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## Chapter 11: Constructions NCERT Solutions

TEXTBOOK’S EXERCISE – 11.1

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

Sol.

Steps of Construction

(i) Let us take a ray AB with initial point A.

(ii) Taking A as a centre and any convenient radius, draw an arc of a circle, which intersects AB at C.

(iii) With C as a centre and the same radius as before, draw an arc, intersecting the previous arc at E.

(iv) With E as a centre and the same radius, as before, draw an arc, which intersects the arc drawn in step (ii) at F.

(v) With E as a centre and any convenient radius, draw an arc.

(vi) With F as a centre and the same radius as before, draw another arc, intersecting the previous arc at G.

(vii) Draw the ray AG.

Then ∠BAG is the required angle of 90°.

Justification : Join AE, CE, EF, FG and GE

AC = CE = AE [By construction]

⇒ ΔACE is an equilateral triangle

⇒ ∠CAE = 60° … (i)

Similarly, ∠AEF = 60° … (ii)

From (i) and (ii), FE || AC … (iii) [Alternate angles are equal]

Also, FG = EG [By construction]

⇒ G lies on the perpendicular bisector of EF

⇒ ∠GIE = 90° … (iv)

∴ ∠GAB = ∠GIE = 90° [Corresponding angles]

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

Sol.

Steps of Construction

(i) Let us take a ray AB with initial point A.

(ii) Draw ∠BAF = 90°, as discussed in Q. 1.

(iii) Taking C as centre and radius more than 1/2 CG, draw an arc.

(iv) Taking G as a centre and the same radius as before, draw another arc, intersecting the previous arc at H.

(v) Draw the ray AH. Then ∠BAH is the required angle of 45°.

Justification: Join GH and CH.

In ΔAHG and ΔAHC, we have

HG = HC [Arcs of equal radii]

AG = AC [Radii of the same arc]

AH = AH [Common]

∴ ΔAHG ≅ ΔAHC [SSS congruence]

⇒ ∠HAG = ∠HAC … (i) [CPCT]

But, ∠HAG + ∠HAC = 90° … (ii) [By construction]

⇒ ∠HAG = ∠HAC = 45° [From (i) and (ii)]

3. Construct the angles of the following measurements.

(i) 30° (ii) 22 12 ° (iii) 15°

Sol. (i) Steps of Construction

(a) Draw a ray AB, with initial point A.

(b) With A as a centre and any convenient radius, draw an arc, intersecting AB at C.

(c) With C as a centre and the same radius as before, draw another arc, intersect­ing the previously drawn arc at D.

(e) Now, taking C and D as centres and with the radius, more than 1/2 DC, draw arcs to intersect each other at E.

(f) Draw ray AE. Then ∠BAE is the required angle of 30°.

(ii) Steps of Construction

(a) Draw a ray AB with initial point A.

(b) Draw ∠BAH = 45° as discussed in Q. 2.

(c) Taking I and C as centres and with the radius, more than 1/2 CI, draw arcs to intersect each other at J.

(d) Draw ray AJ. Then ∠BAJ is the required angle of 22 1/2

(iii) Steps of Construction

(a) Draw ∠BAE = 30° as discussed in part (i).

(b) Taking C and F as centres and with the radius, more than 12 CF, draw arcs to intersect each other at G.

(c) Draw ray AG. Then ∠BAG is the required angle of 15°.