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

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**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.