# Ex.11.1 Q4 Constructions Solution - NCERT Maths Class 9

## Question

Construct the following angles and verify by measuring them by a protractor:

(i) \(75^{\circ}\)

(ii) \(105^{\circ}\)

(iii) \(135^{\circ}\)

## Text Solution

\({\rm{(i)}}\;\;75^{\circ}\)

**Reasoning:**

We need to construct two adjacent angles of \(60^{\circ}\).The second angle should be bisected twice to get a \(15^{\circ}\) angle.

\[\begin{align}75^{\circ}&=60^{\circ}+15^{\circ} \\ 15^{\circ}&=\frac{30^{\circ}}{2}=\frac{60^{\circ}}{2} \div \frac{1}{2}\end{align}\]

**Steps of Construction:**

(i) Draw ray \(PQ\).

(ii) To construct an angle of \(60^{\circ}\) **.**

With \(P\) as centre and any radius draw a wide arc to intersect \(PQ\) at \(R\) With \(R\) as centre and same radius draw an arc to intersect the initial arc at \(S\) .\(\angle \mathrm{SPR}=60^{\circ}\)

(iii) To construct adjacent angle of \(60^{\circ}\) with \(S\) as center and same radius draw an arc to \(T\) intersect the initial arc.

(iv) To bisect \(\angle \mathrm{SPT}\)

With \(T\) and \(S\) as centres and same radius draw arcs to bisect each other at \(U\) Join \(U\) and \(P\). \(\begin{align}\angle \mathrm{UPS}=\frac{1}{2} \angle \mathrm{TPS}=30^{\circ}\end{align}\)

(v) To bisect \(\angle \mathrm{UPS}\)

With \(V\) and \(S\) as centers and radius greater than half of \(VS\) draw arcs to intersect each other at \(X\)

\[\begin{align} \angle {XPS} &=\frac{1}{2} \angle {UPS} \\ &=\frac{1}{2} \times 30^{0} \\ &=15^{\circ} \\ \angle {XPQ} &=\angle {XPS}+\angle {SPQ} \\ &=15^{\circ}+60^{\circ} \end{align}\]

\({\rm{(ii)}}\;\;105^{\circ}\)

**Reasoning:**

We need to construct two adjacent angles of \(60^{\circ}\) . In the second angle we need to bisect it to get two \(30^{\circ}\) angles. The second \(30^{\circ}\) angle should be bisected again to get a \(15^{\circ}\) angle. Together we can make an angle of \(105^{\circ}\) **.**

\[\begin{align} 105^{\circ} &=60^{\circ}+45^{\circ} \\ 45^{\circ} &=60^{\circ}+30^{\circ}+15^{\circ} \end{align}\]

**Steps of Construction:**

(i) Draw ray \(PQ\)

(ii) To construct an angle of \(60^{\circ}\).

With \(P\) as centre and any radius draw a wide arc to intersect \(PQ\) at \(R\). With \(R\) as centre and same radius draw an arc to intersect the initial arc at \(S\). \(\angle \mathrm{SPR}=60^{\circ}\)

(iii) To construct an adjacent angle of \(60^{\circ}\) with \(S\) as the center and same radius as before draw an arc to intersect the initial arc at \(T\). \(\angle \mathrm{TPS}=60^{\circ}\)

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

With \(T\) and \(S\) as centres and same radius draw arcs to bisect each other at \(U\). Join \(U\) and \(P\).

\[\begin{align}\angle \text{UPS = }\frac{1}{2}\angle \text{TPS }=\frac{1}{2}\times {{60}^{0}}\text{=30}{}^\circ \end{align}\]

(v)To bisect \(\angle \text{UPT}\)

With \(T\) and \(V\) as centers and radius greater than half of \(TV,\)draw arcs to intersect each other at \(W\). Join \(P\) and \(W\).

\[\begin{align} \angle {WPU} &=\frac{1}{2} \angle {UTP} \\ &=\frac{1}{2} \times 30^{\circ} \\ &=15^{\circ} \\ \angle {WPR} &=\angle {WPU}+\angle {UPS}+\angle {SPR} \\ &=15^{\circ}+30^{\circ}+60^{\circ} \\ &=105^{\circ} \end{align}\]

\({\rm{(iii)}}\;\;135^{\circ}\)

**Reasoning:**

We need to construct three adjacent angles of \(60^{\circ}\) each. The third angle should be bisected twice successively to get an angle of \(15^{\circ}\) . Together we will get an angle of \(135^{\circ}\) **.**

\[\begin{align} 135^{\circ} &=15^{\circ}+60^{\circ}+60^{\circ} \\ 15^{\circ} &=\frac{60^{\circ}}{2} \div \frac{1}{2} \end{align}\]

**Steps of Construction:**

(i) Draw ray\(PQ\)

(ii) To construct an angle of \(60^{\circ}\) .

With \(P\) as the center and any radius draw an arc to intersect \(PQ\) at \(R\).\( \angle \mathrm{SPR}=60^{\circ}\)

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

With \(S\) as the center and same radius as before draw an arc to intersect the initial arc at \(T\). \(\angle \mathrm{TPS}=60^{\circ}\)

(iv) To construct the second adjacent angle of \(60^{\circ}\)

With \(T\) as centre and some radius as before draw an arc to intersect the initial arc at \(U\)

\[\angle \mathrm{UPT}=60^{\circ}\]

(v) To bisect \(\angle \mathrm{UPT}\)

With \(T\) and \(U\) as centers and same radius as before draw an arc to intersect each other at \(V\) \(\angle \mathrm{VPT}=\angle \mathrm{VPU}=30^{\circ}\)

(vi) To bisect \(\angle \mathrm{VPT}\)

With \(W\) and \(T\) as centers and radius greater than half of \(WT\) draw arcs to intersect each other at \(X\).

\[\begin{align}XPT&=\angle XPV \\&=7\frac{1}{2}\angle V\text{PT}\\&=\frac{1}{2}\times {{30}^\circ}=\,15{}^\circ \\ \\ \angle XPQ &= \angle \text{XPT + }\angle TPS + \angle SPR \\&= 15{}^\circ + 60{}^\circ + 60{}^\circ \\&=135{}^\circ \,\,\, \\\end{align}\]