NCERT Class 9 Solutions: Constructions (Chapter 11) Exercise 11.1 – Part 1 (For CBSE, ICSE, IAS, NET, NRA 2022)

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Construction of point angle at

Construction of Point Angle at 120 Degree

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

Solution:

Ray as with the Initial Point A

To construct an angle at point A

Steps of construction:

  • Draw a ray AS.
  • With its initial point A as centre and any radius, draw an arc BCD, cutting AS at B.
  • With same radius and center at B, cut the arc intersecting the circle with center A at C
  • With centre C and same radius, draw an arc, cutting the circle with center A at D.
  • With D and C as centres, and any convenient radius () draw two arcs intersecting at P
  • Join AP. Then

Justification:

  • AB and AC are the radius and AB = BC by construction, therefore,
  • Therefore is an equilateral triangle. So,
  • Similarly, , therefore is also an equilateral triangle. So
  • AP bisects , so
  • Now

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

Solution:

Ray OA, O is the Centre of the 45° Angle

Steps of Construction:

  • Draw a ray OA
  • With O as centre and any suitable radius draw an arc cutting OA at B.
  • With B as centre and same radius cut the previously drawn arc at C and then with C as centre and same radius cut the arc in step 1 at D.
  • With C as centre and radius more than half CD draw an arc.
  • With D as centre and same radius draw another arc to cut the previous arc at E
  • Join OE. Then (we proved this above) . Let OE cut the circle with center O at F

Now we will draw the bisector OG of Then

  • With B as center and radius more than half of BF draws an arc.
  • With F as center and radius more than half of BF draws another arc intersecting previous arc at G.
  • Now OG is the bisector of angle . Since is the bisector of . So,

Justification:

  • We already proved that constructed this way would be .
  • Now, in and ,
    • OG is common
    • (by construction)
    • (radius of the same circle at O)
  • Therefore, by SSS
  • Therefore,
  • Also,
  • Therefore,

Q-3 Construct the angle of the following measurements:

1.

2.

3.

Solution:

i) Steps of Construction:

∠ QAB = 30°
  • Draw a ray AQ.
  • With its initial point A as centre and any radius, draw an arc, cutting AQ at P.
  • With centre P and same radius. Draw an arc, cutting the arc of step 2 in D
  • With P and D as centres, and any convenient radius () , draw two arcs intersecting at B.
  • Join OB. Then

ii) Step of construction:

POB = 90°
  • Draw an angle
  • Draw the bisector OC of , then
  • Bisect , such that
  • Thus,

iii) Steps of Construction:

Ray AB at ∠ CAB = 15°
  • Construct an
  • Bisect so that
  • Bisect , so that