# Geometrical Constructions

Chapters

- ap-scert-9th-class-Maths-Lesson-1-Real-Numbers
- ap-scert-9th-class-Maths-Lesson-2-Polynomials-and-Factorisation
- ap-scert-9th-class-Maths-Lesson-3-The-Elements-of-Geometry
- ap-scert-9th-class-Maths-Lesson-4-Lines-and-Angles
- ap-scert-9th-class-Maths-Lesson-5-Co-Ordinate-Geometry
- ap-scert-9th-class-Maths-Lesson-6-Linear-Equation-in-Two-Variables
- ap-scert-9th-class-Maths-Lesson-7-Triangles
- ap-scert-9th-class-Maths-Lesson-8-Quadrilaterals
- ap-scert-9th-class-Maths-Lesson-9-Statistics
- ap-scert-9th-class-Maths-Lesson-10-Surface-Areas-and-Volumes
- ap-scert-9th-class-Maths-Lesson-11-Areas
- ap-scert-9th-class-Maths-Lesson-12-Circles
- ap-scert-9th-class-Maths-Lesson-13-Geometrical-Constructions
- ap-scert-9th-class-Maths-Lesson-14-Probability
- ap-scert-9th-class-Maths-Lesson-15-Proofs-in-Mathematics

- ap-scert-9th-class-Maths-Lesson-1-Real-Numbers
- ap-scert-9th-class-Maths-Lesson-2-Polynomials-and-Factorisation
- ap-scert-9th-class-Maths-Lesson-3-The-Elements-of-Geometry
- ap-scert-9th-class-Maths-Lesson-4-Lines-and-Angles
- ap-scert-9th-class-Maths-Lesson-5-Co-Ordinate-Geometry
- ap-scert-9th-class-Maths-Lesson-6-Linear-Equation-in-Two-Variables
- ap-scert-9th-class-Maths-Lesson-7-Triangles
- ap-scert-9th-class-Maths-Lesson-8-Quadrilaterals
- ap-scert-9th-class-Maths-Lesson-9-Statistics
- ap-scert-9th-class-Maths-Lesson-10-Surface-Areas-and-Volumes
- ap-scert-9th-class-Maths-Lesson-11-Areas
- ap-scert-9th-class-Maths-Lesson-12-Circles
- ap-scert-9th-class-Maths-Lesson-13-Geometrical-Constructions
- ap-scert-9th-class-Maths-Lesson-14-Probability
- ap-scert-9th-class-Maths-Lesson-15-Proofs-in-Mathematics

- Solutions
- PDF Download
- Question Papers
- Videos

Solutions

Construct the following angles at the initial point of a given ray and justify the construction.

a) 90°

Solution:

- 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

- 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°

b) 45°

Solution:

Steps:

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

or

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°

Construct the following angles using ruler and compass and verify by measuring them by a protractor.

a) 30°

Solution:

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

b) 22 1°/2

Solution:

- Construct ∠ABD = 90°.
- Bisect ∠ABD such that ∠ABC = ∠CBD = 45°
- Bisect ∠ABC such that

∠ABE = ∠EBC = 22 1°/2

c) 15°

Solution:

Steps of construction : ,

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

d) 75°

Solution:

Steps of construction :

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

e) 105°

Solution:

Steps of construction:

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

f) 135°

Solution:

Steps of construction:

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

Construct an equilateral triangle, given its side of length of 4.5 cm and justify the constraction.

Solution:A.

- 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 = 180°/3 = 60°

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.

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]

Construct AABC in which BC = 7 cm,∠B = 75° and AB + AC =12 cm.

Solution:

A.

Construct ΔPQR in which QR = 8 cm, ∠B = 60° and AB - AC = 3.5 cm.

Read ∠Q = 60°and PQ - PR = 3.5 cm

Solution:

A.

Construct ΔXYZ in which ∠Y = 30 °; ∠Z = 60 ° and XY + YZ + ZX = 10 cm.

Solution:

A.

Construct a right triangle whose base is 7.5 cm and sum of its hypotenuse and otherside is 15 cm.

Solution:

A.

Construct a segment of a circle on a chord of length 5 cm containing the following angles i) 90° ii) 45° iii) 120°

Solution:

i) 90°

A.

ii) 45°

Steps:

- Draw a line segment BC = 5 cm.
- Construct ABOC such that BC = 5 cm, ∠B = 45° = ∠C.
- Draw a circle segment of radius OB or OC with centre ’O’.
- Mark any point A on the segment and join it with B and C.
- ∠BAC = 45°

iii) 120°

Steps:

- Draw a line segment AB = 5 cm. ,
- Construct ΔAOB in which ∠A = 30°; ∠B = 30°; AB = 5 cm.
- With ‘O’ as centre draw a circle segment.
- On the opposite side make any point C and join it with B and C.
- ∠ACB = 120°

PDF Download

Question Papers

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