Construction

Construct an isosceles triangle, quadrilateral and parallelogram having any base and remaining sides.

Summary

Construct an isosceles triangle, quadrilateral and parallelogram having any base and remaining sides.

Things to Remember

  • Construction in geometry means to draw shapes, angles or lines accurately. 
  • These constructions use only compass, straight edge i.e ruler and a pencil. 
  • This is a pure form of geometric construction because no numbers are included. 

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Subjective Questions

Q1:

Discuss instruction cycle in detail.


Type: Long Difficulty: Easy

Q2:

How instruction is fetched and decoded?Explain.


Type: Long Difficulty: Easy

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Construction

Construction

Construction of Quadrilaterals

The construction of square, rectangle, rhombus, parallelogram and trapezium are explained below:

1. Construction of a square when a side is given:

 

square

construction square

Here, ABCD is a square where AB = 8 cm.

Step 1: Draw AB = 8 cm. 

Step 2: Construct angle 900 at both A and B. 

Step 3: Cut AX at 8 cm from A and name it D. Similarly, cut BY at 8 cm from B and name it C.

Step 4: Join C and D. 

Therefore, a square ABCD having a side 8 cm is formed.

2. Construction of a square when a diagonal is given:

ex-2

Here, MNOP is a square whose diagonal NP = 5 cm.

Step 1: Draw NP = 5 cm.

Step 2: Draw 450 angles at N and P.

Step 3: NX and PY intersect at M.

Step 4: Take an arc equal to MN or MP and draw two arcs downwards such that they intersect at C.

Step 5: Now, join N,O and P,O.

Therefore, a square MNOP having a diagonal NP = 5 cm is formed.

3. Construction of a rectangle when two adjacent sides are given:

 

ex-3

Here, ABCD is a rectangle in which BC = 5 cm and AB = 4 cm.

Step 1: Draw BC = 5 cm.

Step 2: Construct 900 angles at both points A and B.

Step 3: Take an arc of 4 cm and cut the BX at A and CY at D.

Step 4: Join A and B.

Therefore, a rectangle ABCD having BC = 5 cm and AB = 4 cm is formed.

4. Construction of a rectangle when diagonal and an angle made by them are given:

 

Ex-4

Here, ABCD is a rectangle in which AC = BD = 5 cm and an angle = 450.

Step 1: Draw AC = 5 cm.

Step 2: Find the mid-point O of the line AC with the help of perpendicular bisector method.

Step 3: Construct 450 angle at O. Produce the line XO straight upto Y on other side.

Step 4: Take an arc of 2.5 cm and cut OX at D and OY at B.

Step 5: Join A and B, B and C, C and D, D and A.

Therefore, a rectangle ABCD having AC = BD = 5 cm and an angle = 450 is formed.

5. Construction of a Rhombus when a side and an angle between two adjacent sides are given:

ex-5

Here, a rhombus MNOP with side MN = 5 cm and∠NMP = 600.

Step 1: Draw MN = 5 cm.

Step 2: Draw ∠NMX = 600 at M.

Step 3: Take measure of 5 cm and cut MX at P.

Step 4: Take a radius of 5 cm from N and P, draw two arcs such that they intersect at O.

Step 5: Join P and O, O and N.

Therefore, a rhombus MNOP is formed.

6. Construction of a Rhombus when the lengths of two diagonals are given:

 

ex-7

Here, a rhombus ABCD having diagonal AC = 5 cm and BD = 6 cm.

Step 1: Draw the diagonal AC = 5 cm.

Step 2: Draw the perpendicular bisector XY of AC such that the mid-point is O.

Step 3: Take radius of 3 cm (half of diagonal BD) and from O, cut OY at D and OX at B.

Step 4: Joint A and B, B and C, C and D, D and A.

Therefore, a rhombus ABCD is formed.

7. Construction of a rhombus when a side and a diagonal are given:

ex-7

Here, a rhombus ABCD where AB = 6 cm and AC = 8 cm.

Step 1: Draw AB = 6 cm.

Step 2: Keeping A center, take an arc of 8 cm.

Step 3: From B, take an arc of 6 cm and cut the previous arc and name it C.

Step 4: Join B and C, A and D.

Step 5: Again, taking A as centre, draw an arc of 5.5 cm upright and take another same arc from C and cut the previous arc and name it.

Step 6: Join C and D, D and A.

Therefore, a rhombus ABCD is formed.

8. Construction of a parallelogram when two adjacent sides and an angle contained by them are given:

ex-8

Here, a parallelogram ABCD in which BC = 6 cm, CD = 5 cm and∠BCD = 1200.

Step 1: Draw BC = 6 cm.

Step 2: Construct an angle of 1200 at C, i.e.∠BCX = 1200.

Step 3: Take an arc of 4.5 cm and cut CX at D.

Step 4: With centre at B, draw an arc upright with radius of 4.5 cm.

Step 5: Likewise, with centre at D, draw another arc with radius of 5 cm.

Step 6: Join A and D, A and B.

Therefore, a parallelogram ABCD is formed.

9. Construction of a parallelogram when a base side, diagonal and angle contained by them are given:

ex-9

Here, a parallelogram MNOP where base MN = 6 cm, diagonal MO = 8 cm and∠NMO = 300.

Step 1: Draw MN = 6 cm.

Step 2: Construct an angle of 300 at M (i.e. ∠NMX = 300).

Step 3: Take a radius of 8 cm and cut MX at O from M.

Step 4: Join N and O.

Step 5: With centre at O, take radius of 6 cm and draw an arc.

Step 6: Likewise with centre at M, take a radius equal to NO and cut the previous arc. Name the intersecting points as P.

Step 7: Join O and P, P and M.

Therefore, a reqiured parallelogram MNOP is formed.

10. Construction of a parallelogram when a side and two diagonals are given:

ex-9

Here, PQ = 6 cm and diagonals PR = 9 cm and QS = 6.5 cm.

Step 1: Draw a base line PQ = 6 cm.

Step 2: Take a radius of 5 cm and draw an arc up from point P.

Step 3: Similarly, take a radius of 5 cm and draw an arc up from Q. Such that it meets the previous arc. Name the point O.

Step 4: Join the point O to P and Q.

Step 5: Produce PO to R such that PO = OR. And produce QO to S such that QO =OS.

Step 6: Join P and S, S and R, R and Q.

Therefore, the required parallelogram PQRS is formed.

11. Construction of a parallelogram when two diagonals and an angle contained by them are given:

ex-11

Here, PR = 6 cm, QS = 8 cm and angle between these two diagonals is 450.

Step 1: Draw PR = 6 cm.

Step 2: Draw the bisector of PR and find the mid-point O.

Step 3: At O, draw an angle of 450.

Step 4: Take a radius of 4 cm (half of QS) and cut OX at S and OY at Q.

Step 5: Join P and Q, Q and R, R and S, S and P.

Therefore, the required parallelogram PQRS is formed.

12. Construction of a trapezium when two adjacent sides and two angles are given:

ex-12

Here, AB = 6 cm, BC = 5 cm, ∠DAB = 600, ∠BCD = 900 and AD||BC.

Step 1: Draw a base line AB = 6 cm.

Step 2: Construct an angle of ∠BAX =600.

Step 3: Since, BC||AD, the angle at B should be 1200. So, construct an angle of 1200. (i.e. ∠ABY = 1200).

Step 4: Taking a radius of 5 cm, mark the point C on BY from the point B.

Step 5: Construct an angle of 900at C. Thus formed line CD meets the pervious line AX at D.

Therefore, the required trapezium ABCD is formed.

13. Construction of a trapezium ABCD when two sides, a diagonal and an angle made by the diagonal with the given one side are given:

 

ex-13

Here, AB = 6 cm, diagonal BD = 8 cm, ∠BAD = 600.

Step 1: Draw a base line AB = 6 cm.

Step 2: Construct an angle of 600 (i.e. ∠BAX) at A.

Step 3: Taking a radius of 8 cm, mark the point D on AX from Q.

Step 4: Join B and D.

Step 5: Since AB||CD, alternate angles are equal. So, construct an angle of 600 at C with base AC to make an equal alternate∠XDC.

Step 6: Take a radius of 7 cm and from point D draw an arc to cut DY at C.

Step 7: Join C and B.

Therefore, the required trapezium ABCD is formed.

14. Construction of a trapezium PETU when three sides and an angle are given:

Ex-14

Here, PE = 8 cm, diagonal ET = 6 cm, TU = 4 cm and ∠PET = 600 such that PE||TU.

Step 1: Draw a base line segment PE = 8 cm.

Step 2: Draw an angle∠PEX = 600 at point E.

Step 3: Taking an arc of 6 cm from E, mark point T on EX.

Step 4: At point T, draw an∠ETY = 1200.

Step 5: Taking an arc of 4 cm from T, mark point U on TY.

Step 6: Join U and P.

Therefore, the required trapezium PETU is formed where PE||TU.

Lesson

Geometry

Subject

Compulsory Maths

Grade

Grade 9

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