# Exercise E11.2 Constructions NCERT Solutions Class 9

Exercise 11.2

## Chapter 11 Ex.11.2 Question 1

Construct a triangle \(ABC\) in which \( BC =\) \(7\;\rm{cm}\) , \(\angle B\) = \(75°\) and \(\begin{align}\mathrm{AB}+\mathrm{AC}=13 \mathrm{cm}\end{align}\)

**Solution**

**Video Solution**

**Steps of Construction:**

(i) Draw base \(BC =7\;\rm{cm}\) and at point \(B\) make an angle of \(75^\circ\) using protractor.

(ii) With \(B\) as center and radius equal to \(13\;\rm{cm}\), draw an arc to intersect ray \(BX\) at \(D\).

(iii) Join \(DC\)

(iv) Measure \(\angle \mathrm{D}\) and make \(\angle \mathrm{ACD}=\angle \mathrm{D}\)

(v) Let \(CY\) intersect \(BD\) at \(A\)

(vi) \(ABC\) is the required triangle

## Chapter 11 Ex.11.2 Question 2

Construct a triangle \(ABC\) in which \({BC}=8\;\mathrm{cm}, \angle B=\;45^{\circ}\) and \(AB-AC=3.5\;\rm{cm}.\)

**Solution**

**Video Solution**

**Steps of Construction:**

(i) Draw base \( BC = 8\;\rm{cm}\) and at point \(B\) make an angle of using \(45^{\circ}\) a protractor.

(ii) With \(B\) as centre and radius \( BD = 3.5\;\rm{cm} \), draw an arc to intersect ray \(BX\) at \(D\).

(iii) Join \(DC\)

(iv) With \(D\) and \(C\) as the centre and radius greater than half of \(DC\). Draw arcs above and below the line to intersect ray \(BX\) at \(A\).

(v) Join \(AC\).\(ABC\) is the required triangle.

## Chapter 11 Ex.11.2 Question 3

Construct a triangle \(PQR\) in which \(QR=6\;\mathrm{cm}, Q=\angle 60^{\circ}\) and \(PR-PQ= 2\;\rm{cm}.\)

**Solution**

**Video Solution**

**Steps of Construction:**

(i) Draw line \(QR =6\;\rm{cm}\) . Make an angle of \(60^{\circ}\) at a point \(Q\) using a protractor and extend it below line \(QR\).

(ii) With \(Q\) as center and radius as \(2\;\rm{cm}\), draw an arc to intersect the ray \(QX\) at \(Z\)

(iii) Join \(RZ\)

(iv) With \(Z\) and \(R\) as centres and radius greater than half of \(ZR\), draw arcs on either side to intersect each other.

(v) Join the intersecting points and extend it to meet the ray \(QY\) at \(P\).

(vi) Join \(P\) and \(R\) .\(PQR\) is the required triangle.

## Chapter 11 Ex.11.2 Question 4

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

**Solution**

**Video 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.

## Chapter 11 Ex.11.2 Question 5

Construct a right triangle whose base is \(12\;\rm{cm}\) and sum of its hypotenuse and other side is \(18\;\rm{cm}\)

**Solution**

**Video Solution**

**Steps of Construction:**

(i) Draw the base \(BC=12\;\rm{cm} \).

(ii) At the point \(B\), make an angle \(CBX=90^\circ\) using a protractor.

(iii) Cut a line segment \(BD=18\;\rm{cm} \) from the ray \(BX\).

(iv) Join \(DC\)

(v) With \(D\) and \(C\) as the centres and radius greater than half of \(DC\) draw arcs on either side of the line to intersect each other. Join the intersecting points and extend the perpendicular bisector to meet \(BD\) at \(A\).

(vi) Join \(A\) and \(C\). \(ABC\) is the required right-angled triangle