Class 8 Exam  >  Class 8 Notes  >  RD Sharma Solutions for Class 8 Mathematics  >  RD Sharma Solutions - Chapter 18 - Practical Geometry (Constructions) (Part - 4), Class 8, Maths

Chapter 18 - Practical Geometry (Constructions) (Part - 4), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics PDF Download

PAGE NO 18.6:

Question 1:

Construct a quadrilateral ABCD in which AB  = 3.8 cm, BC  = 3.0 cm, AD  = 2.3 cm, AC  = 4.5 cm and BD  = 3.8 cm.

ANSWER:

Steps of construction:

Step I: Draw AC = 6 cm.

Step II: With A as the centre and radius 3.8 cm, draw an arc.

Step III : With C as the centre and radius 3.0 cm, draw an arc to intersect the arc drawn in Step II at B.

Step IV: With B as the centre and radius 3.8 cm, draw an arc on the other side of AC. 

Step V :With A as the centre and radius 2.3 cm, draw an arc to intersect the arc drawn in Step IV at D.

Step VI: Join BA, DA, BC and CD to obtained the required quadrilateral.

Chapter 18 - Practical Geometry (Constructions) (Part - 4), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics


Question 2:

Construct a quadrilateral ABCD in which BC  = 7.5 cm, AC  =  AD  = 6 cm, CD  = 5 cm and BD  = 10 cm.

ANSWER:

Steps of construction:

Step I: Draw AC = 6 cm.

Step II: With A as the centre and radius 6 cm, draw an arc.

Step III : With C as the centre and radius 5 cm, draw an arc to intersect the arc drawn in Step II at D.

Step IV: With D as the centre and radius 10 cm, draw an arc on the other side of the line segment AC.

Step V :With C as the centre and radius 7.5 cm, draw an arc to intersect the arc drawn in Step IV at B.

Step VI: Join BA, DA, BC and CD to obtained  the required quadrilateral.

Chapter 18 - Practical Geometry (Constructions) (Part - 4), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics


Question 3:

Construct a quadrilateral ABCD, when AB  = 3 cm, CD  = 3 cm, DA  = 7.5 cm, AC  = 8 cm and BD  = 4 cm.

ANSWER:

If we consider a triangle ABD from the given data, then
AB = 3 cm
BD = 4 cm
AD = 7.5 cm
AB + BD = 3 + 4 = 7 cm
However, we know that the sum of the lengths of two sides of a triangle is always greater than the third side.
Therefore, construction is not possible from the given data.


Question 4:

Construct a quadrilateral ABCD given AD  = 3.5 cm, BC  = 2.5 cm, CD  = 4.1 cm, AC  = 7.3 cm and BD  = 3.2 cm.

ANSWER:

Steps of construction:

Step I: Draw CD = 4.1 cm.

Step II: With C as the centre and radius 7.3 cm, draw an arc.

Step III : With D as the centre and radius 3.5 cm, draw an arc to intersect the arc drawn in Step II at A.

Step IV: With D as the centre and radius 3.2 cm, draw an arc on the other side of AC.

Step V :With C as the centre and radius 2.5 cm, draw an arc to intersect the arc drawn in Step IV at B.

Step VI: Join BA, DA, BC and BD and AC to obtained the required quadrilateral.

Chapter 18 - Practical Geometry (Constructions) (Part - 4), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics


Question 5:

Construct a quadrilateral ABCD given AD  = 5 cm, AB  = 5.5 cm, BC  = 2.5 cm, AC  = 7.1 cm and BD  = 8 cm.

ANSWER:

Steps of construction:

Step I: Draw AB = 5.5 cm.

Step II: With A as the centre and radius 7.1 cm, draw an arc.

Step III : With B as the centre and radius 2.5 cm, draw an arc to intersect the arc drawn in Step II at C.

Step IV: With B as the centre and radius 8 cm, draw an arc.

Step V :With A as the centre and radius 5 cm, draw an arc to intersect the arc drawn in Step IV at D.

Step VI: Join DA, DB, BC, AC and CD to obtained  the required quadrilateral. 

Chapter 18 - Practical Geometry (Constructions) (Part - 4), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics


Question 6:

Construct a quadrilateral ABCD in which BC  = 4 cm, CA  = 5.6 cm, AD  = 4.5 cm, CD  = 5 cm and BD  = 6.5 cm.

ANSWER:

Steps of construction:

Step I: Draw BC = 4 cm.

Step II: With B as the centre and radius 6.5 cm, draw an arc.

Step III : With C as the centre and radius 5 cm, draw an arc to intersect the arc drawn in Step II at D.

Step IV: With C as the centre and radius 5.6 cm, draw an arc on the same side.

Step V :With D as the centre and radius 4.5 cm, draw an arc to intersect the arc drawn in Step IV at A.

Step VI: Join BA, AC, DA, BD and CD to obtained the required quadrilateral.

Chapter 18 - Practical Geometry (Constructions) (Part - 4), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics

The document Chapter 18 - Practical Geometry (Constructions) (Part - 4), 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 - 4), Class 8, Maths RD Sharma Solutions - RD Sharma Solutions for Class 8 Mathematics

1. What are the steps to construct a perpendicular bisector of a line segment?
Ans. To construct a perpendicular bisector of a line segment, follow these steps: 1. Draw the given line segment AB. 2. With A as the center, draw an arc on one side of the line segment. 3. With B as the center, draw an arc on the other side of the line segment, intersecting the previous arc. 4. Draw a straight line connecting the intersection points of the arcs. 5. This line is the perpendicular bisector of the line segment AB.
2. How can I construct a line parallel to a given line passing through a given point?
Ans. To construct a line parallel to a given line passing through a given point, follow these steps: 1. Draw the given line and mark the given point. 2. From the given point, draw a line segment intersecting the given line at any point. 3. With the same length as the line segment, draw an arc above and below the given line, intersecting it at two points. 4. Connect these two points with a straight line. 5. The line passing through the given point and parallel to the given line is now constructed.
3. How can I construct an angle of 60 degrees?
Ans. To construct an angle of 60 degrees, follow these steps: 1. Draw a ray AB. 2. With A as the center, draw an arc cutting the ray at a point P. 3. With P as the center, draw another arc intersecting the previous arc at a point Q. 4. Draw a straight line connecting points A and Q. 5. The angle formed by the ray AB and line AQ is a 60-degree angle.
4. Can I construct a triangle with any three given line segments?
Ans. No, not all combinations of three line segments can form a triangle. According to the triangle inequality theorem, the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. If this condition is not satisfied, a triangle cannot be formed.
5. How can I construct an equilateral triangle?
Ans. To construct an equilateral triangle, follow these steps: 1. Draw a line segment AB. 2. With A as the center, draw an arc of any radius. 3. With B as the center, draw another arc with the same radius, intersecting the previous arc. 4. Connect the intersection points of the arcs with a straight line. 5. The triangle formed by the line segment AB and the arcs is an equilateral triangle, with all sides and angles equal.
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