# Ex.11.1 Q2 Constructions Solution - NCERT Maths Class 9

## Question

Construct an angle of \(45^\circ\) at the initial point of a given ray and justify the construction.

## Text Solution

**Reasoning:**

We need to construct two adjacent angles each of \(60^\circ\) degrees and bisect the second one to construct \(90^\circ\) . Then bisect the \(90^\circ\) angle to get \(45^\circ\)

\[\begin{align}\frac{60^{0}+\frac{60^{0}}{2}}{2}=45^{0}\end{align}\]

**Steps of Construction:**

(i) Draw ray \(PQ\).

(ii) To construct an angle of \(60^{o}\) ** . **With \(P\) as center draw a wide arc of any radius to intersect the ray at \(R\). With \(R\) as center and same radius draw an arc to intersect the initial one at \(S\). \(\angle {SPR}=60^{\circ}\) .

(iii) To construct adjacent angle of \(60^{\circ}\) .

With \(S\) as centre and same radius draw an arc to intersect the previous arc at \(T\)

\[\angle {TPS}=60^{\circ}\]

(iv) To bisect \(\angle {TPS}\)

With \(T\) and \(S\) as centre and same radius, draw arcs to intersect each other at \(U\).

\[\begin{align}\angle {UPS}=\frac{1}{2} \angle {TPS}=30^{\circ}\end{align}\]

(v) Join \(P\) and \(U\) to intersect the initial arc at \(V\).

\[\begin{align} \angle {UPQ} &=\angle {UPS}+\angle {SPR} \\ &=30^{\circ}+60^{\circ} \\ &=90^{\circ} \end{align}\]

(vi) To bisect \(\angle {UPQ}\)

With \(R\) and \(V\) as centers and radius greater than half of \(RV\), draw arc to intersect each other at \(W\). Join \(PW\). \(PW\) is the angle bisector of \(\angle U P Q\)

\[\begin{align} \angle {WPQ} &=\frac{1}{2} \angle {UPQ} \\ &=\frac{1}{2} \times 90^{\circ} \\ &=45^{\circ} \end{align}\]

(vii) Ray \(PW\) forms an angle of \(45^{\circ}\) with ray \(PQ\) at the initial point.