**Chapter 4 Constructions of Triangles Practice Set 4**

Chapter 4 Constructions of Triangles Practice Set 4

**Question 1.Construct âˆ†XYZ, such that XY + XZ = 10.3 cm, YZ = 4.9 cm, âˆ XYZ = 45Â°.Solution:**

As shown in the rough figure draw segYZ = 4.9cm

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

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

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

âˆ´ YX + XW=10.3cm â€¦..(i)

Also, XY + Xâˆ 10.3cm â€¦â€¦(ii) [Given]

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

âˆ´ XW = XZ

âˆ´ Point X is on the perpendicular bisector of seg WZ

âˆ´ 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 âˆ ZYT = 75Â°.

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, âˆ†XYZ is the required triangle.

**Question 2.Construct âˆ†ABC, in which âˆ B = 70Â°, âˆ C = 60Â°, 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 â€¦â€¦(i)

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

âˆ´ AB + BC + AC = DE â€¦..(ii)

Also,

AB + BC + AC= 11.2 cm â€¦.(iii) [Given]

âˆ´ DE = 11.2 cm [From (ii) and (iii)]

ii. In âˆ†ADB

AB = BD [From (i)]

âˆ´ âˆ BAD = âˆ BDA = xÂ° â€¦.(iv) [Isosceles triangle theorem]

In âˆ†ABD, âˆ ABC is the exterior angle.

âˆ´ âˆ BAD + âˆ BDA = âˆ ABC [Remote interior angles theorem]

x + x = 70Â° [From (iv)]

âˆ´ 2x = 70Â° x = 35Â°

âˆ´ âˆ ADB = 35Â°

âˆ´ âˆ D = 35Â°

Similarly, âˆ E = 30Â°

iii. Now, in âˆ†ADE

âˆ D = 35Â°, âˆ E = 30Â° and DE = 11.2 cm

Elence, âˆ†ADE can be drawn.

iv. Since, AB = BD

âˆ´ Point B lies on perpendicular bisector of seg AD.

Also AC = CE

âˆ´ Point C lies on perpendicular bisector of seg AE.

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

âˆ´ âˆ†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 35Â°.

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

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, âˆ†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

âˆ´ In âˆ†ABC,

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

Perimeter of triangle = 14.4 cm

âˆ´ AB + BC + AC= 14.4

âˆ´ 9x = 14.4

âˆ´ x = 1.6

âˆ´ AB = 2x = 2x 1.6 = 3.2 cm

âˆ´ AC = 3x = 3 x 1.6 = 4.8 cm

âˆ´ BC = 4x = 4 x 1.6 = 6.4 cm

**Question 4.Construct âˆ†PQR, in which PQ â€“ PR = 2.4 cm, QR = 6.4 cm and âˆ PQR = 55Â°.Solution:**

Here, PQ â€“ PR = 2.4 cm

âˆ´ PQ > PR

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

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

Take a point S on ray QT, such that QS = 2.4 cm.

Now, PQ â€“ PS = QS [Q-S-P]

âˆ´ PQ â€“ PS = 2.4 cm â€¦(i)

Also, PQ â€“ PR = 2.4 cm â€¦.(ii) [Given]

âˆ´ PQ â€“ PS = PQ â€“ PR [From (i) and (ii)]

âˆ´ PS = PR

âˆ´ Point P is on the perpendicular bisector of seg RS

âˆ´ 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 âˆ RQT = 55Â°.

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, âˆ†PQR is the required triangle.