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Chapter 18 - Practical Geometry (Constructions) (Part - 1), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics PDF Download

PAGE NO 18.10:

Question 1:

Construct a quadrilateral ABCD, in which AB  = 6 cm, BC  = 4 cm, CD  = 4 cm, ∠B  = 95° and ∠C  = 90°.

ANSWER:

Steps of construction:

Step I: Draw BC = 4 cm.

Step II: Construct ∠ABC = 95° at B.

Step III : With B as the centre and radius 6 cm, cut off BA = 6 cm.

Step IV: Construct ∠BCD = 90° at C. 

Step V :With C as the centre and radius 4 cm, cut off BA = 4 cm.

Step VI: Join CD.The quadrilateral so obtained is the required quadrilateral.

Chapter 18 - Practical Geometry (Constructions) (Part - 1), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics

 


Question 2:

Construct a quadrilateral ABCD, where AB  = 4.2 cm, BC  = 3.6 cm, CD  = 4.8 cm, ∠B  = 30° and ∠C  = 150°.

ANSWER:

Steps of construction:

Step I: Draw BC = 3.6 cm.

Step II: Construct ∠ABC = 30° at B.

Step III : With B as the centre and radius 4.2 cm, cut off BA = 4.2 cm.

Step IV: Construct ∠BCD = 150° at C.

Step V :With C as the centre and radius 4.8 cm, cut off CD = 4.8 cm.

Step VI: Join AD.The quadrilateral so obtained is the required quadrilateral.

Chapter 18 - Practical Geometry (Constructions) (Part - 1), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics

 


Question 3:

Construct a quadrilateral PQRS, in which PQ  = 3.5 cm, QR  = 2.5 cm, RS  = 4.1 cm, ∠Q  = 75° and ∠R  = 120°.

ANSWER:

Steps of construction:

Step I: Draw QR = 2.5 cm.

Step II: Construct ∠PQR = 75° at Q.

Step III : With Q as the centre and radius 3.5 cm, cut off QP = 3.5 cm.

Step IV: Construct ∠QRS = 120° at  R.

Step V :With R as the centre and radius 4.1 cm, cut off RS = 4.1 cm.

Step VI: Join PSThe quadrilateral so obtained is the required quadrilateral.

Chapter 18 - Practical Geometry (Constructions) (Part - 1), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics

 

PAGE NO 18.11:

Question 4:

Construct a quadrilateral ABCD given BC  = 6.6 cm, CD  = 4.4 cm, AD  = 5.6 cm and ∠D  = 100° and ∠C  = 95°.

ANSWER:

Steps of construction:

Step I: Draw DC = 4.4 cm.

Step II: Construct ∠ADC = 100° at D.

Step III : With D as the centre and radius 5.6 cm, cut off DA = 5.6 cm.

Step IV: Construct ∠BCD = 95° at C. 

Step V :With C as the centre and radius 6.6 cm, cut off CB = 6.6 cm.

Step VI: Join  AB.The quadrilateral so obtained is the required quadrilateral.

Chapter 18 - Practical Geometry (Constructions) (Part - 1), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics

 


Question 5:

Construct a quadrilateral ABCD, in which AD  = 3.5 cm, AB  = 4.4 cm, BC  = 4.7 cm, ∠A  = 125° and ∠B  = 120°.

ANSWER:

Steps of construction:

Step I: Draw AB = 4.4 cm.

Step II: Construct ∠BAD = 125° at A.

Step III : With A as the centre and radius 3.5 cm, cut off  AD = 3.5 cm.

Step IV: Construct ∠ABC = 120° at B.

Step V :With B as the centre and radius 4.7 cm, cut off  BC = 4.7 cm.

Step VI: Join  CD.The quadrilateral so obtained is the required quadrilateral.

Chapter 18 - Practical Geometry (Constructions) (Part - 1), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics

 


Question 6:

Construct a quadrilateral PQRS, in which ∠Q = 45°, ∠R  = 90°, QR  = 5 cm, PQ  = 9 cm and Rs = 7 cm.

ANSWER:

Steps of construction:

Step I: Draw QR = 5 cm.

Step II: Construct ∠PQR = 45° at Q.

Step III : With Q as the centre and radius 9 cm, cut off QP = 9 cm.

Step IV: Construct ∠QRS = 90° at R. 

Step V :With R as the centre and radius 7 cm, cut off RS = 7 cm.

Since, the line segment PQ and RS intersect each other, the quadrilateral cannot be constructed.

Chapter 18 - Practical Geometry (Constructions) (Part - 1), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics


Question 7:

Construct a quadrilateral ABCD, in which AB  =  BC  = 3 cm, AD  = 5 cm, ∠A  = 90° and ∠B  = 105°.

ANSWER:

Steps of construction:Step I: Draw AB = 3 cm.

Step II: Construct ∠DAB = 90° at A.

Step III : With A as the centre and radius 5 cm, cut off  AD = 5 cm.

Step IV: Construct ∠ABC = 105° at B.

Step V: With B as the centre and radius 3 cm, cut off BC = 3 cm.

Step VI: Join CD.The quadrilateral so obtained is the required quadrilateral.

 

Chapter 18 - Practical Geometry (Constructions) (Part - 1), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics

 


Question 8:

Construct a quadrilateral BDEF, where DE  = 4.5 cm, EF  = 3.5 cm, FB  = 6.5 cm, ∠F  = 50° and ∠E  = 100°.

ANSWER:

Steps of construction:

Step I: Draw EF = 3.5 cm.

Step II: Construct ∠DEF = 100° at E.

Step III : With E as the centre and radius 4.5 cm, cut off  DE = 4.5 cm.

Step IV: Construct ∠EFB = 50° at F.

Step V :With F as the centre and radius 6.5 cm, cut off FB = 6.5 cm.

Step VI: Join BD. The quadrilateral so obtained is the required quadrilateral.

Chapter 18 - Practical Geometry (Constructions) (Part - 1), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics

The document Chapter 18 - Practical Geometry (Constructions) (Part - 1), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics is a part of the Class 8 Course RD Sharma Solutions for Class 8 Mathematics.
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FAQs on Chapter 18 - Practical Geometry (Constructions) (Part - 1), Class 8, Maths RD Sharma Solutions - RD Sharma Solutions for Class 8 Mathematics

1. What are practical geometrical constructions?
Ans. Practical geometrical constructions refer to the methods and techniques used to construct different geometrical figures using a compass, ruler, and other tools. These constructions are based on specific rules and steps to ensure accurate results.
2. What are the basic tools required for practical geometrical constructions?
Ans. The basic tools required for practical geometrical constructions are a compass, a ruler, a protractor, and a pencil. The compass is used to draw circles and arcs, while the ruler is used to draw straight lines. The protractor helps in measuring angles accurately.
3. How are angles constructed using practical geometry?
Ans. Angles can be constructed using practical geometry by following these steps: 1. Draw a line segment as one side of the angle. 2. Place the compass at one endpoint of the line segment and draw an arc that intersects the line. 3. Without changing the compass width, place the compass at the other endpoint of the line segment and draw another arc that intersects the previous arc. 4. Draw a line connecting the intersection point of the two arcs to the endpoint of the line segment. 5. This line represents the other side of the constructed angle.
4. How can we construct a perpendicular bisector using practical geometry?
Ans. To construct a perpendicular bisector using practical geometry, follow these steps: 1. Draw a line segment. 2. Place the compass at one endpoint of the line segment and draw arcs on both sides of the line. 3. Without changing the compass width, place the compass at the other endpoint and draw arcs intersecting the previous arcs. 4. Draw a straight line connecting the intersection points of the arcs. 5. This line is the perpendicular bisector of the given line segment.
5. Can practical geometry be used to construct similar triangles?
Ans. Yes, practical geometry can be used to construct similar triangles. Two triangles are said to be similar if their corresponding angles are equal and their corresponding sides are in proportion. By using the compass and ruler, we can construct similar triangles by applying the corresponding angles and side ratios.
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