NCERT Solutions for Class 9 Maths Exercise 11.2

NCERT Solutions for Class 9 Maths Exercise 11.2 book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 9 Maths chapter wise NCERT solution for Maths Book for all the chapters can be downloaded from our website and myCBSEguide mobile app for free.

NCERT solutions for Class 9 Maths Constructions Download as PDF

NCERT Solutions for Class 9 Mathematics Constructions

1. Construct a triangle ABC in which BC = 7 cm, B = and AB + AC = 13 cm.

Ans. Given : Base BC = 7 cm, B = and sum of two sides AB + AC = 13 cm.

To construct : A triangle ABC.

Steps of construction:

(a) Draw a ray BX and cut off a line segment BC = 7 cm from it.

(b) At B, construct YBX = .

(c) With B as centre and radius = 13 cm ( AB + AC = 13 cm) draw an arc to meet BY at D.

(d) Join CD.

(e) Draw perpendicular bisector PQ of CD intersecting BD at A.

(f) Join AC.

Then ABC is required triangle.

Justification:

A lies on perpendicular bisector of CD.

AC = AD

And AB = BD -AD AB = BD -AC

AB + AC = BD = 13 cm

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NCERT Solutions for Class 9 Maths Exercise 11.2

2. Construct a triangle ABC in which BC = 8 cm, B = and AB -AC = 3.5 cm.

Ans. Given: Base BC = 8 cm, One Base angle B = and AB -AC = 3.5 cm

To construct: A triangle ABC.

Steps of construction:

(a) Draw a ray BX and cut off a line segment BC = 8 cm from it.

(b) Cut YBC =

(c) Cut off a line segment BD = 3.5 cm

( AB -AC = 3.5 cm) from BY.

(d) Join CD.

(e) Draw perpendicular bisector PQ of CD intersecting BY at a point A.

(f) Join AC.

Then ABC is the required triangle.

Justification:

A lies on the perpendicular bisector of CD.

AD = AC

Now BD = AB -AD

BD = AB -AC = 3.5 cm

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NCERT Solutions for Class 9 Maths Exercise 11.2

3. Construct a triangle PQR in which QR = 6 cm, Q = and PR -PQ = 2 cm.

Ans. Given: Base QR = 6 cm, one base angle Q = and PR -PQ = 2 cm.

To construct: A triangle PQR.

Steps of construction:

(a) Draw a ray QX and cut off a line segment QR = 6 cm from it.

(b) Construct a ray QY making an angle of with QR and produce YQ to form a line YQY’.

(c) Cut off a line segment QO = 2 cm ( PR -PQ = 2 cm) from QY’.

(d) Join OR.

(e) Draw perpendicular bisector MN of OR.

(f) Join PR.

Then PQR is the required triangle.

Justification:

P lies on perpendicular bisector of OR.

PO = PR

PQ + QO = PR

QO = PR -PQ = 2 cm

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NCERT Solutions for Class 9 Maths Exercise 11.2

4. Construct a triangle XYZ in which Y = Z = and XY + YZ + ZX = 11 cm.

Ans. Given: Base angles Y = and Z = and XY + YZ + ZX = 11 cm.

To construct: XYZ

Steps of construction:

(a) Draw a line segment PQ = 11 cm.

(b) Draw KPQ = and LQP =

(c) Bisect the KPQ and LQP. Let these intersect at point X.

(d) Draw perpendicular bisectors, MN of PX and RS of XQ.

(e) Let MN intersects PQ at Y and RS intersects PQ at Z.

(f) Join XY and XZ.

Then XYZ is the required triangle.

Justification:

Y lies on perpendicular bisector MN of PX.

PY = XY and similarly QZ = XZ

This gives XY + YZ + XZ = PY + YZ + QZ = PQ = 11 cm

Again YXP = XPY [Since XY = PY]

XYZ = YXP + XPY = 2XPY = KPQ

XYZ =

Similarly, XZY = LQP

XZY =

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NCERT Solutions for Class 9 Maths Exercise 11.2

5. Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm.

Ans. Given: Base BC = 12 cm and AB + AC = 18 cm.

To construct: A right angled triangle ABC right angled at B.

Steps of construction:

(a) Draw a ray BX and cut off a line segment BC = 12 cm from it.

(b) Draw an angle XBY =

(c) From BY cut off a line segment BD = 18 cm.

(d) Join CD.

(e) Draw the perpendicular bisector of CD intersecting BD at A.

(f) Join AC.

Then ABC is the required right angled triangle.

Justification:

A lies on the perpendicular bisector of CD.

AC = AD

And then AB = BD -AD

AB = BD -AC

AB + AC = BD = 18 cm.

NCERT Solutions for Class 9 Maths Exercise 11.2

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