AP State Syllabus AP Board 8th Class Maths Solutions Chapter 3 Construction of Quadrilaterals Ex 3.1 Textbook Questions and Answers.

## AP State Syllabus 8th Class Maths Solutions 3rd Lesson Construction of Quadrilaterals Exercise 3.1

Construct the quadrilaterals with the measurements given below:

Question (a).

Quadrilateral ABCD with AB = 5.5 cm, BC = 3.5 cm, CD = 4 cm, AD = 5 cm and ∠A = 45°.

Solution:

In Quadrilateral ABCD with AB = 5.5 cm, BC = 3.5 cm, CD = 4 cm, AD = 5 cm and ∠A = 45°.

Construction Steps:

- Construct a line segment \(\overline{\mathrm{AB}}\) with radius 5.5 cm
- With the centre A draw a ray and an arc which are equL1 to 45° and 5 cm.
- These intersecting point is keep as ‘D’.
- With centres D, B draw two arcs equal to radius 4 cm, 3.5 cm respectively.
- The intersecting point of these two arcs is keep as ‘C’.
- Join DC and BC. A F
- ∴ The required quadrilateral ABCD is formed.

Question (b).

Quadrilateral BEST with BE = 2.9 cm, ES = 3.2 cm, ST = 2.7 cm, BT = 3.4 cm and ∠B=75°.

Solution:

In Quadrilateral BEST

BE = 2.9 cm, ES = 3.2 cm, ST = 2.7 cm, BT = 3.4 cm and ∠B=75°.

Construction Steps:

- Draw a line segment \(\overline{\mathrm{BE}}\) with radius 2.9 cm.
- With the centre B, draw a ray of 75° and draw 2.9

an arc with radius 3.4 cm, keep the intersecting point of these two as T. - With the centres T, E draw arcs with radius 2.7 cm, 3.2 cm respectively. These intersection point is keep as S’.
- Join T, S and E,S.
- ∴ The required quadrilateral BEST is formed.

Question (c).

Parallelogram PQRS with PQ = 4.5 cm, QR =3 cm and ∠PQR = 60°.

Solution:

In a parallelogram PQRS

PQ = 4.5 cm, QR = 3 cm, ZPQR = 60°.

=RS4.5cmzPS=3crn

[: Opposite sides of a I)aralielograrn are equal]

Construction Steps:

- Draw a line segment ¡i with radius 4.5 cm.
- With the centre Q draw a ray and an arc equal to 60° and 3 cm.
- The intersecting point of these two keep as R’.
- With the centres R, P draw arcs with 4.5 cm, 3 cm respectively. Keep ‘S’ as the intersecting point of these two arcs.
- Join P, S and R, S.
- ∴ The required parallelogram PQRS is formed.

Question (d).

Rhombus MATH with AT =4 cm, ∠MAT =120°.

Solution:

Construction Steps:

- Draw a line segment \(\overline{\mathrm{MA}}\) with radius 4 cm.
- With the centre A draw a ray and an arc equal to 120°, 4 cm. These two intersecting point be keep as T.
- With the centres M, T draw arcs equal to 4 cms.

These two arcs intersected at the point ‘H’. - Join M, H and T, H.
- ∴ The required rhombus MATH is formed.

Question (e).

Rectangle FLAT with FL =5 cm, LA= 3 cm.

Solution:

In a rectangle FLAT

FL=AT=5cm, LA = TF = 3cm, ∠F = ∠L = ∠A = ∠T = 90°

Construction Steps:

- Draw a line segment \(\overline{\mathrm{FL}}\) with radius 5 cm.
- With the centre F draw a ray and an arc equal to 900, 3 cm.

These to meet at point T. - With the centres T, L draw arcs equal to 5 cm, 3 cm respectively.
- These two arcs meet at the point ‘A’.
- Join T, A and L, A.
- ∴ The required rectangle FLAT is formed.

Question (f).

Square LUDO with LU = 4.5 cm.

Solution:

In a square LUDO

LU = UD = DO = OL = 4.5 cm

∠L = ∠U = ∠D = ∠O = 90°

Construction Steps: 45

- Draw a line segment \(\overline{\mathrm{LU}}\) with radius 4.5 cm.
- With the centre ‘L’, draw a ray of 90° and an arc with radius 4.5 cm. These two meet at the point ‘O’.
- Now with the centre U’, draw another ray of 90° and an arc with radius 4.5 cm. These two meet at the point “D”.
- Join O, D.
- ∴ The required square LUDO is formed.