# Ex.10.2 Q1 Practical Geometry Solution- NCERT Maths Class 7

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## Question

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

Video Solution

Practical Geometry

Ex 10.2 | Question 1

## Text 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.