**Chapter 4 Constructions of Triangles Practice Set 4.3**

Chapter 4 Constructions of Triangles Practice Set 4.3

**Question 1.Construct âˆ†PQR, in which âˆ Q = 70Â°, âˆ R = 80Â° 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 â€¦.(i)

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

âˆ´ PQ + QR + PR = TS â€¦..(ii) [From (i)]

Also,

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

âˆ´ TS = 9.5 cm

ii. In âˆ†PQT

PQ = QT [From (i)]

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

In âˆ†PQT, âˆ PQR is the exterior angle.

âˆ´ âˆ QPT + âˆ QTP = âˆ PQR [Remote interior angles theorem]

âˆ´ x + x = 70Â° [From (iv)]

âˆ´ 2x = 70Â° x = 35Â°

âˆ´ âˆ PTQ = 35Â°

âˆ´ âˆ T = 35Â°

Similarly, âˆ S = 40Â°

iii. Now, in âˆ†PTS

âˆ T = 35Â°, âˆ S = 40Â° and TS = 9.5 cm Hence, âˆ†PTS can be drawn.

iv. Since, PQ = TQ,

âˆ´ Point Q lies on perpendicular bisector of seg PT.

Also, RP = RS

âˆ´ 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.

âˆ´ âˆ†PQR can be drawn.

Steps of construction:

i. Draw seg TS of length 9.5 cm.

ii. From point T draw ray making angle of 35Â°.

iii. From point S draw ray making angle of 40Â°.

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

v. Draw the perpendicular bisector of seg PT and seg PS intersecting seg TS in Q and R respectively.

vi. Join PQ and PR.

Hence, âˆ†PQR is the required triangle.

**Question 2.Construct âˆ†XYZ, in which âˆ Y = 58Â°, âˆ X = 46Â° and perimeter of triangle is 10.5 cm.Solution:**

i. As shown in the figure, take point W and V on line YX, such that

YW = ZY and XV = ZX â€¦â€¦(i)

YW + YX + XV = WV [W-Y-X, Y-X-V]

âˆ Y + YX + âˆ X = WV â€¦â€¦(ii) [From (i)]

Also,

âˆ Y + YX + âˆ X = 10.5 cm â€¦..(iii) [Given]

âˆ´ WV = 10.5 cm [From (ii) and (iii)]

ii. In âˆ†ZWY

âˆ Y = YM [From (i)]

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

In âˆ†ZYW, âˆ ZYX is the exterior angle.

âˆ´ âˆ YZW + âˆ YWZ = âˆ ZYX [Remote interior angles theorem]

âˆ´ x + x = 58Â° [From (iv)]

âˆ´ 2x = 58Â°

âˆ´ x = 29Â°

âˆ´ âˆ ZWY = 29Â°

âˆ´ âˆ W = 29Â°

âˆ´ Similarly, âˆ V = 23Â°

iii. Now, in âˆ†ZWV

âˆ W = 29Â°, âˆ V = 23Â° and

WV= 10.5 cm

Hence, âˆ†ZWV can be drawn.

iv. Since, ZY = YW

âˆ´ Point Y lies on perpendicular bisector of seg ZW.

Also, ZX = XV

âˆ´ Point X lies on perpendicular bisector of seg ZV.

âˆ´ Points Y and X can be located by drawing the perpendicular bisector of ZW and ZV respectively.

âˆ´ âˆ†XYZ can be drawn.

Steps of construction:

i. Draw seg WV of length 10.5 cm.

ii. From point W draw ray making angle of 29Â°.

iii. From point V draw ray making angle of 23Â°.

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

v. Draw the perpendicular bisector of seg WZ and seg VZ intersecting seg WV in Y and X respectively.

vi. Join XY and XX.

Hence, âˆ†XYX is the required triangle

**Question 3.Construct âˆ†LMN, in which âˆ M = 60Â°, âˆ N = 80Â° and LM + MN + NL = 11 cm.Solution:**

i. As shown in the figure, take point S and T on line MN, such that

MS = LM and NT = LN â€¦..(i)

MS + MN + NT = ST [S-M-N, M-N-T]

âˆ´ LM + MN + LN = ST â€¦..(ii)

Also,

LM + MN + LN = 11 cm â€¦.(iii)

âˆ´ ST = 11 cm [From (ii) and (iii)]

ii. In âˆ†LSM

LM = MS

âˆ´ âˆ MLS = âˆ MSL = xÂ° â€¦..(iv) [isosceles triangle theorem]

In âˆ†LMS, âˆ LMN is the exterior angle.

âˆ´ âˆ MLS + âˆ MSL = âˆ LMN [Remote interior angles theorem]

âˆ´ x + x = 60Â° [From (iv)]

âˆ´ 2x = 60Â°

âˆ´ x = 30Â°

âˆ´ âˆ LSM = 30Â°

âˆ´ âˆ S = 30Â°

Similarly, âˆ T = 40Â°

iii. Now, in âˆ†LST

âˆ S = 30Â°, âˆ T = 40Â° and ST = 11 cm

Hence, ALST can be drawn.

iv. Since, LM = MS

âˆ´ Point M lies on perpendicular bisector of seg LS.

Also LN = NT

âˆ´ Point N lies on perpendicular bisector of seg LT.

âˆ´ Points M and N can be located by drawing the perpendicular bisector of LS and LT respectively.

âˆ´ âˆ†LMN can be drawn.

Steps of construction:

i. Draw seg ST of length 11 cm.

ii. From point S draw ray making angle of 30Â°.

iii. From point T draw ray making angle of 40Â°.

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

v. Draw the perpendicular bisector of seg LS and seg LT intersecting seg ST in M and N respectively.

vi. Join LM and LN.

Hence, âˆ†LMN is the required triangle.