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.

Frank Solutions for Class 9 Maths Chapter 14 Construction of Triangles Download PDF

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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

FRANK Solutions Class 9 Maths Chapter 14 - 1

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

FRANK Solutions Class 9 Maths Chapter 14 - 2

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

FRANK Solutions Class 9 Maths Chapter 14 - 3

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 ∠C = 600

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

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

Solution:

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

FRANK Solutions Class 9 Maths Chapter 14 - 4

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 ∠X = 750

FRANK Solutions Class 9 Maths Chapter 14 - 5

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 ∠MXN thus formed to draw a ray XP.

5. Again, bisect the ∠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 ∠Q = 300

FRANK Solutions Class 9 Maths Chapter 14 - 6

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 ∠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, ∠B = 600 and ∠C = 450

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

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

Solution:

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

FRANK Solutions Class 9 Maths Chapter 14 - 7

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 ∠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, ∠P = 1050 and ∠Q = 450

FRANK Solutions Class 9 Maths Chapter 14 - 8

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 ∠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 ∠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, ∠D = 750 and ∠E = 600

FRANK Solutions Class 9 Maths Chapter 14 - 9

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 ∠RDQ thus formed to draw a ray XD

5. Let DY be the bisector to bisect the ∠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 ∠E = 300, ∠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 ∠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

FRANK Solutions Class 9 Maths Chapter 14 - 10

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

Steps of Construction:

1. Draw a line segment DE of length 6 cm

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

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

4. Ray DP and ray EM intersect at point F

Therefore,

DEF is the required triangle

FRANK Solutions Class 9 Maths Chapter 14 - 11

(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 = 450 and a ray RS such that ∠PRS = 450

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

Therefore,

PQR is the required triangle

FRANK Solutions Class 9 Maths Chapter 14 - 12

Now,

In △PQR,

QP = QR …… (given)

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

Here, hypotenuse PR = 7cm,

So,

∠PQR = 900

Hence,

∠QPR + ∠QRP = 900

∠QPR = ∠QRP = 450

5. Construct an isosceles triangle in which:

(a) AB = AC, BC = 6 cm and ∠B = 750

(b) XY = XZ, YZ = 5.5 cm and ∠X = 600

(c) PQ = QR, PR = 4.5 cm and ∠R = 600

Solution:

(a) AB = AC, BC = 6 cm and ∠B = 750

In △ABC,

AB = AC …… (given)

∠ACB = ∠ABC = 750

Steps of Construction:

1. Draw a line segment BC of length 6 cm

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

3. Ray CM and ray BN meet at point A

Therefore,

ABC is a required triangle

FRANK Solutions Class 9 Maths Chapter 14 - 13

(b) XY = XZ, YZ = 5.5 cm and ∠X = 600

In △XYZ,

XY = XZ ….. (given)

∠XZY = ∠XYZ ….. (1)

∠X = 600 …… (given)

Now,

∠X + ∠Y + ∠Z = 1800

600 + ∠Y + ∠Y = 1800 ……… [from (1)]

We get,

2∠Y = 1800 – 600

2∠Y = 1200

We get,

∠Y = 600 = ∠Z

Steps of Construction:

1. Draw a line segment YZ of length 5.5 cm

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

3. Ray ZP and YQ meet at a point X

Therefore,

XYZ is the required triangle

FRANK Solutions Class 9 Maths Chapter 14 - 14

(c) PQ = QR, PR = 4.5 cm and ∠R = 600

In △PQR,

PQ = QR ………. (given)

∠PRQ = ∠RPQ = 600

Steps of Construction:

1. Draw a line segment PR of length 4.5 cm

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

3. Ray RU and PV meet at a point Q

Therefore,

PQR is the required triangle

FRANK Solutions Class 9 Maths Chapter 14 - 15

6. Construct an isosceles triangle using the given data:

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

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

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

Solution:

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

FRANK Solutions Class 9 Maths Chapter 14 - 16

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 ∠MRP = ∠MRQ = (1 / 2) × 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 ∠A = 900

FRANK Solutions Class 9 Maths Chapter 14 - 17

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 ∠DAB = ∠DAC = (1 / 2) × 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 ∠PAD and ∠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 ∠X = 300

FRANK Solutions Class 9 Maths Chapter 14 - 18

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 ∠TXY = ∠TXZ = (1 / 2) × 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 ∠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 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 △UVW be the isosceles right-angled triangle,

Right angled at point U

Hypotenuse VW = 6 cm

UV = UW

∠UWV = ∠UVW

∠U = 900

∠UWV + ∠UVW = 900

2 ∠UWV = 900

We get,

∠UWV = ∠UVW = 450

Steps of Construction:

1. Draw a line segment of length 6 cm

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

3. Ray VY and ray WX meet at point U

Hence,

UVW is the required triangle

FRANK Solutions Class 9 Maths Chapter 14 - 19

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

FRANK Solutions Class 9 Maths Chapter 14 - 20

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

FRANK Solutions Class 9 Maths Chapter 14 - 21

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

FRANK Solutions Class 9 Maths Chapter 14 - 22

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 ∠X = 450

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

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

Solution:

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

FRANK Solutions Class 9 Maths Chapter 14 - 23

Steps of Construction:

1. Draw a line segment XZ of length 4.5 cm

2. Taking X as the centre, construct ∠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 ∠R = 450

FRANK Solutions Class 9 Maths Chapter 14 - 24

Steps of Construction:

1. Draw a line segment QR of length 4.8 cm

2. Taking Q as the centre, construct ∠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 ∠E = 750

FRANK Solutions Class 9 Maths Chapter 14 - 25

Steps of Construction:

1. Draw a line segment EF of length 6.4 cm

2. Taking E as centre, construct ∠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 – PR = 1.5 cm, QR = 6.0 cm and ∠Q = 750

(ii) AB – AC = 1.2 cm, BC = 6.0 cm and ∠B = 600

(iii) XY – XZ = 1.5 cm, YZ = 3.4 cm and ∠X = 450

Solution:

(i) PQ – PR = 1.5 cm, QR = 6.0 and ∠Q = 450

FRANK Solutions Class 9 Maths Chapter 14 - 26

Steps of Construction:

1. Draw a line segment QR of length 6 cm

2. Taking Q as the centre, draw ∠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 – AC = 1.2 cm, BC = 6.0 cm and ∠B = 600

FRANK Solutions Class 9 Maths Chapter 14 - 27

Steps of Construction:

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

2. Taking B as the centre, draw ∠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 – XZ = 1.5 cm, YZ = 3.4 cm and ∠X = 450

FRANK Solutions Class 9 Maths Chapter 14 - 28

Steps of Construction:

1. Draw a line segment YZ of length 3.4 cm

2. Taking Y as the centre, draw ∠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

FRANK Solutions Class 9 Maths Chapter 14 - 29

Steps of Construction:

1. Draw DE of measure 6.4 cm

2. Draw DP and EQ such that ∠PDE = 450 and ∠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

FRANK Solutions Class 9 Maths Chapter 14 - 30

Steps of Construction:

1. Draw DE of measure 9 cm

2. Draw DP and EQ, such that ∠PDE = 450 and ∠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

FRANK Solutions Class 9 Maths Chapter 14 - 31

Steps of Construction:

1. Draw DE of measure 10.6 cm

2. Draw DP and EQ such that ∠PDE = 900 and ∠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 △XYZ with YZ = 7.5 cm, ∠Y = 600 and ∠Z = 450. Draw the bisectors of ∠Y and ∠Z. If these bisectors meet at O, measure angle YOZ.

Solution:

FRANK Solutions Class 9 Maths Chapter 14 - 32

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 ∠UZY. Let TZ be the bisector. TZ makes an angle of 450 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 = 127.50

13. Construct a △RST with side ST = 5.4 cm, ∠RST = 600 and the perpendicular from R on ST = 3.0 cm.

Solution:

FRANK Solutions Class 9 Maths Chapter 14 - 33

Steps of Construction:

1. Draw a line segment ST of length 5.4 cm

2. Taking S as the centre, draw XS such that ∠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, ∠RST is the required triangle with angle 600

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

Solution:

FRANK Solutions Class 9 Maths Chapter 14 - 34

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

8. Now, join PQ and PR

9. PQR is the required triangle

10. 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 600.

Solution:

FRANK Solutions Class 9 Maths Chapter 14 - 35

Steps of Construction:

1. Draw DE of length 10 cm

2. Draw DP and EQ such that ∠PDE = 900 and ∠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.

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