# Ex.10.2 Q4 Practical Geometry Solution - NCERT Maths Class 7

## Question

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

Video Solution

Practical Geometry

Ex 10.2 | Question 4

## Text 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 \(∠B =\) \(90^\circ.\)

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