# Ex.10.3 Q1 Practical Geometry Solution- NCERT Maths Class 7

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## Question

Construct \(ΔDEF\) such that \(DE = 5\,\rm{ cm}\), \(DF = 3\,\rm{ cm}\) and \(m∠EDF = 90^\circ \).

Video Solution

Practical Geometry

Ex 10.3 | Question 1

## Text Solution

**What is known?**

Lengths of sides of a \(ΔDEF\) and measure of one of the angles.

**To construct:**

A triangle \(ΔDEF\) such that \(DE = 5\,\rm{ cm}\), \(DF = 3\,\rm{ cm}\) and \(∠EDF = 90^\circ \).

**Reasoning:**

To construct a \(ΔDEF\) first, we draw a rough sketch with the given measure such that \(DE = 5\,\rm{ cm}, DF = 3\,\rm{ cm}\) and \(∠EDF = 90^\circ\), then follow the steps given below.

**Steps:**

__Steps of construction:__

- Draw a line segment \(DE\) of length \(5\,\rm{cm}.\)
- At \(D,\) draw \(DX\) making \(90^\circ\) with \(DE\).
- With \(D\) as centre, draw an arc of radius \(3\,\rm{cm}\). It cuts \(DX\) at the point \(F.\)
- Join \(EF\) to get the required triangle.

Thus,\(ΔDEF\) is the required triangle.