KSEEB Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.2
KSEEB Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.2
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
- 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.