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

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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 \(∠EDF = 90^\circ \).


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 of construction:

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

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

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