The document Facts that Matter, Ex 11.1 NCERT Solutions- Constructions Class 9 Notes | EduRev is a part of the Class 9 Course Class 9 Mathematics by VP Classes.

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**BASIC CONSTRUCTIONS**

A geometrical construction means to draw geometrical figures, such as an angle, a circle, a triangle, a quadrilateral, and a polygon, etc. We normally use all or some of the following instruments for drawing geometrical figures:

(i) A protractor

(ii) A pair of compasses

(iii) A pair of set squares

(iv) A pair of dividers

(v) A graduated scale.**EXERCISE 11.1 ****Ques 1: Construct an angle of 90Âº at the initial point of a given ray and justify the construction.Sol:** Steps of construction:

I. Draw a ray OA.

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

III. Keeping the radius same, divide the semicircle into three equal parts such that

IV. Draw

V. Draw, the bisector of âˆ COD.

Thus, âˆ AOF = 90Âº.

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

âˆ´

â‡’ âˆ BOC = âˆ COD = âˆ DOE

[âˆµ Equal chords subtend equal angles at the centre]

âˆµ âˆ BOC + âˆ COD + âˆ DOE = 180Âº

â‡’ âˆ BOC + âˆ BOC + âˆ BOC = 180Âº

â‡’ 3âˆ BOC = 180Â°

âˆ´ âˆ BOC = 60Âº

Similarly, âˆ COD = 60Âº and âˆ DOE = 60Âº

âˆµ OF is the bisector of âˆ COD.

âˆ´ = 30Âº

Now, âˆ BOC + âˆ COF = 60Âº + 30Âº

â‡’ âˆ BOF = 90Âº

or

âˆ AOF = 90Âº

I. Draw a ray.

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

III. Taking B as centre and keeping the same radius, cut the semicircle at C. Similarly cut the semicircle at D and E, such that . Join OC and produce.

IV. Divideinto two equal parts, such that

V. Draw OG, the angle bisector of âˆ FOC.

Thus, âˆ BOG = 45Âº

or

âˆ AOG = 45Âº**Justification: **

âˆµ

âˆ´ âˆ BOC = âˆ COD = âˆ DOE

[âˆµ Equal chords subtend equal angles at the centre]

Since âˆ BOC + âˆ COD + âˆ DOE = 180Âº

â‡’ âˆ BOC = 60Âº

âˆµ is the bisector of âˆ BOC

âˆ´ .. (1)

Also, is the bisector of âˆ COF

âˆ´ ... (2)

Adding (1) and (2), we get

âˆ COF + âˆ FOG = 30Âº + 15Âº = 45Âº

âˆ BOF + âˆ FOG = 45Âº [âˆµ âˆ COF = âˆ BOF]

â‡’ âˆ BOG = 45Âº**Ques 3: Construct the angles of the following measurements: ****(i) 30Âº ****(ii) ****(iii) 15Âº****Sol:** (i) Angle of 30Âº

Steps of construction:

I . Draw a ray OA.

II . With O as centre and a suitable radius, draw an arc, cutting at B.

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

IV. Join and produce, such that âˆ BOC = 60Âº.

V. bisector of âˆ BOC, such that

Thus, âˆ BOD = 30Âº

(ii) Angle of** **

Steps of construction:

I . Draw a ray

II. Draw an angle âˆ AOB = 90Âº

III. Draw OC, the bisector of âˆ AOB, such that

IV. Now, draw OD, the bisector of âˆ AOC, such that

Thus,

(iii) Angle of 15ÂºSteps of construction:

I. Draw a ray.

II. Construct âˆ AOB = 60Âº.

III. Draw the bisector of âˆ AOB, such that

i.e. âˆ AOC = 30Âº

IV. Draw the angle bisector of âˆ AOC such that

Thus, âˆ AOD = 15Âº**Ques 4: Construct the following angles and verify by measuring them by a protractor: ****(i) 75Âº ****(ii) 105Âº ****(iii) 135Âº****Sol: **(i) Angle of 75Âº:**Hint:** 75Âº = 60Âº + 15Âº

Steps of construction:

I. Draw .

II. With O as centre and having a suitable radius, draw an arc which meets

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

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

V. Draw the bisector of such that âˆ COP

VI. Draw , the bisector of âˆ COP, such that âˆ COQ = 15Âº

Thus, âˆ BOQ = 60Âº + 15Âº = 75Âº

or âˆ AOQ = 75Âº.

(ii) Angle of 105Âº:

Hint: 105Âº = 90Âº + 15Âº

Steps of construction:

I. Draw

II. With centre O and having a suitable radius, draw an arc which meet OA at B.

III. With centre B and keeping the same radius, mark a point C on the arc of step II.

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

V. Draw OP, the bisector of

VI. Draw OQ, the bisector of.

Thus, âˆ AOQ = 105Âº

(iii) Angle of 135Âº:

Hint: 120Âº + 15Âº = 135Âº

Steps of construction:

I. Draw a ray

II. With centre O and having a suitable radius draw an arc to meet OP at A.

III. Keeping the same radius and starting from A, mark points Q, R and S on the arc of step II.

IV. Draw the bisector of

V. Draw the bisector of

Thus, âˆ POM = 135Âº.**Ques 5: Construct an equilateral triangle, given its side and justify the construction.Sol: **Let us construct an equilateral triangle, each of whose side = PQ Steps of construction:

I. Draw a ray.

II. Taking O as centre and radius equal to PQ, draw an arc to cut OA at B such that OB = PQ

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

IV. Join OC and OB.

Thus, Î”OBC is the required equilateral triangle.

**Justification: **

âˆµ The are drawn with the same radius.

âˆ´

â‡’ [Chords corresponding to equal arcs are equal]

âˆµ OC = OB = BC

âˆ´ Î”OBC is an equilateral triangle.

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