# Draw ∆PQR with PQ = 4 cm, QR = 3.5 cm and PR = 4 cm. What type of triangle is this?

**Solution:**

We use the basic rules of construction to solve the question given. We will use scale and compass to do the construction required.

We will draw a rough sketch of ΔPQR with the given measures for our reference. This will help us in deciding how to proceed. Then follow the steps given below.

Steps of construction -

- Draw a line segment QR of length 3.5 cm.
- From Q, we need a point P is at a distance of 4 cm. So, with Q as the center, draw an arc of radius 4 cm (Now P will be somewhere on this arc. Our job is to find where exactly P is).
- From R, we know that point P is at a distance of 4 cm. So, with R as the center, draw an arc of radius 4cm (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 intersection of arcs. Mark the point of intersection of arcs as P
- Join PQ and PR.
- Thus, PQR is the required triangle.

ΔPQR is an isosceles triangle as two of its sides are equal, i.e., PQ = PR = 4 cm

**Video Solution:**

## Draw ∆PQR with PQ = 4 cm, QR = 3.5 cm and PR = 4 cm. What type of triangle is this?

### Class 7 Maths NCERT Solutions - Chapter 10 Exercise 10.2 Question 3

**Summary:**

∆PQR is constructed with PQ = 4 cm, QR = 3.5 cm and PR = 4 cm. ΔPQR is an isosceles triangle as two of the sides are equal.