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