Practical Geometry Ex-10.2

NCERT solutions for Maths Practical Geometry 

NCERT Solutions for Class 7 Maths Practical Geometry

Class –VII Mathematics (Ex. 10.2)

Question 1.Construct $\Delta$XYZ in which XY = 4.5 cm, YZ = 5 cm and ZX = 6 cm.

Answer:

To construct: $\Delta$XYZ, where XY = 4.5 cm, YZ = 5 cm and ZX = 6 cm.

Steps of construction:

(a) Draw a line segment YZ = 5 cm.

(b) Taking Z as centre and radius 6 cm, draw an arc.

(c) Similarly, taking Y as centre and radius 4.5 cm, draw another arc which intersects first arc at point X.

(d) Join XY and XZ.

It is the required $\Delta$XYZ.

NCERT Solutions for Class 7 Maths Exercise 10.2

Question 2.Construct an equilateral triangle of side 5.5 cm.

Answer:

To construct: A $\Delta$ABC where AB = BC = CA = 5.5 cm

Steps of construction:

(a) Draw a line segment BC = 5.5 cm

(b) Taking points B and C as centers and radius 5.5 cm, draw arcs which intersect at point A.

(c) Join AB and AC.

It is the required $\Delta$ABC.

NCERT Solutions for Class 7 Maths Exercise 10.2

Question 3.Draw $\Delta$PQR with PQ = 4 cm, QR = 3.5 cm and PR = 4 cm. What type of triangle is this?

Answer:

To construction: $\Delta$PQR, in which PQ = 4 cm, QR = 3.5 cm and PR = 4 cm.

Steps of construction:

(a) Draw a line segment QR = 3.5 cm.

(b) Taking Q as centre and radius 4 cm, draw an arc.

(c) Similarly, taking R as centre and radius 4 cm, draw an another arc which intersects first arc at P.

(d) Join PQ and PR.

It is the required isosceles $\Delta$PQR.

NCERT Solutions for Class 7 Maths Exercise 10.2

Question 4.Construct $\Delta$ABC such that AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm. Measure $\angle$B.

Answer:

To construct :$\Delta$ABC in which AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm.

Steps of construction:

(a) Draw a line segment BC = 6 cm.

(b) Taking B as centre and radius 2.5 cm, draw an arc.

(c) Similarly, taking C as centre and radius 6.5 cm, draw another arc which intersects first arc at point A.

(d) Join AB and AC.

(e) Measure angle B with the help of protractor.

It is the required $\Delta$ABC where $\angle$B = $80^\circ .$