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NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.2

CBSE students can refer to NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.2 Textbook Questions and Answers are provided by experts in order to help students secure good marks in exams.

Class 7 Maths NCERT Solutions Chapter 10 Practical Geometry Ex 10.2

Ex 10.2 Class 7 Maths Question 1.
Construct ∆XYZ in which XY = 45 cm, YZ = 5 cm and ZX = 6 cm.
Solution:
Steps of Construction :
NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry 4

  1. Draw a line segment YZ = 5 cm.
  2. With Y centre and draw an arc radius cm.
  3. With Z as a centre and draw another arc intersecting the arc radius = 6 cm, at X.
  4. Join XY and XZ to obtain the required triangle.

Ex 10.2 Class 7 Maths Question 2.
Construct an equilateral triangle of side 5.5 cm.
Solution:
Steps of Construction :
NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry 5

  1. Draw a line segment BC = 5.5 cm.
  2. With centre B and radius = 5.5 cm, draw an arc.
  3. With centre C and radius = 5.5 cm, draw another arc intersecting the arc drawn in step 2 at A.
  4. Join AB and AC to obtain the required triangle.

Ex 10.2 Class 7 Maths Question 3.
Draw ∆PQR with PQ = 4, QR = 3.5 cm and PR = 4 cm. What type of triangle is this?
Solution:
Steps of Construction :
NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry 6

  1. Draw a line segment QR = 3.5 cm.
  2. With centre Q and radius = 4 cm, draw an arc.
  3. With R as a centre and radius = 4 cm, draw another arc intersecting the arc drawn in step 2 at P.
  4. Join PQ and PR to obtain the required triangle. ∆PQR is an isosceles triangle.

Ex 10.2 Class 7 Maths Question 4.
Construct AABC such that AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm. Measure ∠B.
Solution:
Steps of Construction :

  1. Draw a line segment BC = 6 c.m
    NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry 7
  2. With centre B and radius = 2.5 cm, draw an arc.
  3. With centre C and radius =6.5 cm, draw another arc intersecting the arc drawn in step 2 at A.
  4. Join AB and AC to obtain the required triangle.
  5. On measuring, we find that ∠B = 90°

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