NCERT Solutions for Class 9 Maths Chapter 11 Constructions

## NCERT Solutions for Class 9 Maths Chapter 11 Constructions Ex 11.1

Ex 11.1 Class 9 Maths Question 1.

Construct an angle of 90° at the initial point of a given ray and justify the construction.

Solution:

Steprf of Construction:

Step I : Draw AB¯¯¯¯¯¯¯¯.

Step II : Taking O as centre and having a suitable radius, draw a semicircle, which cuts OA¯¯¯¯¯¯¯¯ at B.

Step III : Keeping the radius same, divide the semicircle into three equal parts such that BC˘=CD˘=DE˘ .

Step IV : Draw OC¯¯¯¯¯¯¯¯ and OD¯¯¯¯¯¯¯¯.

Step V : Draw OF¯¯¯¯¯¯¯¯, the bisector of ∠COD.

Thus, ∠AOF = 90°

Justification:

∵ O is the centre of the semicircle and it is divided into 3 equal parts.

∴ \breve { BC } =\quad \breve { CD } \quad =\quad \breve { DE }

BC˘=CD˘=DE˘

⇒ ∠BOC = ∠COD = ∠DOE [Equal chords subtend equal angles at the centre]

And, ∠BOC + ∠COD + ∠DOE = 180°

⇒ ∠BOC + ∠BOC + ∠BOC = 180°

⇒ 3∠BOC = 180°

⇒ ∠BOC = 60°

Similarly, ∠COD = 60° and ∠DOE = 60°

∵ OF¯¯¯¯¯¯¯¯ is the bisector of ∠COD

∴ ∠COF = 1/2 ∠COD = 1/2 (60°) = 30°

Now, ∠BOC + ∠COF = 60° + 30°

⇒ ∠BOF = 90° or ∠AOF = 90°

Ex 11.1 Class 9 Maths Question 2.

Construct an angle of 45° at the initial point of a given ray and justify the construction.

Solution:

Steps of Construction:

Stept I : Draw OA¯¯¯¯¯¯¯¯.

Step II : Taking O as centre and with a suitable radius, draw a semicircle such that it intersects OA¯¯¯¯¯¯¯¯. at B.

Step III : Taking B as centre and keeping the same radius, cut the semicircle at C. Now, taking C as centre and keeping the same radius, cut the semicircle at D and similarly, cut at E, such that BC˘=CD˘=DE˘

Step IV : Draw OC¯¯¯¯¯¯¯¯ and OD¯¯¯¯¯¯¯¯.

Step V : Draw OF¯¯¯¯¯¯¯¯, the angle bisector of ∠BOC.

Step VI : Draw OG¯¯¯¯¯¯¯¯, the ajngle bisector of ∠FOC.

Thus, ∠BOG = 45° or ∠AOG = 45°

Justification:

∵ BC˘=CD˘=DE˘

∴ ∠BOC = ∠COD = ∠DOE [Equal chords subtend equal angles at the centre]

Since, ∠BOC + ∠COD + ∠DOE = 180°

⇒ ∠BOC = 60°

∵ OF¯¯¯¯¯¯¯¯ is the bisector of ∠BOC.

∴ ∠COF = 1/2 ∠BOC = 1/2(60°) = 30° …(1)

Also, OG¯¯¯¯¯¯¯¯ is the bisector of ∠COF.

∠FOG = 1/2∠COF = 1/2(30°) = 15° …(2)

Adding (1) and (2), we get

∠COF + ∠FOG = 30° + 15° = 45°

⇒ ∠BOF + ∠FOG = 45° [∵ ∠COF = ∠BOF]

⇒ ∠BOG = 45°

Ex 11.1 Class 9 Maths Question 3.

Construct the angles of the following measurements

(i) 30°

(ii) 22 1/2∘

(iii) 15°

Solution:

(i) Angle of 30°

Steps of Construction:

Step I : Draw OA¯¯¯¯¯¯¯¯.

Step II : With O as centre and having a suitable radius, draw an arc cutting OA¯¯¯¯¯¯¯¯ at B.

Step III : With centre at B and keeping the same radius as above, draw an arc to cut the previous arc at C.

Step IV : Join OC¯¯¯¯¯¯¯¯ which gives ∠BOC = 60°.

Step V : Draw OD¯¯¯¯¯¯¯¯, bisector of ∠BOC, such that ∠BOD = 1/2∠BOC = 1/2(60°) = 30°

Thus, ∠BOD = 30° or ∠AOD = 30°

(ii) Angle of 22 1/2∘

Steps of Construction:

Step I : Draw OA¯¯¯¯¯¯¯¯.

Step II : Construct ∠AOB = 90°

Step III : Draw OC¯¯¯¯¯¯¯¯, the bisector of ∠AOB, such that

∠AOC = 1/2∠AOB = 1/2(90°) = 45°

Step IV : Now, draw OD, the bisector of ∠AOC, such that

∠AOD = 1/2∠AOC = 1/2(45°) = 22 1/2∘

Thus, ∠AOD = 22 1/2∘

(iii) Angle of 15°

Steps of Construction:

Step I : Draw OA¯¯¯¯¯¯¯¯.

Step II : Construct ∠AOB = 60°.

Step III : Draw OC, the bisector of ∠AOB, such that

∠AOC = 1/2∠AOB = 1/2(60°) = 30°

i.e., ∠AOC = 30°

Step IV : Draw OD, the bisector of ∠AOC such that

∠AOD = 1/2∠AOC = 1/2(30°) = 15°

Thus, ∠AOD = 15°

Ex 11.1 Class 9 Maths Question 4.

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

(i) 75°

(ii) 105°

(iii) 135°

Solution:

Step I : Draw OA¯¯¯¯¯¯¯¯.

Step II : With O as centre and having a suitable radius, draw an arc which cuts OA¯¯¯¯¯¯¯¯ at B.

Step III : With centre B and keeping the same radius, mark a point C on the previous arc.

Step IV : With centre C and having the same radius, mark another point D on the arc of step II.

Step V : Join OC¯¯¯¯¯¯¯¯ and OD¯¯¯¯¯¯¯¯, which gives ∠COD = 60° = ∠BOC.

Step VI : Draw OP¯¯¯¯¯¯¯¯, the bisector of ∠COD, such that

∠COP = 1/2∠COD = 1/2(60°) = 30°.

Step VII: Draw OQ¯¯¯¯¯¯¯¯, the bisector of ∠COP, such that

∠COQ = 1/2∠COP = 1/2(30°) = 15°.

Thus, ∠BOQ = 60° + 15° = 75°∠AOQ = 75°

(ii) Steps of Construction:

Step I : Draw OA¯¯¯¯¯¯¯¯.

Step II : With centre O and having a suitable radius, draw an arc which cuts OA¯¯¯¯¯¯¯¯ at B.

Step III : With centre B and keeping the same radius, mark a point C on the previous arc.

Step IV : With centre C and having the same radius, mark another point D on the arc drawn in step II.

Step V : Draw OP, the bisector of CD which cuts CD at E such that ∠BOP = 90°.

Step VI : Draw OQ¯¯¯¯¯¯¯¯, the bisector of BC˘ such that ∠POQ = 15°

Thus, ∠AOQ = 90° + 15° = 105°

(iii) Steps of Construction:

Step I : Draw OP¯¯¯¯¯¯¯¯.

Step II : With centre O and having a suitable radius, draw an arc which cuts OP¯¯¯¯¯¯¯¯ at A

Step III : Keeping the same radius and starting from A, mark points Q, R and S on the arc of step II such that AQ˘=QR˘=RS˘ .

StepIV :Draw OL¯¯¯¯¯¯¯, thebisector of RS˘ which cuts the arc RS˘ at T.

Step V : Draw OM¯¯¯¯¯¯¯¯¯, the bisector of RT˘.

Thus, ∠POQ = 135°

Ex 11.1 Class 9 Maths Question 5.

Construct an equilateral triangle, given its side and justify the construction.

Solution:

pt us construct an equilateral triangle, each of whose side = 3 cm(say).

Steps of Construction:

Step I : Draw OA¯¯¯¯¯¯¯¯.

Step II : Taking O as centre and radius equal to 3 cm, draw an arc to cut OA¯¯¯¯¯¯¯¯ at B such that OB = 3 cm

Step III : Taking B as centre and radius equal to OB, draw an arc to intersect the previous arc at C.

Step IV : Join OC and BC.

Thus, ∆OBC is the required equilateral triangle.

Justification:

∵ The arcs OC˘ and BC˘ are drawn with the same radius.

∴ OC˘ = BC˘

⇒ OC = BC [Chords corresponding to equal arcs are equal]

∵ OC = OB = BC

∴ OBC is an equilateral triangle.

### NCERT Solutions for Class 9 Maths Chapter 11 Constructions Ex 11.2

## Constructions Class 9 Extra Questions Maths Chapter 11

**Extra Questions for Class 9 Maths Chapter 11 Constructions**

**IMPORTANT QUESTIONS**

**VERY SHORT ANSWER TYPE QUESTIONS**

**1. Construct an angle of 90° at the initial point of the given ray. [CBSE-15-6DWMW5A]**

**Answer.**

**2. Draw a line segment PQ = 8.4 cm. Divide PQ into four equal parts using ruler and compass. [CBSE-14-ERFKZ8H], [CBSE – 14-GDQNI3W], [CBSE-14-17DIG1U]**

**Answer.** **Steps of construction :**

- Draw a line segment PQ = 8.4 cm.
- With P and Q as centres, draw arcs of radius little more than half of PQ. Let his line intersects PQ in M.
- With M and Q as centres, draw arcs of radius little more than half of MQ. Let this line intersects PQ in N.
- With P and M as centres, draw arcs of radius little more than half of PM. Let this line intersects PQ in L. Thus, L, M and N divide the line segment PQ in four equal parts.

**3. Draw any reflex angle. Bisect it using compass. Name the angles so obtained. [CBSE-15-NS72LP7]**

**Answer.**

**4. Why we cannot construct a ΔABC, if ****∠A=60°, AB — 6 cm, AC + BC = 5 cm but construction of A ABC is possible if ∠A=60°, AB = 6 cm and AC – BC = 5 cm. [CBSE-14-GDQNI3W]**

**Answer.** We know that, by triangle inequality property, construction of triangle is possible if sum of two sides of a triangle is greater than the third side.

Here, AC + BC = 5 cm which is less than AB ( 6 cm)

Thus, ΔABC is not possible.

Also, by triangle inequality property, construction of triangle is possible, if difference of two

sides of a triangle is less than the third side

Here, AC – BC = 5 cm, which is less than AB (6 cm)

Thus, ΔABC is possible.

**5. Construct angle of [52 1/2]0 using compass only. [CBSE-14-17DIG1U]**

**Answer.**

**SHORT ANSWER QUESTIONS TYPE-I**

**6. Using ruler and compass, construct 4∠XYZ, if ∠XYZ= 20° [CBSE-14-ERFKZ8H]**

**Answer.**

**7. Construct an equilateral triangle LMN, one of whose side is 5 cm. Bisect ∠ M of the triangle. [CBSE March 2012]**

**Answer. Steps of construction :**

- Draw a line segment LM = 5 cm.
- Taking L as centre and radius 5 cm draw an arc.
- Taking M as centre and radius draw an other arc intersecting previous arc at N.
- Join LN and MN. Thus, ΔLMN is the required equilateral triangle.
- Taking M as centre and any suitable radius, draw an arc intersecting LM at P and MN at Q.
- Taking P and Q as centres and same radii, draw arcs intersecting at S.
- Join MS and produce it meet LN at R. Thus, MSR is the required bisector of
**∠**M.

**SHORT ANSWER QUESTIONS TYPE-II**

**8. Construct a A ABC with BC = 8 cm, ∠B= 45° and AB – AC = 3.1 cm. [CBSE-15-NS72LP7]**

**Answer.**

**9. Construct an isosceles triangle whose two equal sides measure 6 cm each and whose base is 5 cm. Draw the perpendicular bisector of its base and show that it passes through the opposite vertex [CBSE-15-6DWMW5A]**

**Answer. Steps of construction :**

- Draw a line segment AB = 5 cm.
- With A and B as centres, draw two arcs of radius 6 cm and let they intersect each other in C.
- Join AC and BC to get ΔABC.
- With A and B as centres, draw two arcs of radius little more than half of AB. Let they intersect each other in P and Q. Join PQ and produce, to pass through C.

**10. Construct a right triangle whose base is 8 cm and sum of the hypotenuse and other side is 16 cm.**

**Answer. Given :** In ΔABC, BC = 8 cm, ∠B= 90° and AB + AC = 16 cm.

**Required :** To construct ΔABC.

**Steps of construction:**

- Draw a line segment BC = 8 cm.
- At B, Draw ∠CBX = 90°.
- From ray BX, cut off BE = 16 cm.
- Join CE .
- Draw the perpendicular bisector of EC meeting BE at A.
- Join AC to obtain the required ΔABC.

**11. To construct an isosceles ΔABC in which base BC = 4 cm, sum of the perpendicular from A to BC and side AB = 6.5 cm.**

**Answer.** **Given :** In ΔABC, BC = 4 cm and sum of the perpendicular from A to BC and side AB = 6.5 cm.

**Required :** To construct ΔABC.

**Steps of construction :**

- Draw any line segment BC = 4 cm.
- Draw ‘p’ the perpendicular bisector of BC and let it intersect BC in R
- Cut off PQ = 6.5 cm.
- Join QB.
- Draw the perpendicular bisector of BQ and let it intersect PQ in A.
- Join AB and AC. Thus, ΔABC is the required triangle.

**12. Construct an equilateral triangle of altitude 6 cm. [CBSE-15-6DWMW5A]**

**Answer. Steps of construction :**

- Draw any line l.
- Take any point M on it and draw a line p perpendicular to l.
- With M as centre, cut off MC = 6 cm
- At C, with initial line CM construct angles of measures 30° on both sides and let these lines intersect line l in A and B. Thus, ΔABC is the required triangle.

**13. Draw a line segment QR = 5 cm. Construct perpendiculars at point Q and R to it. Name them as QX and RY respectively. Are they both parallel ? [CBSE-15-NS72LP7] [CBSE-14-ERFKZ8H]**

**Answer. Steps of construction :**

- Draw a line segment QR = 5 cm.
- With Q as centre, construct an angle of 90° and let this line through Q is QX.
- With R as centre, construct an angle of 90° and let this line through R is RY. Yes, the perpendicular lines QX and- RY are parallel.

**LONG ANSWER TYPE QUESTIONS**

**14. Construct a triangle ABC in which BC = 4.7 cm, AB + AC = 8.2 cm and ∠C = 60°. [CBSE March 2012]**

**Answer.** **Given :** In ΔABC, BC= 4.7 cm, AB + AC = 8.2 cm and ∠C= 60°.

**Required :** To construct ΔABC.

**15. To construct a triangle, given its perimeter and its two base angles, e.g., construct a triangle with perimeter 10 cm and base angles 60° and 45°. [CBSE March 2012]**

**Answer.**

**16. Construct ΔXYZ, if its perimeter is 14 cm, one side of length 5 cm and ∠X= 45°. [CBSE-14-ERFKZ8H]**

**Answer.** Here, perimeter of ΔXYZ = 14 cm and one side XY = 5 cm

.-. YZ + XZ = 14 – 5 = 9 cm and ∠X = 45°.

**Steps of construction :**

- Draw a line segment XY = 5 cm.
- Construct an ∠YXA = 45° with the help of compass and ruler.
- From ray XA, cut off XB – 9 cm.
- Join BY.
- Draw perpendicular bisector of BY and let it intersect XB in Z.
- Join ZY. Thus, ΔXYZ is the required triangle.