75°
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 – PR = 1.5 cm, QR = 6.0 cm and Q = 75°

(ii) AB – AC = 1.2 cm, BC = 6.0 cm and B = 60°

(iii) XY – XZ = 1.5 cm, YZ = 3.4 cm and X = 45°

(i) PQ – PR = 1.5 cm, QR = 6.0 and ∠Q = 45°

Steps of Construction:

1. Draw a line segment QR of length 6 cm
2. Taking Q as centre, draw ∠TQR = 45°
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 – AC = 1.2 cm, BC = 6.0 cm and ∠B = 60°

Steps of Construction:

1. Draw a line segment BC of length = 6 cm
2. Taking B as centre, draw ∠TBC = 60°
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 – XZ = 1.5 cm, YZ = 3.4 cm and ∠X = 45°

Steps of Construction:

1. Draw a line segment YZ of length 3.4 cm
2. Taking Y as centre, draw ∠TYZ = 45°
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 60° and 45°

(ii) Perimeter of triangle is 9 cm, and the base angles are 60° and 45°

(iii) Perimeter of triangle is 10.6 cm, and the base angles are 60° and 90°

(i) Perimeter of triangle is 6.4 cm, and the base angles are 60° and 45°

Steps of Construction:

1. Draw DE of measure 6.4 cm
2. Draw DP and EQ such that ∠PDE = 45° and ∠QED = 60°
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 60° and 45°

Steps of Construction:

1. Draw DE of measure 9 cm
2. Draw DP and EQ, such that ∠PDE = 45° and ∠QED = 60°
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 60° and 90°

Steps of Construction:

1. Draw DE of measure 10.6 cm
2. Draw DP and EQ such that ∠PDE = 90° and ∠QED = 60°
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 XYZ with YZ = 7.5 cm, Y = 60° and Z = 45°. Draw the bisectors of Y and Z. If these bisectors meet at O, measure angle YOZ.

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 60° 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 ∠UZY. Let TZ be the bisector. TZ makes an angle of 45° with YZ
8. Bisect ∠SYZ and ∠TZY
9. Consider the point as O where the bisectors of ∠SYZ and ∠TZY meet
10. On measuring we get, ∠YOZ = 127.5°

13. Construct a RST with side ST = 5.4 cm, RST = 60° and the perpendicular from R on ST = 3.0 cm

Steps of Construction:

1. Draw a line segment ST of length 5.4 cm
2. Taking S as centre, draw XS such that ∠XST = 60°
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, ∠RST is the required triangle with angle 60°

14. Construct a PQR with Q = 60°, R = 45° and the perpendicular from P to QR be 3.5 cm. Measure PQ.

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 30° and 45° angles with PY respectively.
8. (In triangle PQY, ∠PQY = 60°, ∠QYP = 900, hence ∠QPY = 30° and in triangle PYR, ∠YRP = 45°, ∠RYP = 90°, hence ∠YPR = 45°)
9. Now, join PQ and PR

PQR is the required triangle

On measuring, PQ = 4.1 cm.

15. Construct a ABC, right angled at B with a perimeter of 10 cm and one acute angle of 60°.

Steps of Construction:

1. Draw DE of length 10 cm
2. Draw DP and EQ such that ∠PDE = 90° and ∠QED = 60°.
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.

16. Construct a PQR with Q = 60°, R = 45° and the perpendicular from P to QR be 3.5 cm. Measure PQ.