Maharashtra Board Class 9 Math Part 2 Chapter 4 Constructions of Triangles Solution

Maharashtra Board Class 9 Math Part 2 Solution Chapter 4 – Constructions of Triangles

Balbharati Maharashtra Board Class 9 Math Part 2 Solution Chapter 4: Constructions of Triangles. Marathi or English Medium Students of Class 9 get here Constructions of Triangles full Exercise Solution.

Std

Maharashtra Class 9
Subject

Math Part 2 Solution

Chapter

Constructions of Triangles

Practice Set 4.1

(1) Construct PQR, in which QR = 4.2 cm, m Q = 40° and PQ + PR = 8.5 cm

Solution:

Steps –

(1) Construct QR = 4.2 cm and ∠Q = 40

(2) Cut an arc QA on QD such that AQ = 8.5 cm

(3) Join AR and Draw perpendicular bisector that meet AQ at P

(4) Joining PR we get the △PQR

 

(2) Construct XYZ, in which YZ = 6 cm, XY + XZ = 9 cm. XYZ = 50°

Solution:

Steps –

(1) Construct YZ = 6 cm and ∠Y = 50°

(2) Cut an arc on YP such that YD = 9 cm

(3) Join DZ and draw it’s perpendicular bisector which intersects YP at x.

(4) Join XZ and △XYZ is the required triangle.

 

(3) Construct ABC, in which BC = 6.2 cm, ACB = 50°, AB + AC = 9.8 cm

Solution:

Steps –

(1) Construct BC = 6.2 cm and ∠C = 50°

(2) Cut an arc on CP such that CD = 9.8 cm

(3) Join BD and draw it’s perpendicular bisector which intersects CD at A

(4) Joining AB we get △ABC

 

(4) Construct ABC, in which BC = 3.2 cm, ACB = 45° and perimeter of D ABC is 10 cm

 

Solution:

Perimeter of △ABC = 10 cm

= AB + BC + AC = 10cm

= AB + AC = 10 – 3.2

= 6.8 cm

Steps –

(1) Construct BC = 3.2 cm and ∠C = 45°

(2) Cut on arc of CP such that CD = 6.8 cm

(3) Join BD and draw it’s perpendicular bisector which intersects CD at A

(4) Joining AB, we get △ABC

Practice Set 4.2

(1) Construct XYZ, such that YZ = 7.4 cm, XYZ = 45° and XY – XZ = 2.7 cm.

Solution:

Steps –

(1) Construct YZ = 7.4 cm and ∠Y = 45°

(2) Cut are on YP such that YD = 3 cm

(3) Join DZ and draw it’s perpendicular bisector which intersects YP at x

(4) Joining XZ we get the required △XYZ

 

(2) Construct PQR, such that QR = 6.5 cm, PQR = 40° and PQ – PR = 2.5 cm.

Solution:

Steps –

(1) Construct QR = 6.5 cm and ∠Q = 40°

(2) Cut arc on Q such that QD = 2.5 cm

(3) Join DR and draw its perpendicular bisector which intersects QM at P

(4) Joining PR we get △PQR

 

(3) Construct ABC, such that BC = 6 cm, ABC = 100° and AC – AB = 2.5 cm.

Solution:

Steps –

(1) Construct BC = 6 cm & ∠B = 100°

(2) Cut arc on BS such that BD = 2.5 cm

(3) Join CD and draw it’s perpendicular bisector which intersects BP at A

(4) Join AC we get △ABC

Practice Set – 4.3

(1) Construct PQR, in which Q = 70°, R = 80° and PQ + QR + PR = 9.5 cm.

Solution:

Given, ∠Q= 70

∴ ∠PAQ = 1/2 ∠Q = 35°

And ∠R = 80°

∠PBR = ½ = CR = 40°

Steps –

(1) Construct AB = 9.5 cm and ∠A = 35° and ∠B = 40o

(2)  Let, the line from A and B meet at P

(3) Draw perpendicular bisectors of AP and PB which intersects AB at Q and R respectively.

(4) Join PQ and QR, use get △PQR

 

(2) Construct XYZ, in which Y = 58°, X = 46° and perimeter of triangle is 10.5 cm.

Solution:

Given ∠Y = 58° ∴ ∠A = 58°/2 = 29°

∠X = 46° ∴ ∠B = 46°/2 = 23°

Steps –

(1) Construct AB = 10.5 cm and ∠A = 29° & ∠B = 23°

(2) Let, lines from A & B meet at Z

(3) Draw perpendicular bisectors of AZ and BZ. Which intersects AB at Y and X respectively.

(4) Join XZ and YZ we, get △XYZ

 

(3) Construct LMN, in which M = 60°, N = 80° and LM + MN + NL = 11 cm.

Solution:

Given, ∠M = 60°, ∠A = 30°

∠N = 80°, ∠B = 40°

Steps –

(1) Construct AB = 11cm and ∠A = 30° & ∠B = 40°

(2) Let, lines from A & B meet at L

(3) Draw perpendicular bisectors of AL and BL which intersects AB at M and N respectively

(4) Join LM and LN, we get △LMN

Problem Set 4

(1) Construct XYZ, such that XY + XZ = 10.3 cm, YZ = 4.9 cm, XYZ = 45°.

Solution:

Steps –

(1) Construct YZ = 4.9 cm and ∠Y = 45°

(2) Cut arc at YP such that YD = 10.3 cm

(3) Join DZ and dress it perpendicular bisector which intersect YD at X

(4) Joining XZ we get △XYZ

 

(2) Construct ABC, in which B = 70°, C = 60°, AB + BC + AC = 11.2 cm.

Solution:

∠B = 70°, ∠P = 35°

∠C = 60°, ∠Q = 30°

Steps –

(1) Draw PQ = 11.2 and ∠P = 35°, ∠Q = 30°

(2) Let lines from P and Q meet at A

(3) Draw perpendicular bisectors of AP & AQ which intersect PQ at B & C respectively

(4) Join AB and AC, we get △ABC

 

(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:

Given ratio of length of sides is 2:3:4
Let, The sides be 2x, 3x & 4x

∴ 2x + 3x + 4x = 14.4

= x = 14.4/9 = 1.6 cm

∴ The sides are: 3.2, 4.8 and 6.4

Steps –

(1) Construct BC = 6.4 cm

(2) With B as centre and length 4.8 cut an arc

(3) With c as centre and 3.2 as length draw an arc which cuts previous arc at A

(4) ABC is required triangle

 

(4) Construct PQR, in which PQ – PR = 2.4 cm, QR = 6.4 cm and PQR = 55°.

Solution:

Steps –

(1) Construct QR = 6.4 cm & ∠Q = 55°

(2) Cut arc on QS such that QD = 2.4 cm

(3) Join DR and draw its perpendicular bisector which intersects QM at P

(4) Join PR and we get △PQR

Updated: September 25, 2021 — 7:19 pm

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