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Chapter 18 - Practical Geometry (Constructions) (Part - 5), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics PDF Download

PAGE NO 18.8:

Question 1:

Construct a quadrilateral ABCD in which AB  = 3.8 cm, BC  = 3.4 cm, CD  = 4.5 cm, AD  = 5 cm and ∠B  = 80°.

ANSWER:

Steps of construction:

Step I: Draw AB = 3.8 cm.

Step II: Construct ∠ABC = 80°.

Step III : With B as the centre and radius 3.4 cm, cut off BC = 3.4 cm.

Step IV: With C as the centre and radius 4.5 cm, draw an arc.

Step V :With A as the centre and radius 5.3 cm, draw an arc to intersect the arc drawn in Step IV at D.

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

 

Chapter 18 - Practical Geometry (Constructions) (Part - 5), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics


Question 2:

Construct a quadrilateral ABCD, given that AB  = 8 cm, BC  = 8 cm, CD  = 10 cm, AD  = 10 cm and ∠A  = 45°.

ANSWER:

Steps of Construction:

Step I: Draw AB = 8 cm.

Step II: Construct ∠BAD = 45°.

Step III : With A as the centre and radius 10 cm, cut off AD = 10 cm.

Step IV: With D as the centre and radius 10 cm, draw an arc.

Step V :With B as the centre and radius 8 cm, draw an arc to intersect the arc drawn in Step IV at C.

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

 

Chapter 18 - Practical Geometry (Constructions) (Part - 5), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics


Question 3:

Construct a quadrilateral ABCD in which AB  = 7.7 cm, BC  = 6.8 cm, CD  = 5.1 cm, AD  = 3.6 cm and ∠C  = 120°.

ANSWER:

Steps of construction:

Step I: Draw DC = 5.1 cm.

Step II: Construct ∠DCB = 120°.

Step III : With C as the centre and radius 6.8 cm, cut off BC = 6.8 cm.

Step IV: With B as the centre and radius 7.7 cm, draw an arc.

Step V :With D as the centre and radius 3.6 cm, draw an arc to intersect the arc drawn in Step IV at A.

Step VI: Join AB and AD to obtained the required quadrilateral.

Chapter 18 - Practical Geometry (Constructions) (Part - 5), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics

 


Question 4:

Construct a quadrilateral ABCD in which AB  =  BC  = 3 cm, AD  =  CD  = 5 cm, and ∠B  = 120°.

ANSWER:

Steps of construction:

Step I: Draw AB = 3 cm.

Step II: Construct ∠ABC = 120°.

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

Step IV: With C as the centre and radius 5 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 AD and CD to obtained the required quadrilateral.

Chapter 18 - Practical Geometry (Constructions) (Part - 5), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics

 

Question 5:

Construct a quadrilateral ABCD in which AB  = 2.8 cm, BC  = 3.1 cm, CD  = 2.6 cm, and DA  = 3.3 cm and ∠A  = 60°.

ANSWER:

Steps of construction:

Step I: Draw AB = 2.8 cm.

Step II: Construct ∠BAD = 60°.

Step III : With A as the centre and radius 3.3 cm, cut off AD = 3.3 cm.

Step IV: With D as the centre and radius 2.6 cm, draw an arc.

Step V :With B as the centre and radius 3.1 cm, draw an arc to intersect the arc drawn in Step IV at C.

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

Chapter 18 - Practical Geometry (Constructions) (Part - 5), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics


Question 6:

Construct a quadrilateral ABCD in which AB  =  BC  = 6 cm, AD  =  DC  = 4.5 cm and ∠B  = 120°.

ANSWER:

Steps of construction:

Step I: Draw AB = 6 cm.

Step II: Construct ∠ABC = 120°.

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

Now, we can see that AC is about 10.3 cm 

which is greater than AD+CD = 4.5+4.5 = 9 cm.

We know that sum of the lengths of two sides of triangle is always greater than the third side but here, the sum of AD and CD is less than AC.

So, construction of the given quadrilateral is not possible.

Chapter 18 - Practical Geometry (Constructions) (Part - 5), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics

The document Chapter 18 - Practical Geometry (Constructions) (Part - 5), 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 - 5), Class 8, Maths RD Sharma Solutions - RD Sharma Solutions for Class 8 Mathematics

1. How do I construct a perpendicular bisector of a line segment?
Ans. To construct a perpendicular bisector of a line segment, follow these steps: 1. Draw a line segment and label its endpoints as A and B. 2. With A as the center, draw an arc that intersects the line segment at two points, labeled as C and D. 3. With C and D as centers, draw two arcs of the same radius that intersect each other. Label the point of intersection as E. 4. Draw a straight line passing through points E and A. This line is the perpendicular bisector of the line segment AB.
2. Can we construct a triangle with any three given line segments?
Ans. No, we cannot construct a triangle with any three given line segments. In order to construct a triangle, the sum of the lengths of any two sides must be greater than the length of the third side. This is known as the Triangle Inequality Theorem. If this condition is not satisfied, a triangle cannot be constructed using the given line segments.
3. How do I construct an angle bisector?
Ans. To construct an angle bisector, follow these steps: 1. Draw an angle and label its vertex as V. 2. With V as the center, draw an arc that intersects both sides of the angle, creating two points of intersection labeled as A and B. 3. With A and B as centers, draw two arcs of the same radius that intersect each other. Label the point of intersection as C. 4. Draw a straight line passing through point C and vertex V. This line is the angle bisector of the given angle.
4. How do I construct a triangle given its three sides?
Ans. To construct a triangle given its three sides, follow these steps: 1. Draw a line segment and label its endpoints as A and B, representing one side of the triangle. 2. Using a compass, measure the length of the second side of the triangle and draw an arc with A as the center. 3. Keeping the same compass width, measure the length of the third side of the triangle and draw another arc with B as the center. 4. The intersection of the two arcs will be the third vertex of the triangle. Connect this vertex with points A and B to complete the triangle.
5. How do I construct a line parallel to a given line through a given point?
Ans. To construct a line parallel to a given line through a given point, follow these steps: 1. Draw the given line and label it as l. 2. Choose a point on the given line and label it as P. 3. With P as the center, draw an arc that intersects the given line at two points, labeled as A and B. 4. With A and B as centers, draw two arcs of the same radius that intersect each other. Label the point of intersection as C. 5. Draw a straight line passing through point C and point P. This line will be parallel to the given line l and passes through the given point P.
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