Exercise 10.3 Practical-Geometry -NCERT Solutions Class 7

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.