# Ex.11.2 Q4 Constructions Solution - NCERT Maths Class 9

## Question

Construct a triangle \(XYZ\) in which \(\angle Y=30^{\circ}, \angle Z=90^{\circ}\) and \(XY + YZ + ZX = 11 \;\rm{cm.}\)

## Text Solution

**Steps of Construction:**

(i) Draw line \(BC=11\;\rm{cm} \).

(ii) Make angle of \( 30^\circ\) at \(B\) and \(90^\circ\) at \(C\) using a protractor.

(iii) Bisect angle \(B\).

With \(B\) as centre and any radius draw a wide arc to intersect both the arms of angle \(B\).

(iv) With intersecting points as the centre and same radius draw two arcs to intersect each other at \(P\) Draw line joining \(B\) and \(P\) and extend it beyond \(P\)

(v) Bisect angle \(C\).

With \(C\) as the centre and radius draw two arcs to intersect each other at \(Q\). Join \(Q\) and \(C\)such that it intersects ray \(BP\) at \(X\).

(vi) Perpendicular bisector of \(BX\).

With \(B\) and \(X\) as centres and radius greater than half of \(BX\) draw arcs on either side of line \(BX\) to intersect each other. Join the intersecting lines such that the perpendicular bisector intersects \(BC\) at \(Y\).

(vii) Perpendicular bisector of \(CX\)

With \(C\) and \(X\) as centres and radius greater than half of \(CX\) draw arcs on Join the intersecting lines such that the perpendicular bisector intersects \(BC\) at \(Z\).

(viii) Join \(XY\) and \(XZ\). \(XYZ\) is the required triangle.