Ex.10.5 Q1 Practical Geometry Solution- NCERT Maths Class 7

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Construct the right angled \(ΔPQR\), where \(m∠Q = 90^\circ\), \(QR = 8 \rm{cm}\) and \(PR = 10\, \rm{cm}.\)

 Video Solution
Practical Geometry
Ex 10.5 | Question 1

Text Solution

What is known?

Length of two side of a right angled \(ΔPQR\).

To construct:

A right angled \(ΔPQR\), where \(∠Q = 90^\circ\), \(QR = 8 \rm{cm}\) and \(PR = 10\, \rm{cm}.\) 


To construct a right-angled \(ΔPQR\), where \(∠Q = 90^\circ\), \(QR = 8\,\rm{cm}\) and \(PR = 10 \,\rm{cm}\), draw a rough sketch and mark the measures. Remember to mark the right angle and follow the steps given below.


Steps of construction :

  1. Draw \(QR\) of length \( 8\, \rm{cm}.\)
  2. At \(Q\), draw \(QX\,\text{perpendicular}\,QR.\)
  3. With \(R\) as centre, draw an arc of radius \(10\,\rm{ cm}\) which should intersect \(QX\) at point \(P\).
  4. Join \(P\) and \(R\).

\(ΔPQR\) is the required triangle.

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