Students can Download Chapter 10 Practical Geometry Ex 10.2, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

## Karnataka State Syllabus Class 7 Maths Chapter 10 Practical Geometry Ex 10.2

Question 1.

Construct ∆ XYZ in which XY = 4.5 cm, YZ = 5 cm and ZX = 6 cm.

Solution:

Steps of Construction

- Draw a line segment YZ of length 5 cm with Y as centre, draw an arc of radius 4.5 cm.
- With Z as centre, draw an arc of radius 6cms which cuts the first arc. Mark the point of intersection of arcs as X.
- Join XY and XZ. ∆ XYZ is the required triangle.

Question 2.

Construct an equilateral triangle of side 5.5 cm.

Solution:

Steps of Constructions

a

- Draw a line segment BC of length 5.5 cms. with ‘B’ as centre draw an arc of radius 5.5 cm.
- With ‘C’ as centre, Cut the first arc with radius 5.5 cms. Mark the point of intersection of arcs as ‘A’
- Join AB and AC. ∆ ABC is the required equilateral triangle.

Question 3.

Draw ∆ PQR with PQ = 4 cm, QR = 3.5 cm and PR = 4 cm. What type of triangle is this ?

Solution:

Steps of Construction

- Draw a line segment QR of length 3.5 cm with Q as centre, draw an arc of radius 4 cm.
- With R as centre, cut the first arc with a radius 4 cm. Mark the point of intersection of arcs as P.
- Join PQ & PR A PQR is an isosceles triangle PQ = PR.

Question 4.

Construct ∆ ABC such that AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm. Measure <B.

Solution:

Steps of Construction

- Draw a line segment BC of length 6 cm with ‘B’ as centre, draw an arc of radius 2.5 cm.
- With ‘C’ as centre cut the first arc with a radius 6.5 cm.
- Mark the point of intersection of arcs as A. Join AB & AC. ∆ ABC is the required triangle.