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

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## Question

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 \(PR = 10\, \rm{cm}.\)

**Reasoning:**

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:**

** Steps of construction** :

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

\(ΔPQR\) is the required triangle.