Class 6 Exam  >  Class 6 Notes  >  RD Sharma Solutions for Class 6 Mathematics  >  RD Sharma Solutions -Ex-19.6, Geometrical Constructions, Class 6, Maths

Ex-19.6, Geometrical Constructions, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics PDF Download

Q.1 Construct an angle of 600 with the help of compasses and bisect it by paper folding.

Sol.1 : Draw a ray OA.

With convenient radius and centre O, draw an arc cutting the ray OA at P.

With the same radius and centre at P, draw another arc cutting the previous arc at Q.

Draw OQ and extend it to B.

∠AOB is the required angle of 600.

Ex-19.6, Geometrical Constructions, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics

We cut the part of paper as sector OPQ.

Now, fold the part of paper such that line segments OP and OQ get coincided.

Angle made at point O is the required angle, which is half of angle ∠AOB.

Ex-19.6, Geometrical Constructions, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics

 

Q.2 Construct the following angles with the help of ruler and compasses only.

(i) 300

(ii) 900

(iii) 450

(iv) 1350

(v) 1500

(vi) 1050

Sol.2 :

(i) 300

Draw a ray OA.

With a convenient radius and centre at O, draw an arc, which cuts OA at P.

With the same radius and centre at P, draw an arc cutting the previous arc at P.

Taking P and Q as centres and radius more than half of PQ, draw two arcs, which cuts each other at R.

Draw OR and extend it to B.

∠AOB is the required angle of 300.

Ex-19.6, Geometrical Constructions, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics

(ii) 900

Draw a ray OA.

With a convenient radius and centre at O, draw an arc cutting the ray OA at P.

With the same radius and centre at P, draw another arc, which cuts the first arc at Q.

With the same radius and centre at Q, draw another arc, which cuts the first arc at R.

With Q and R as centres and radius more than half of QR, which cuts each other at S.

Draw OS and extend it to B from the ray OB.

∠AOB is required angle of 900.

Ex-19.6, Geometrical Constructions, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics

(iii) 450

To construct an angle of 450, construct an angle of 900 and bisect it.

Construct the angle ∠AOB = 900, where rays OA and OB intersect the arc at points P and T as shown in figure.

With P and T as centres and radius more than half of PT, draw two arcs, which cut each other at X

Draw OX and extend it to C to form the ray OC.

∠AOC is the required angle of 450.

Ex-19.6, Geometrical Constructions, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics

(iv) 1350

Draw the line AB and take the point O at the middle of AB.

With a convenient radius and centre at O, draw an arc, which cuts AB at P and Q, respectively.

Draw an angle of 900 on the ray OB as ∠BOC = 900, where ray OC cuts the arc at R.

With Q and R as centres and radius more than half of QR, draw two arcs, which cuts each other at S.

Draw OS and extend it to form the ray OD.

∠BOD is required angle of 1350.

Ex-19.6, Geometrical Constructions, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics

(v) 1500

Draw a line AB and take point O at the middle of AB.

With a convenient radius and centre at O, draw an arc, which cuts the line AB at P and Q.

With the same radius and centre at Q, draw an arc, which cuts the first arc at R.

With the same radius and centre at R, draw an arc, which cuts the first arc at S.

With the centres P and S and radius more than half of PS, draw two arcs, which cut each other at T.

Draw OT and extend it to C to form the ray OC.

∠BOC is required angle of 1500.

Ex-19.6, Geometrical Constructions, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics

(vi) 1050

Draw a ray OA and make an angle ∠AOB = 900 and ∠AOC = 1200

Now bisect ∠BOC and get the ray OD.

∠AOD is the required angle of 1050

Ex-19.6, Geometrical Constructions, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics

 

Q.3 Construct a rectangle whose adjacent sides are 8 cm and 3 cm.

Sol.3 : Draw a line segment AB of length 8 cm.

Construct ∠BAX = 90at point A and ∠ABY = 900 at point B.

Using a compass and ruler, mark a point D on the ray AX such that AD = 3 cm.

Similarly mark the point C on the ray Y such that BC = 3 cm.

Draw the line segment CD.

ABCD is the required rectangle.

Ex-19.6, Geometrical Constructions, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics

The document Ex-19.6, Geometrical Constructions, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics is a part of the Class 6 Course RD Sharma Solutions for Class 6 Mathematics.
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FAQs on Ex-19.6, Geometrical Constructions, Class 6, Maths RD Sharma Solutions - RD Sharma Solutions for Class 6 Mathematics

1. What are geometrical constructions in mathematics?
Ans. Geometrical constructions in mathematics are methods or techniques used to create precise geometric figures using only a compass and a straightedge. These constructions involve drawing lines, angles, triangles, and other shapes without the use of measurements or calculations.
2. What is the importance of geometrical constructions in mathematics?
Ans. Geometrical constructions are important in mathematics as they help in developing spatial awareness, logical thinking, and problem-solving skills. They also aid in understanding geometric concepts and properties by allowing students to visualize and manipulate shapes. Additionally, constructions are frequently used in engineering, architecture, and other fields to create accurate diagrams and designs.
3. How do you construct an equilateral triangle using geometrical constructions?
Ans. To construct an equilateral triangle using geometrical constructions, follow these steps: 1. Draw a line segment AB. 2. Place the compass on point A and draw an arc that intersects the line segment AB at point C. 3. Without changing the compass width, place the compass on point C and draw another arc. 4. The intersection of these two arcs will be point D. 5. Draw line segments AD and BD. 6. The triangle ABD is an equilateral triangle.
4. Can you construct a square using geometrical constructions?
Ans. Yes, a square can be constructed using geometrical constructions. To construct a square, follow these steps: 1. Draw a line segment AB. 2. Place the compass on point A and draw an arc that intersects the line segment AB at point C. 3. Without changing the compass width, place the compass on point C and draw another arc. 4. The intersection of these two arcs will be point D. 5. Draw line segments AD and BD. 6. Place the compass on point D and adjust its width to the distance from point D to point A. 7. Draw an arc that intersects line segment AD at point E. 8. Without changing the compass width, place the compass on point E and draw another arc. 9. The intersection of these two arcs will be point F. 10. Draw line segments DF, FE, and EB. 11. The quadrilateral ADFE is a square.
5. How do you construct a perpendicular bisector using geometrical constructions?
Ans. To construct a perpendicular bisector using geometrical constructions, follow these steps: 1. Draw a line segment AB. 2. Place the compass on point A and draw an arc that intersects the line segment AB at point C. 3. Without changing the compass width, place the compass on point B and draw another arc. 4. The intersection of these two arcs will be point D. 5. Draw line segments CD and DB. 6. The line segment CD is the perpendicular bisector of line segment AB.
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