# 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:__

- Draw a line segment \(DE\) of length \(5\,\rm{cm}.\)
- At \(D,\) draw \(DX\) making \(90^\circ\) with \(DE\).
- With \(D\) as centre, draw an arc of radius \(3\,\rm{cm}\). It cuts \(DX\) at the point \(F.\)
- 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 :__

- Draw a line segment \(QR\) of length \(6.5\,\rm{cm}.\)
- At \(Q,\) draw \(QX\) making an angle of \(110^\circ\) with \(QR.\)
- With \(Q\) as centre, draw an arc of radius \(6.5\,\rm{cm}.\) It cuts \(QX\) at the point \(P.\)
- 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** :

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

Triangle \(ABC\) is the required triangle.