AP Board 8th Class Maths Solutions Chapter 3 Construction of Quadrilaterals Ex 3.1

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:

1. Construct a line segment \(\overline{\mathrm{AB}}\) with radius 5.5 cm
2. With the centre A draw a ray and an arc which are equL1 to 45° and 5 cm.
3. These intersecting point is keep as ‘D’.
4. With centres D, B draw two arcs equal to radius 4 cm, 3.5 cm respectively.
5. The intersecting point of these two arcs is keep as ‘C’.
6. Join DC and BC. A F
7. ∴ 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:

1. Draw a line segment \(\overline{\mathrm{BE}}\) with radius 2.9 cm.
2. 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.
3. With the centres T, E draw arcs with radius 2.7 cm, 3.2 cm respectively. These intersection point is keep as S’.
4. Join T, S and E,S.
5. ∴ 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:

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

Question (d).
Rhombus MATH with AT =4 cm, ∠MAT =120°.
Solution:

Construction Steps:

1. Draw a line segment \(\overline{\mathrm{MA}}\) with radius 4 cm.
2. With the centre A draw a ray and an arc equal to 120°, 4 cm. These two intersecting point be keep as T.
3. With the centres M, T draw arcs equal to 4 cms.
   These two arcs intersected at the point ‘H’.
4. Join M, H and T, H.
5. ∴ 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:

1. Draw a line segment \(\overline{\mathrm{FL}}\) with radius 5 cm.
2. With the centre F draw a ray and an arc equal to 900, 3 cm.
   These to meet at point T.
3. With the centres T, L draw arcs equal to 5 cm, 3 cm respectively.
4. These two arcs meet at the point ‘A’.
5. Join T, A and L, A.
6. ∴ 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

1. Draw a line segment \(\overline{\mathrm{LU}}\) with radius 4.5 cm.
2. With the centre ‘L’, draw a ray of 90° and an arc with radius 4.5 cm. These two meet at the point ‘O’.
3. 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”.
4. Join O, D.
5. ∴ The required square LUDO is formed.