Class nine Maths Chap 11 Constructions NCERT Sol
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.
(d) Draw ray AD.
(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°.
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