Ex.10.4 Q2 Practical Geometry Solution - NCERT Maths Class 7

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Construct \(ΔPQR\)  if  \(PQ = 5 \rm{cm}\), \(m∠PQR = 105^\circ \) and \(m∠QRP = 40^\circ \). (Hint: Recall angle-sum property of a triangle).


What is known?

In triangle \(PQR\) , \(PQ = 5 \rm{cm}\), \(∠PQR = 105^\circ \) and \(∠QRP = 40^\circ \).

To construct:

A triangle \(ΔPQR\) if \(PQ = 5 \rm{cm}\), \(∠PQR = 105^\circ \) and \(∠QRP = 40^\circ \).


We will use angle-sum property of a triangle to find the measure of \(∠RPQ\)
By angle sum property of a triangle,
\(∠PQR + ∠QRP + ∠RPQ = 180^\circ.\)
\(105^\circ + 40^\circ + ∠RPQ = 180^\circ.\)
So, \(∠RPQ = 35^\circ\)
Now, let’s Construct \(ΔPQR\) such that \(PQ = 5\, {\rm{cm}} , ∠PQR = 105^\circ\) and \(∠RPQ = 35^\circ\), with the steps given below


Steps of construction

  1. Draw a line segment \(PQ\) of length \(5 \, \rm{cm}.\)
  2. At \(P\), draw \(PX\) making \(35^\circ\) with \(PQ.\)
  3. At \( Q,\) draw \(QY\) making an angle of \(105^\circ\) with \(PQ.\)
  4. \(PX\) and \(QY\) will intersect at point \(R.\)

\(PQR\) is the required triangle.

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