Solutions-Class-9-Mathematics-2-Chapter-4-Constructions of Triangles-Maharashtra Board

Practice set 4.1 :

Question 1.1. Construct Δ PQR, in which QR = 4.2 cm, m∠Q = 40° and PQ + PR = 8.5 cm

Solution :

As shown in the rough figure, first we draw seg QR = 4.2 cm of length.

Draw a ray QD making an angle of 40° with seg QR.

Mark point S on QD such that QS = 8.5 cm

Now, PQ + PS = QS ... [Q-P-S]

∴ PQ + PS = 8.5 cm ... (i)

Given PQ + PR =8.5 cm ... (ii)

∴ PQ + PS = PQ+ PR ... [From (i) and (ii)]

∴ PS = PR

∴ P is on the perpendicular bisector of SR. ... (Perpendicular bisector theorem)

∴ The point of intersection of ray QD and perpendicular bisector of seg SR is point P.

Steps of constructions:

Draw seg QR of length 4.2 cm.
Draw ray QD such that ∠RQD = 40°.
Mark point S on ray QD such that QS = 8.5 cm
Draw seg RS.
Construct the perpendicular bisector of seg RS.
Draw perpendicular bisector of SR which intersects ray QD. Mark the point as P.
Draw seg PR.

Therefore, Δ PQR is the required triangle.

Question1.2. Construct Δ XYZ, in which YZ = 6 cm, XY + XZ = 9 cm. ∠XYZ = 50°

Solution :

As shown in the rough figure, first we draw seg YZ = 6 cm of length.

Draw a ray YN making an angle of 50° with seg YZ.

Mark point M on YN such that YM = 9 cm

Now, YX + XM = YM ... [Y-X-M]

∴ YX + XM = 9 cm ... (i)

Given XY + XZ = 9 cm ... (i) Given

∴ YX + XM = XY + XZ ... [From (i) and (ii)]

∴ XM = XZ

∴ X is on the perpendicular bisector of MZ. ... (Perpendicular bisector theorem)

∴ The point of intersection of ray YN and perpendicular bisector of seg MZ is point X.

Steps of construction:

Draw seg YZ of length 6 cm.
Draw ray YN such that ∠ZYN = 50°.
Mark point M on ray YN such that YM = 9 cm
Draw seg MZ.
Construct the perpendicular bisector of seg MZ.
Name the point of intersection of ray YM and the perpendicular bisector of MZ as X.
Draw seg XZ.

Therefore, Δ XYZ is the required triangle.

Question 1.3. Construct Δ ABC, in which BC = 6.2 cm, ∠ACB = 50°, AB + AC = 9.8 cm

Solution :

As shown in the rough figure draw seg CB = 6.2 cm

Draw a ray CT making an angle of 50° with CB

Take a point D on ray CT, such that

CM = 9.8 cm

Now, AC + AM = CM ... [C-A-M]

∴ AC + AM = 9.8 cm ... (i)

Also, AB + AC = 9.8 cm ... (ii) [Given]

∴ AC + AM = AB + AC ... [From (i) and (ii)]

∴ AM = AB

∴ Point A is on the perpendicular bisector of seg MB. ... (Perpendicular bisector

theorem)

∴ The point of intersection of ray CT and perpendicular bisector of seg MB is point A.

Steps of constructions:

Draw seg BC of length 6.2 cm.
Draw ray CT, such that ∠BCT = 50°.
Mark point M on ray CT such that l(CM) = 9.8 cm.
Join points M and B.
Draw perpendicular bisector of seg MB intersecting ray CT. Name the point as A.
Join the points A and B.

Therefore, Δ ABC is the required triangle.

Question 1.4. Construct Δ ABC, in which BC = 3.2 cm, ∠ACB = 45° and perimeter of Δ ABC is 10 cm.

Solution :

BC = 3.2 cm

Perimeter of A ACB = 10 cm

∴ AB + AC + BC = 10

∴ AB + BC + 3.2 = 10

∴ AB + AC = 10-3.2

∴ AB + AC = 6.8 cm

Now, In Δ ABC

BC= 3.2 cm, ∠ACB = 45° and AB + AC = 6.8 cm ... (i)

As shown in the rough figure draw seg BC = 3.2 cm

Draw a ray CX making an angle of 45° with CB

Take a point M on ray CX, such that

CM = 6.8 cm

Now, CA + AM = CM ... [C-A-M]

∴ CA + AM = 6.8 cm .(ii)

Also, AB + CA = 6.8 cm ... (iii) [From (i)]

∴ CA + AM = AB + CA ….[From (ii) and (iii)]

∴ AM = AB

∴ Point A is on the perpendicular bisector of seg MB. ... (Perpendicular bisector

theorem)

∴ The point of intersection of ray CX and perpendicular bisector of seg MB is point A.

Therefore, Δ ABC is the required triangle.

Practice set 4.2 :

Question 2.1. Construct Δ XYZ, such that YZ = 7.4 cm, ∠XYZ = 45° and XY - XZ = 2.7 cm.

Solution :

Explanation and steps of construction :

As shown in the rough figure, first we draw seg YZ of length 7.4 cm. We draw ray YL such that ∠LYZ = 45°.

Now, we have to locate point X on ray YL.

Take a point W on ray YL such that YW = 2.7 cm

XY = XW + YW

∴ XY = XW + 2.7 cm

∴ XY - XW = 2.7 cm

Now, XY - XZ = 2.7 cm

∴ XY - XW = XY - XZ

∴ XW = XZ

∴ X is equidistant from the points W and Z of seg WZ.

∴ X lies on the perpendicular bisector of seg WZ ... (Perpendicular bisector theorem)

∴ point X is the point of intersection of ray YL and perpendicular bisector of seg WZ.

Steps of construction :

(1) Draw seg YZ of length 7.4 cm.

(2) Draw ray YL such that ∠LYZ = 45°.

(3) Take a point W on ray YL such that YW = 2.7 cm.

(4) Construct the perpendicular bisector of seg WZ.

(5) Name the point of intersection of ray YL and the perpendicular bisector of seg WZ as X.

(6) Draw seg XZ.

Therefore, Δ XYZ is the required triangle.

Question 2.2. Construct Δ PQR, such that QR = 6.5 cm, ∠PQR = 40° and PQ - PR = 2.5 cm.

[ Note : Considering ∠PQR = 60° as given in the textbook, the triangle formed is very large. Hence, ∠PQR = 40° is considered to make it small. There will not be any difference in conceptual understanding of the construction for the student.]

Solution :

Here, PQ - PR = 2.5 cm

∴ PQ > PR

As shown in the rough figure draw seg QR = 6.5 cm

Draw a ray QX making on angle of 40° with QR

Take a point S on ray QX, such that QS = 2.5 cm

Now, PQ - PS = QS ... [Q-S-P]

∴ PQ - PS = 2.5 cm ... (i) [Given]

Also, PQ - PR = 2.5 cm ... (ii) [From (i) and (ii)]

∴ PQ – PS = PQ - PR

∴ PS = PR

∴ Point P is on the perpendicular bisector of seg RS.

∴ Point P is the intersection of ray QX and the perpendicular bisector of seg RS

Steps of construction:

Draw seg QR of length 6.5 cm.
Draw ray QX such that ∠RQX = 40°
Take point S on ray QX such that QS = 2.5 cm.
Join points S and R.
Draw perpendicular bisector of seg SR intersecting ray QX. Name the point as P.
Draw seg PR.

Therefore, Δ PQR is required triangle.

Question 2.3. Construct Δ ABC, such that BC = 6 cm, ∠ABC = 100° and AC - AB = 2.5 cm.

Solution :

Here, AC - AB = 2.5 cm

∴ AC > AB

As shown in the rough figure draw seg BC = 6 cm

Draw a ray BT making an angle of 100° with BC.

Take a point D on opposite ray of BT, such that BD 2.5 cm.

Now, AD - AB = BD ... [A-B-D]

∴ AD – AB = 2.5 cm ... (i)

Also, AC - AB = 2.5 cm ... (ii) [Given]

∴ AD - AB = AC – AB ...[From (i) and (ii)]

∴ AD = AC

∴ Point A is on the perpendicular bisector of seg DC.

∴ Point A is the intersection of ray BT and the perpendicular bisector of seg DC.

Steps of construction:

Draw seg BC of length 6 cm.
Draw ray BT, such that ∠CBT = 100°.
Take point D on opposite ray of BT such that l(BD) = 2.5 cm.
Join the points D and C.
Draw the perpendicular bisector of seg DC intersecting ray BT. Name the point as A.
Join the points A and C.

Therefore, Δ ABC is required triangle.

Practice set 4.3 :

Question 3.1. Construct Δ PQR, in which ∠Q = 70°, ∠R = 80° and PQ + QR + PR = 9.5 cm.

Solution :

(i) As shown in the figure, take point T and S on line QR, such that

QT = PQ and RS = PR ... (i)

QT + QR + RS = TS ... [T-Q-R, Q-R-S]

∴ PQ + QR + PR = TS ... (ii) [From (i)]

Also,

PQ + QR + PR = 9.5 cm ... (iii) [Given]

∴ TS = 9.5 cm

(ii) In Δ PQT

PQ = QT ... [From (i)]

∴ ∠QPT = ∠QTP = x° ... (iv) [Isosceles triangle theorem]

In Δ PQT, PQR is the exterior angle.

∴ ∠QPT + ∠QTP = ∠PQR ..[Remote interior angles theorem]

∴ x + x = 70° ... [From (iv)]

∴ 2x = 70° ∴ x = 35°

∴ ∠PTQ = 35°

∴ ∠T = 35°

Similarly, ∠S = 40°

(iii) Now, in Δ PTS

∠T = 35°, ∠S = 40° and TS = 9.5 cm

Hence, Δ PTS can be drawn.

(iv) Since, PQ = TQ,

: Point Q lies on perpendicular bisector of seg PT.

Also, RP = RS

∴ Point R lies on perpendicular bisector of seg PS.

Points Q and R can be located by drawing the perpendicular bisector of PT and PS respectively.

∴ Δ PQR can be drawn.

Steps of construction:

Draw seg TS of length 9.5 cm.
From point T draw ray making angle of 35°.
From point S draw ray making angle of 40°.
Name the point of intersection of two rays as P.
Draw the perpendicular bisector of seg PT and seg PS intersecting seg TS in Q and R respectively.
Join PQ and PR.

Therefore, Δ PQR is the required triangle.

Question 3.2. Construct Δ XYZ, in which ∠Y = 58°, ∠X = 46° and perimeter of triangle is 10.5 cm.

Solution :

Question 3.3. Construct Δ LMN, in which ∠M = 60°, ∠N = 80° and LM + MN + NL = 11 cm.

Solution :

Problem set 4 :

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

Solution :

As shown in the rough figure draw seg YZ = 4.9 cm

Draw a ray YT making an angle of 45° with YZ

Take a point D on ray YT, such that YD = 10.3 cm

Now, YX + XD =YD ... [Y-X-D]

∴ YX + XD = 10.3 cm ... (i)

Also, XY + YZ = 10.3 cm ... (ii) [Given]

∴ YX + XD = XY + XZ .. [From (i) and (ii)]

∴ XD = XZ

Point X is on the perpendicular bisector of seg DZ

∴ The point of intersection of ray YT and perpendicular bisector of seg DZ is point X.

Steps of construction:

Draw seg YZ of length 4.9 cm.
Draw ray YT, such that ∠ZYT = 75°.
Mark point D on ray YT such that l(YD) = 10.3 cm.
Join points D and Z.
Draw perpendicular bisector of seg DZ intersecting ray YT. Name the point as X.
Join the points X and Z.

Therefore, Δ XYZ is the required triangle.

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

Solution :

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

Solution :

Let the given triangle be ABC.

AB : BC : AC =2 : 3 : 4

Let the common multiple of the given ratio be x.

Then AB = 2x cm, BC = 3x cm, AC = 4x cm.

Perimeter of A ABC = 14.4 cm

∴ AB + BC + AC = 14.4

∴ 2x + 3x + 4x =14.4

∴ 9x = 14.4

∴ x = 14.4/9 = 1.6

AB = 2x = 2 x 1.6 = 3.2 cm;

BC = 3x = 3 x 1.6 = 4.8 cm;

AC = 4x = 4 x 1.6 = 6.4 cm

Question 4. Construct Δ PQR, in which PQ - PR = 2.4 cm, QR = 6.4 cm and ∠PQR = 55°.

Solution :

Here, PQ - PR = 2.4 cm

∴ PQ > PR

As shown in the rough figure, draw seg QR = 6.4 cm

Draw a ray QT making on angle of 55° with QR

Take a point D on ray QE, such that QD = 2.4 cm.

Now, PQ - PD = QD ... [Q-D-P]

∴ PQ – PD = 2.4 cm .(i)

Also, PQ-PR = 2.4 cm ... (ii) [Given]

∴ PQ – PD = PQ – PR ... [From (i) and (ii)]

∴ PD = PR

∴ Point P is on the perpendicular bisector of seg RD.

∴ Point P is the intersection of ray QE and the perpendicular bisector of seg RD.

Steps of construction:

Draw seg QR of length 6.4 cm.
Draw ray QE, such that ∠RQE = 55°.
Take point D on ray QE such that l(QD) = 2.4 cm.
Join the points D and R.
Draw perpendicular bisector of seg SR intersecting ray QE. Name that point as P.
Join the points P and R.