KSEEB Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.3

   

Students can Download Chapter 10 Practical Geometry Ex 10.3, 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 10.3

Question 1.
Construct ∆ DEF such that DE = 5 cm, DF = 3 cm and ∠EDF = 90°.
Solution:
KSEEB Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.3 1
Steps of Construction

  1. Draw a line segment DE of length 5 cm. At D draw DM making 90° with DE.
  2. With ‘D’ as centre, draw an arc of radius 3 cm, It cuts DM at the point F.
  3. Join FE. The required triangle is obtained. That is ∆ DEF.

KSEEB Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.3

Question 2.
Construct an Isosceles triangle in which the lengths of each of its equal sides is 6.5 cm and the angle between them is 110°.
KSEEB Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.3 2
Solution:
Steps of Construction

  1. Draw a line segment AB of length 6.5 cm. At A draw AM making 110° with AB using a protractor.
  2. With ‘A’ as centre draw an arc of radius 6.5 cm and it cuts AM at C.
  3. Join BC. The required triangle is ∆ ABC.

KSEEB Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.3

Question 3.
Construct ∆ ABC with BC = 7.5 cm. AC = 5 cm and ∠C = 60°.
Solution:
KSEEB Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.3 3
Steps of Construction

  1. Draw a line segment BC of length 7.5 cm. At C draw CM making 60° with BC.
  2. With ‘C’ as centre, draw an arc of radius 5cm, it cuts at CM at A.
  3. Join AB. ∆ ABC is the required triangle.
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