Constructions – Exercise 11.1 – (MATHEMATICS) – 9th Class

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Exercise 11.1 

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

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Sol. Steps of construction:

(i) Let the given ray be OA with initial point 0.
(ii) Taking 0 as centre and any suitable radius, draw an arc to intersect the ray at P.
(iii) Taking P as centreand the same radius, draw an arc to intersect the arc drawn in step (ii) at A1.
(iv) Taking A1 as centre and the same radius, draw an arc to cut the arc drawn in step (ii) at A2.
(v) Taking A1 and A2 as centres and radius greater than {{{A_1}{A_2}} \over 2} ,draw two arcs to intersect each other at L on the same side of the line segment A1A2.
(vi) Join OL and produce it along OL.
Hence, ∠LOA = 90°.

Justification: ∠A1OA = 60°, and ∠A2OA1 = 60°.

Bisector of ∠A2OA1 = ∠LOA1, = {{60^\circ } \over 2} = 30°.

Then ∠LOA = ∠LOA1 + ∠A1OA = 30° + 60° = 90°.


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

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Sol. Steps of construction:

 



(a) Draw an angle AOP = 90° as in Ans.1.
(b) Taking A and P as centres and radius greater than {{AP} \over 2} draw two arcs to intersect each other at B as shown in the adjoining figure.
(c) Join OB and produce it. Hence, acute ∠BOX = 45°.

Justification: ∠AOP = 900
∠AOB + ∠BOP = 90°                                         [ ∠AOB = ∠BOP]
2∠BOP = 90°
∠BOP = 45°.


3. Construct the angles of the following measurements:

(i) 30°                 (ii) 22{{1^\circ } \over 2}                      (iii) 15°.

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Sol.
(i) Steps of construction:
(a) Draw a ray OX with initial point O.
(b) Taking O as centre and suitable radius, draw an arc to cut OX at P.

(c) Taking P as centre and same radius draw an arc to cut the arc drawn in step (b) at A.
(d) Taking P and A as centres and radius greater than {{AP} \over 2} draw two arcs to intersect each other at A1.
(e) Join OA1 and produce it along OA1. Hence, acute angle A1,OX = 30°.

(ii) Steps of construction:
(a) Draw an angle of 45° on a ray OX such that ∠AOX = 45° as in Q. 2.

(b) Taking B and P as centres and radius greater than {{BP} \over 2}, draw arcs to intersect each other at C.
(c) Join OC and produce it to Y. Hence, acute \angle XOY = 22{{1^\circ } \over 2}

(iii) Steps of construction:
(a) Draw an ∠A1OX = 30° on a ray OX as in Part (i).
(b) Taking A2 and P as centres and radius greater than {{{A_2}P} \over 2} , draw two arcs to intersect each other at A3.

(c) Join 0A3 and produce it to B. Hence, acute angle BOX = 15°.


4. Construct the following angles and verify by measuring them by a protractor.
(i) 75°                            (ii) 105°                                       (iii) 135°.

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Sol. (1) Steps of construction:

(a) Draw a ray OX with initial point 0.
(b) Taking 0 as centre and suitable radius, draw an arc to intersect OX at P.


(c) Taking P as centre and same radius, draw an arc to intersect the arc drawn in step (b) at A.
(d) Taking A as centre and same radius, draw an arc to intersect the arc drawn in step (b) at B.
(e) Taking A and B as centres and radius greater than {{AB} \over 2} draw two arcs to intersect each other at C.
(f) Join OC to intersect the arc drawn in step (b) at D.
(g) Taking A and D as centres and radius greater than {{AD} \over 2} draw two arcs to intersect each other at E.
(h) Join OE and produce it to M.
Hence, acute angle ∠MOX = 75°.

Verification:
On measuring by a protractor, we get ∠MOX = 75°.

(ii) Steps of construction:

(a) Draw a ray OX with initial point 0.
(b) Taking 0 as centre and any suitable radius, draw an arc to intersect OX at P.
(c) Taking P as centre and same radius, draw an arc to intersect the arc drawn in step (b) at A.
(d) Taking A as centre and same radius, draw an arc to intersect the arc in step (h) at B.
(e) Taking A and B as centres and radius greater than {{AB} \over 2}, draw two arcs to intersect each other at C.
(f) Join OC to intersect the arc drawn in step (b) at D.
(g) Taking B and D as centres and any suitable radius greater than  {{BD} \over 2} , draw two arcs to intersect each other at E.
(h) Join OE and produce it to M.
Hence, obtuse ∠MOX = 105°.

Verification: On measuring by a protractor, we find ∠MOX = 105°.

(iii) Steps of construction:

(a) Draw a ray OX with initial point 0.
(h) Taking O as centre and an suitable radius, draw an arc to intersect the ray OX at P and XO produced at D.

(c) Taking P as centre and same radius, draw an arc to intersect the arc drawn in step (b) at A.
(d) Taking A as centre and same radius, draw another arc to intersect the arc drawn in step (b) at B.
(e) Draw the angle bisector of ∠DOB, which intersects the arc drawn in step (b) at C.
(f) Again, draw the angle bisector of ∠COB, which intersects the arc drawn in step (b) at H.
(g) Join OH and produce it to M. Hence, obtuse angle MOX = 135°.
Verification: On measuring by the protractor, ∠MOX is of measure 135°.


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

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Sol.
Let the given side of an equilateral triangle is of length 5 cm.

Steps of construction:
(i) Draw a line segment AB of length 5 cm.

(ii) Draw two angles BAC and ABC, each of measure 60°, on the same side of AB at the points A and B respectively such that their non-common arms intersect each other at C.
Hence, ΔABC is an equilateral triangle.
Justification: ∠ABC = ∠BCA = ∠CAB                        [Each 60°]
⇒                 AC = AB = BC.
[Sides opposite to equal angles are equal.]


 


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