# Ex.10.3 Q3 Practical Geometry Solution- NCERT Maths Class 7

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## Question

Construct \(\Delta ABC\) with \(BC = 7.5\,\rm{ cm}\), \(AC = 5\,\rm{ cm}\) and \(m∠C = 60^\circ\).

## Text Solution

**What is known?**

Lengths of sides of a triangle are \(BC = 7.5\,\rm{ cm}\), \(AC = 5\,\rm{ cm}\) and \(∠C = 60^\circ\).

**To construct:**

A triangle \(\Delta ABC\) with \(BC = 7.5\,\rm{ cm}\), \(AC = 5\,\rm{ cm}\) and \(∠C = 60^\circ\).

**Reasoning:**

To construct a triangle \(ΔABC\) with \(BC = 7.5\,{\rm{ cm}}, AC = 5\,\rm{ cm}\) and \(∠C = 60^\circ\), follow the steps given below.

**Steps:**

** Steps of construction** :

- Draw a line segment \(BC\) of length \(7.5\,\rm{cm}.\)
- At \(C\), draw \(CX\) making \(60^\circ\) with \(BC\).
- With \(C\) as centre, draw an arc of radius \(5\,\rm{cm}.\) It cuts \(CX\) at the point \(A\).
- Join \(AB.\)

Triangle \(ABC\) is the required triangle.