AP Board 9th Class Maths Solutions Chapter 13 Geometrical Constructions Ex 13.1

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 13 Geometrical Constructions Ex 13.1 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 13th Lesson Geometrical Constructions Exercise 13.1

AP Board 9th Class Maths Solutions Chapter 13 Geometrical Constructions Ex 13.1

Question 1.
Construct the following angles at the initial point of a given ray and justify the construction.
a) 90°
Solution:
AP Board 9th Class Maths Solutions Chapter 13 Geometrical Constructions Ex 13.1 1

  • Let AB be the given ray.
  • Produce BA to D.
  • Taking A as centre draw a semi circle with some radius.
  • With X and Y as Center draw two intersecting arcs of same radius.

Or

AP Board 9th Class Maths Solutions Chapter 13 Geometrical Constructions Ex 13.1 2|

  • Let \(\overrightarrow{\mathrm{AB}}\) be the given ray.
  • With A as centre draw an arc of any radius.
  • Mark off two equal arcs from X as shown in the figure with the same radius taken as before.
  • Bisect the second segment.
  • Join the point of intersection of above arcs, with A.
  • ∠BAC is the required right angle.
  • Join the point of intersection ‘C’ and ‘A’.
  • ∠BAC = 90°

In ΔAXY; ∠YAX = 60° and
in ΔAYC ∠YAC = 30° ∠BAC = 90°

AP Board 9th Class Maths Solutions Chapter 13 Geometrical Constructions Ex 13.1

b) 45°
Solution:
Steps:

  • Construct 90° with the given ray AB.
  • Bisect it from ∠BAD = 45°

AP Board 9th Class Maths Solutions Chapter 13 Geometrical Constructions Ex 13.1 3
Or

AP Board 9th Class Maths Solutions Chapter 13 Geometrical Constructions Ex 13.1 4

Steps:

  • Construct ∠BAC = 60°
  • Bisect ∠BAC = ∠DAC = 30°
  • Bisect ∠DAC such that ∠DAE = ∠FAC = 15°
  • ∠BAE=45°

ΔAXZ is equilateral
and ∠YAZ = 15°
∴∠XAY = 45°

AP Board 9th Class Maths Solutions Chapter 13 Geometrical Constructions Ex 13.1

Question 2.
Construct the following angles using ruler and compass and verify by measuring them by a protractor.
a) 30°
Solution:
AP Board 9th Class Maths Solutions Chapter 13 Geometrical Constructions Ex 13.1 5

  • Construct ∠ABY = 60°
  • Bisect ∠ABY = 60°
  • Such that ∠ABC = ∠CBY = 30°

b) 22\(\frac{1}{2}^{\circ}\)
Solution:
AP Board 9th Class Maths Solutions Chapter 13 Geometrical Constructions Ex 13.1 6

  • Construct ∠ABD = 90°.
  • Bisect ∠ABD such that ∠ABC = ∠CBD = 45°
  • Bisect ∠ABC such that
    ∠ABE = ∠EBC = 22\(\frac{1}{2}^{\circ}\)

AP Board 9th Class Maths Solutions Chapter 13 Geometrical Constructions Ex 13.1

c) 15°
Solution:
AP Board 9th Class Maths Solutions Chapter 13 Geometrical Constructions Ex 13.1 7
Steps of construction : ,

  • Construct ∠BAE = 60°
  • Bisect ∠BAE such that ∠BAC = ∠CAE = 30°
  • Bisect ∠BAC such that ∠BAF = ∠FAC = 15°

d) 75°
Solution:
AP Board 9th Class Maths Solutions Chapter 13 Geometrical Constructions Ex 13.1 8
Steps of construction :

  • Construct ∠BAC = 60°
  • Construct ∠CAD = 60°
  • Bisect ∠CAD such that ∠BAE = 90°
  • Bisect ∠CAE such that ∠BAF = 75°

AP Board 9th Class Maths Solutions Chapter 13 Geometrical Constructions Ex 13.1

e) 105°
Solution:
AP Board 9th Class Maths Solutions Chapter 13 Geometrical Constructions Ex 13.1 9
Steps of construction:

  • Construct ∠ABC = 90°
  • Construct ∠CBE = 30°
  • Bisect ∠CBE such that the angle formed ∠ABD = 105°

f) 135°
Solution:
AP Board 9th Class Maths Solutions Chapter 13 Geometrical Constructions Ex 13.1 10
Steps of construction:

  • Construct ∠ABC = 120°
  • Construct ∠CBD = 30°
  • Bisect ∠CBD such that the angle formed ∠ABE = 135°

AP Board 9th Class Maths Solutions Chapter 13 Geometrical Constructions Ex 13.1

Question 3.
Construct an equilateral triangle, given its side of length of 4.5 cm and justify the constraction.
Solution:
A.
AP Board 9th Class Maths Solutions Chapter 13 Geometrical Constructions Ex 13.1 11

  • Draw a line segment AB = 4.5 cm.
  • With B and A as centres draw two arcs of radius 4.5 cm meeting at C.
  • Join C to A and B.
  • ΔABC is the required triangle.

Justification:
In ΔABC
AB = ∠C ⇒ ∠C = ∠B
Also AB = BC ⇒ ∠C = ∠A
Hence ∠A = ∠B = ∠C
But ∠A + ∠B + ∠C = 180°
∴ ∠A = ∠B = ∠C = \(\frac{180^{\circ}}{3}\) = 60°

AP Board 9th Class Maths Solutions Chapter 13 Geometrical Constructions Ex 13.1

Question 4.
Construct an isosceles triangle, given its base and base angle and justify the construction. [Hint: You can take any measure of side and angle]
Solution:
A.
AP Board 9th Class Maths Solutions Chapter 13 Geometrical Constructions Ex 13.1 12
Steps:

  • Draw a line segment AB of any given length.
  • Construct ∠BAX and ∠ABY at A and B such that ∠A = ∠B.
  • \(\overrightarrow{\mathrm{AX}}\) and \(\overrightarrow{\mathrm{BY}}\) will intersect at C.
  • ΔABC is the required triangle.

Justification:
Drop a perpendicular CM to AB from C.
Now in ΔAMC and ΔBMC
∠AMC = ∠BMC [Right angle]
∠A = ∠B [Construction]
CM = CM (Common)
∴ ΔAMC ≅ ΔBMC
⇒ AC = BC [CPCT]

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