# Construct ∆ABC such that AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm. Measure ∠B.

**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 ΔABC 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 BC of length 6cm.
- From B, we need a point A is at a distance of 2.5cm. So, with B as center, draw an arc of radius 2.5cm. (now A will be somewhere on this arc and our job is to find where exactly A is).
- From C, point A is at a distance of 6.5cm. So, with C as the center, draw an arc of radius 6.5cm. (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 the arcs as A.
- Join AB and AC.
- Thus, ABC is the required triangle.
- Measure angle B with the help of a protractor. It is a right-angled triangle ABC, where ∠B = 90
^{∘}.

**Video Solution:**

## Construct ∆ABC such that AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm. Measure ∠B.

### Class 7 Maths NCERT Solutions - Chapter 10 Exercise 10.2 Question 4

**Summary:**

∆ABC is constructed such that AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm. The value of ∠B = 90^{∘}.