Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Problem Set 4 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 4 Constructions of Triangles.

## Problem Set 4 Geometry 9th Std Maths Part 2 Answers Chapter 4 Constructions of Triangles

Question 1.

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

Solution:

As shown in the rough figure draw segYZ = 4.9cm

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

Take a point W on ray YT, such that YW= 10.3 cm

Now,YX + XW = YW [Y-X-W]

∴ YX + XW=10.3cm …..(i)

Also, XY + X∠10.3cm ……(ii) [Given]

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

∴ XW = XZ

∴ Point X is on the perpendicular bisector of seg WZ

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

Steps of construction:

i. Draw seg YZ of length 4.9 cm.

ii. Draw ray YT, such that ∠ZYT = 75°.

iii. Mark point W on ray YT such that l(YW) = 10.3 cm.

iv. Join points W and Z.

v. Draw perpendicular bisector of seg WZ intersecting ray YT. Name the point as X.

vi. Join the points X and Z.

Hence, ∆XYZ is the required triangle.

Question 2.

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

Solution:

i. As shown in the figure, take point D and E on line BC, such that

BD = AB and CE = AC ……(i)

BD + BC + CE = DE [D-B-C, B-C-E]

∴ AB + BC + AC = DE …..(ii)

Also,

AB + BC + AC= 11.2 cm ….(iii) [Given]

∴ DE = 11.2 cm [From (ii) and (iii)]

ii. In ∆ADB

AB = BD [From (i)]

∴ ∠BAD = ∠BDA = x° ….(iv) [Isosceles triangle theorem]

In ∆ABD, ∠ABC is the exterior angle.

∴ ∠BAD + ∠BDA = ∠ABC [Remote interior angles theorem]

x + x = 70° [From (iv)]

∴ 2x = 70° x = 35°

∴ ∠ADB = 35°

∴ ∠D = 35°

Similarly, ∠E = 30°

iii. Now, in ∆ADE

∠D = 35°, ∠E = 30° and DE = 11.2 cm

Elence, ∆ADE can be drawn.

iv. Since, AB = BD

∴ Point B lies on perpendicular bisector of seg AD.

Also AC = CE

∴ Point C lies on perpendicular bisector of seg AE.

∴ Points B and C can be located by drawing the perpendicular bisector of AD and AE respectively.

∴ ∆ABC can be drawn.

Steps of construction:

i. Draw seg DE of length 11.2 cm.

ii. From point D draw ray making angle of 35°.

iii. From point E draw ray making angle of 30°.

iv. Name the point of intersection of two rays as A.

v. Draw the perpendicular bisector of seg DA and seg EA intersecting seg DE in B and C respectively.

vi. Join AB and AC.

Hence, ∆ABC is the required triangle.

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 common multiple be x

∴ In ∆ABC,

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

Perimeter of triangle = 14.4 cm

∴ AB + BC + AC= 14.4

∴ 9x = 14.4

∴ x = \(\frac { 14.4 }{ 9 }\)

∴ x = 1.6

∴ AB = 2x = 2x 1.6 = 3.2 cm

∴ AC = 3x = 3 x 1.6 = 4.8 cm

∴ BC = 4x = 4 x 1.6 = 6.4 cm

Question 4.

Construct ∆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 S on ray QT, such that QS = 2.4 cm.

Now, PQ – PS = QS [Q-S-P]

∴ PQ – PS = 2.4 cm …(i)

Also, PQ – PR = 2.4 cm ….(ii) [Given]

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

∴ PS = PR

∴ Point P is on the perpendicular bisector of seg RS

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

Steps of construction:

i. Draw seg QR of length 6.4 cm.

ii. Draw ray QT, such that ∠RQT = 55°.

iii. Take point S on ray QT such that l(QS) = 2.4 cm.

iv. Join the points S and R.

v. Draw perpendicular bisector of seg SR intersecting ray QT.

Name that point as P.

vi. Join the points P and R.

Hence, ∆PQR is the required triangle.