Maharashtra Board Class 9 Math Part 2 Solution Chapter 4 – Quadrilaterals
Balbharati Maharashtra Board Class 9 Math Part 2 Solution Chapter 4: Quadrilaterals. Marathi or English Medium Students of Class 9 get here Quadrilaterals full Exercise Solution.
|
Std |
Maharashtra Class 9 |
|
Subject |
Math Part 2 Solution |
| Chapter |
Quadrilaterals |
Practice Set 4.1
(1) Construct △PQR, in which QR = 4.2 cm, m∠Q = 40° and PQ + PR = 8.5 cm
Steps –
(1) Construct QR = 4.2 cm and ∠Q = 40°
(2) Cut an are QA on QD such that AQ = 8.5 cm
(3) Join AR and draw perpendicular bisector that must AQ at P
(4) Joining PR we get the △PQR
(2) Construct △XYZ, in which YZ = 6 cm, XY + XZ = 9 cm. ∠XYZ = 50°
Steps –
(1) Construct YZ = 6cm and ∠Y = 50°
(2) Cut an arc on YP such that YD = 9cm
(3) Join DZ and draw its 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
Steps –
(1) Construct BC = 6.2 cm and ∠C = 50°
(2) Cut an are on CP such that CD = 9.8 cm
(3 Join BD and draw its 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 △ABC is 10 cm
Solution:
Perimeter of △ABC = 10cm
= AB + BC + AC = 10cm
= AB + AC = 10 – 3.2
= 6.8 cm
Steps –
(1) Construct BC = 3.2 cm and ∠C = 45°
(2) Cut an arc of CP such that CD = 6.8 cm
(3) Join BD and draw its 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 QM such that QD = 2.5 cm
(3) Join DR and draw it’s 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 = ½ CQ = 35° and ∠R = 80°
∠PBR = ½ ∠R = 40°
Steps –
(1) Construct AB = 9.5 cm and ∠A = 35° and ∠B = 40°
(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, we 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
(3) Draw perpendicular bisectors of AZ and BZ. Which intersects A 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
