# Exercise 10.2 Practical-Geometry -NCERT Solutions Class 7

## Chapter 10 Ex.10.2 Question 1

Construct \(ΔXYZ\) in which \(XY =4.5\,\rm{cm,}\) \(YZ = 5\,\rm{cm}\) and \(ZX =6\,\rm{cm}.\)

**Solution**

**Video Solution**

**What is known?**

Lengths of sides of a triangle \(XYZ\) are \(XY =4.5\,\rm{cm,}\) \(YZ = 5\,\rm{cm}\) and \(ZX =6\,\rm{cm}.\)

**To construct:**

A triangle \(XYZ\) in which \(XY =4.5\,\rm{cm},\) \(YZ = 5\,\rm{cm}\) and \( ZX =6\,\rm{cm}.\)

**Reasoning:**

We will draw a rough sketch of \(ΔXYZ\) with the given measure. This will help us in deciding how to proceed. Then follow the steps given below.

**Steps:**

__Steps of construction__

- Draw a line segment \(YZ\) of length \(5\) \(\rm{cm.}\)
- From \(Y\), point \(X\) is at a distance of \(4.5\) \(\rm{cm}.\). So, with \(Y\) as centre, draw an arc of radius \(4.5\) \(\rm{cm}.\) (now \(X\) will be somewhere on this arc & our job is to find where exactly \(X\) is).
- From \(Z\), point \(X\) is at a distance of \(6\) \(\rm{cm}.\). So, with \(Z\) as centre, draw an arc of radius \(6\) \(\rm{cm}.\) (now \(X\) will be somewhere on this arc, we have to fix it).
- \(X\) has to be on both the arcs drawn. So, it is the point of intersection of arcs. Mark the point of intersection of arcs as \(X\). Join \(XY \) and \( XZ.\)

Thus, \(XYZ\) is the required triangle.

## Chapter 10 Ex.10.2 Question 2

Construct an equilateral triangle of side \(5.5\,\rm{cm} \).

**Solution**

**Video Solution**

**What is known?**

Lengths of sides of an equilateral triangle are \(5.5\,\rm{cm}\) each.

**To construct:**

An equilateral triangle of side \(5.5\,\rm{cm} \).

**Reasoning:**

To construct an equilateral triangle of side \(5.5\,\rm{cm} \). Follow the steps given below

**Steps:**

__Steps of construction__ :

- Draw a line segment \(BC\) of length \(5.5\,\rm{cm} \).
- From \(B\), point \(A\) is at a distance of \(5.5\,\rm{cm} \).So, with \(B\) as centre, draw an arc of radius \(5.5\,\rm{cm} \) (Now \(A\) will be somewhere on this arc. Our job is to find where exactly \(A\) is).
- From \(C\), point \(A\) is at a distance of \(5.5\,\rm{cm} \).So, with \(C\) as centre, draw an arc of radius \(5.5\,\rm{cm} \) (Now \(A\) will be somewhere on this arc, we have to fix it).
- \(A\) has to be on both the arcs drawn. So, it is the point of intersection of arcs. Mark the point of intersection of arcs as \(A\) join \(AB \) and \(AC\).

Thus, \(ABC\) is the required triangle.

## Chapter 10 Ex.10.2 Question 3

Draw \( ΔPQR\) with \(PQ = 4\,\rm{cm} \), \(QR = 3.5\,\rm{cm} \) and \(PR = 4\,\rm{cm} \). What type of triangle is this?

**Solution**

**Video Solution**

**What is known?**

Lengths of sides of a triangle are \(PQ = 4\,\rm{cm} \), \(QR = 3.5\,\rm{cm} \) and \(PR = 4\,\rm{cm} \).

**To construct:**

A \(ΔPQR\) with \(PQ = 4\,\rm{cm} \), \(QR = 3.5\,\rm{cm} \) and \(PR = 4\,\rm{cm} \).

**Reasoning:**

To construct a \(ΔPQR\) with \(PQ = 4 \,\rm{cm,}\, QR = 3.5\,\rm{ cm}\) and \(PR = 4 \,\rm{cm}\), follow the steps given below.

**Steps:**

__Steps of construction__ –

- Draw a line segment \(QR\) of length \(3.5\, \rm{cm}.\)
- From \(Q\), point \(P\) is at a distance of \(4\,\rm{cm}\). So, with \(Q\) as centre, draw an arc of radius \(4\,\rm{cm}\) (now \(P\) will be somewhere on this arc & our job is to find where exactly \(P\) is).
- From \(R\), point \(P\) is at a distance of \(4\,\rm{cm}\). So, with \(R\) as centre, draw an arc of radius \(4\,\rm{cm}\) (now \(P\) will be somewhere on this arc, we have to fix it).
- \(P\) has to be on both the arcs drawn. So, it is the point of intersetion of arcs. Mark the point of intersection of arcs as \(P\) join \(PQ\) and \(PR.\)

Thus, \(PQR\) is the required triangle.

\(\Delta PQR\) is an isosceles triangle as two of the sides are equal.

## Chapter 10 Ex.10.2 Question 4

Construct \(ΔABC\) such that \(AB = 2.5\,\rm{cm}\)., \(BC = 6\,\rm{cm}\). and \(AC = 6.5\,\rm{cm}\).. Measure \(∠B.\)

**Solution**

**Video Solution**

**What is known?**

Lengths of sides of a triangle are \(AB = 2.5 \,\rm{cm}\)., \(BC = 6\,\rm{ cm}\). and \(AC = 6.5\,\rm{ cm}\)..

**To construct:**

A \(ΔABC\) such that \(AB = 2.5\,\rm{cm}\)., \(BC = 6\,\rm{cm}\). and \(AC = 6.5\,\rm{cm}\).. Measure \(∠B.\)

**Reasoning:**

To Construct \(ΔABC\) such that \(AB = 2.5 \,\rm{cm}, \,BC = 6 \,\rm{cm}\) and \(AC = 6.5\,\rm{ cm}\). and to measure \(∠B\), follow the steps given below.

**Steps:**

**Steps of construction:**

- Draw a line segment \(BC\) of length \(6\,\rm{ cm}\)..
- From \(B\), point \(A\) is at a distance of \(2.5\,\rm{ cm}\).. So, with \(B\) as centre, draw an arc of radius \(2.5\,\rm{ cm}\). (now \(A\) will be somewhere on this arc & our job is to find where exactly \(A\) is).
- From \(C\), point \(A\) is at a distance of \(6.5\,\rm{ cm}\).. So, with \(C\) as centre, draw an arc of radius \(6.5\,\rm{ cm}\). (now \(A\) will be somewhere on this arc, we have to fix it).
- \(A\) has to be on both the arcs drawn, so it is the point of intersection of arcs. Mark the point of intersection of arcs as \(A\) join \(AB\) and \(AC.\)

Thus, \(ABC\) is the required triangle.

Measure angle \(B\) with the help of protractor. It is the right-angled triangle \(ABC\), where \(∠B =\) \(90^\circ.\)