Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 4.3 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 4 Constructions of Triangles.

## Practice Set 4.3 Geometry 9th Std Maths Part 2 Answers Chapter 4 Constructions of Triangles

Question 1.

Construct ∆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 ∆PQT

PQ = QT [From (i)]

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

In ∆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 ∆PTS

∠T = 35°, ∠S = 40° and TS = 9.5 cm Hence, ∆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.

∴ ∆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 35°.

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

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, ∆PQR is the required triangle.

Question 2.

Construct ∆XYZ, in which ∠Y = 58°, ∠X = 46° 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 ……(i)

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

∠Y + YX + ∠X = WV ……(ii) [From (i)]

Also,

∠Y + YX + ∠X = 10.5 cm …..(iii) [Given]

∴ WV = 10.5 cm [From (ii) and (iii)]

ii. In ∆ZWY

∠Y = YM [From (i)]

∴ ∠YZW = ∠YWZ = x° …..(iv) [Isosceles triangle theorem]

In ∆ZYW, ∠ZYX is the exterior angle.

∴ ∠YZW + ∠YWZ = ∠ZYX [Remote interior angles theorem]

∴ x + x = 58° [From (iv)]

∴ 2x = 58°

∴ x = 29°

∴ ∠ZWY = 29°

∴ ∠W = 29°

∴ Similarly, ∠V = 23°

iii. Now, in ∆ZWV

∠W = 29°, ∠V = 23° and

WV= 10.5 cm

Hence, ∆ZWV can be drawn.

iv. Since, ZY = YW

∴ Point Y lies on perpendicular bisector of seg ZW.

Also, ZX = XV

∴ Point X lies on perpendicular bisector of seg ZV.

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

∴ ∆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 29°.

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

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, ∆XYX is the required triangle

Question 3.

Construct ∆LMN, in which ∠M = 60°, ∠N = 80° 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 …..(i)

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

∴ LM + MN + LN = ST …..(ii)

Also,

LM + MN + LN = 11 cm ….(iii)

∴ ST = 11 cm [From (ii) and (iii)]

ii. In ∆LSM

LM = MS

∴ ∠MLS = ∠MSL = x° …..(iv) [isosceles triangle theorem]

In ∆LMS, ∠LMN is the exterior angle.

∴ ∠MLS + ∠MSL = ∠LMN [Remote interior angles theorem]

∴ x + x = 60° [From (iv)]

∴ 2x = 60°

∴ x = 30°

∴ ∠LSM = 30°

∴ ∠S = 30°

Similarly, ∠T = 40°

iii. Now, in ∆LST

∠S = 30°, ∠T = 40° and ST = 11 cm

Hence, ALST can be drawn.

iv. Since, LM = MS

∴ Point M lies on perpendicular bisector of seg LS.

Also LN = NT

∴ Point N lies on perpendicular bisector of seg LT.

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

∴ ∆LMN can be drawn.

Steps of construction:

i. Draw seg ST of length 11 cm.

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

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

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, ∆LMN is the required triangle.