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

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

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**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 \(∠QRP = 40^\circ \).

**Reasoning:**

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:**

** Steps of construction** –

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

\(PQR\) is the required triangle.

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