Exercise 10.3 Practical-Geometry -NCERT Solutions Class 7

Go back to  'Practical Geometry'

Chapter 10 Ex.10.3 Question 1

Construct \(ΔDEF\) such that \(DE = 5\,\rm{ cm}\), \(DF = 3\,\rm{ cm}\) and \(m∠EDF = 90^\circ \).

Solution

Video Solution

What is known?

Lengths of sides of a \(ΔDEF\) and measure of one of the angles.

To construct:

A triangle \(ΔDEF\) such that \(DE = 5\,\rm{ cm}\), \(DF = 3\,\rm{ cm}\) and \(∠EDF = 90^\circ \).

Reasoning:

To construct a \(ΔDEF\) first, we draw a rough sketch with the given measure such that \(DE = 5\,\rm{ cm}, DF = 3\,\rm{ cm}\) and \(∠EDF = 90^\circ\), then follow the steps given below.

Steps:

Steps of construction:

  1. Draw a line segment \(DE\) of length \(5\,\rm{cm}.\)
  2. At \(D,\) draw \(DX\) making \(90^\circ\) with \(DE\).
  3. With \(D\) as centre, draw an arc of radius \(3\,\rm{cm}\). It cuts \(DX\) at the point \(F.\)
  4. Join \(EF\) to get the required triangle.

Thus,​​​​​​\(ΔDEF\)  is the required triangle.     

Chapter 10 Ex.10.3 Question 2

Construct an isosceles triangle in which the lengths of each of its equal sides is \(6.5\,\rm{ cm}\) and the angle between them is \(110^\circ\)

Solution

Video Solution

What is known?

Lengths of each of its equal sides is \(6.5\,\rm{ cm}\) and the angle between them is \(110^\circ.\)

To construct:

An isosceles triangle in which the lengths of each of its equal sides is \(6.5\,\rm{ cm}\) and the angle between them is \(110^\circ.\) 

Reasoning:

To construct an isosceles triangle in which the lengths of each of its equal sides is \(6.5\,\rm{ cm}\) and the angle between them is \(110^\circ,\)  follow the steps given below.

Steps:

Steps of construction :

  1. Draw a line segment \(QR\) of length \(6.5\,\rm{cm}.\)
  2. At \(Q,\) draw \(QX\) making an angle of \(110^\circ\) with \(QR.\)
  3. With \(Q\) as centre, draw an arc of radius \(6.5\,\rm{cm}.\) It cuts \(QX\) at the point \(P.\)
  4. Join \(PR\). Triangle \(PQR\) is the required isosceles triangle.

Chapter 10 Ex.10.3 Question 3

Construct \(\Delta ABC\) with \(BC = 7.5\,\rm{ cm}\), \(AC = 5\,\rm{ cm}\) and \(m∠C = 60^\circ\).

Solution

Video Solution

What is known?

Lengths of sides of a triangle are \(BC = 7.5\,\rm{ cm}\), \(AC = 5\,\rm{ cm}\) and \(∠C = 60^\circ\).

To construct:

A triangle \(\Delta ABC\)  with \(BC = 7.5\,\rm{ cm}\), \(AC = 5\,\rm{ cm}\) and \(∠C = 60^\circ\).

Reasoning:

To construct a triangle \(ΔABC\) with \(BC = 7.5\,{\rm{ cm}}, AC = 5\,\rm{ cm}\) and \(∠C = 60^\circ\), follow the steps given below.

Steps:

Steps of construction :

  1. Draw a line segment \(BC\) of length \(7.5\,\rm{cm}.\)
  2. At \(C\), draw \(CX\) making \(60^\circ\) with \(BC\).
  3. With \(C\) as centre, draw an arc of radius \(5\,\rm{cm}.\) It cuts \(CX\) at the point \(A\).
  4. Join \(AB.\)

Triangle \(ABC\) is the required triangle.

  
Download Cuemath NCERT App
Related Sections
Related Sections

Learn from the best math teachers and top your exams

Learn from the best

math teachers and top

your exams


Personalized Curriculum
Instant Doubts clarification
Cover latest CBSE Syllabus
Unlimited Mock & Practice tests
Covers CBSE, ICSE, IB curriculum

Instant doubt clearing with Cuemath Advanced Math Program
0