Class 8 Exam  >  Class 8 Notes  >  RD Sharma Solutions for Class 8 Mathematics  >  RD Sharma Solutions - Chapter 18 - Practical Geometry (Constructions) (Part - 3), Class 8, Maths

Chapter 18 - Practical Geometry (Constructions) (Part - 3), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics PDF Download

PAGE NO 18.4:

Question 1:

Construct a quadrilateral ABCD in which AB  = 4.4 cm, BC  = 4 cm, CD  = 6.4 cm, DA  = 3.8 cm and BD  = 6.6 cm.

ANSWER:

First, we draw a rough sketch of the quadrilateral ABCD and write down its dimensions along the sides.

We may divide the quadrilateral into two constructible triangles ABD and BCD.

Steps of Construction:

Step I: Draw BD = 6.6 cm

Step II: With B as the centre and radius BC = 4 cm, draw an arc.

Step III: With D as the centre and radius 6.4 cm, draw an arc to intersect the arc drawn in Step II at  C.

Step IV: With B as the centre and radius 4.4 cm, draw an arc on the side BD opposite to that of C.

Step V: With D as the centre and radius 3.8 cm, draw an arc to intersect the arc drawn in Step IV at  A.

Step VI: Join BA, DA, BC and CD

The quadrilateral ABCD so obtained is the required quadrilateral.

Chapter 18 - Practical Geometry (Constructions) (Part - 3), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics


Question 2:

Construct a quadrilateral ABCD such that AB  =  BC  = 5.5 cm, CD  = 4 cm, DA  = 6.3 cm and AC  = 9.4 cm. Measure BD.

ANSWER:

Steps of construction:

Step I: Draw AB = 5.5 cm

Step II: With B as the centre and radius BC = 5.5 cm, draw an arc.

Step III: With A as the centre and radius AC = 9.4 cm,

 draw an arc to intersect the arc drawn in Step II at  C.

Step IV: With C as the centre and radius CD = 4 cm, draw an arc.

Step V: With A as the centre and radius AD = 6.3 cm, 

draw an arc to intersect the arc drawn in Step IV at  D.

Step VI: Join DA, BC, AC and CD.

The quadrilateral ABCD so obtained is the required quadrilateral.

Chapter 18 - Practical Geometry (Constructions) (Part - 3), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics


Question 3:

Construct a quadrilateral XYZW in which XY  = 5 cm, YZ  = 6 cm, ZW  = 7 cm, WX  = 3 cm and XZ  = 9 cm.

ANSWER:

Steps of construction:

Step I: Draw XZ = 9 cm

Step II: With X as the centre and radius 5 cm, draw an arc above XZ.

Step III: With Z as the centre and radius 6 cm, draw an arc to intersect the arc drawn in Step II at Y above XZ.

Step IV: With Z as the centre and radius 7 cm, draw an arc below XZ.

Step V: With X as the centre and radius 3 cm, draw an arc to intersect the arc drawn in Step IV at  W below XZ.

Step VI: Join XY, YZ, ZW and XW.

The quadrilateral WXYZ so obtained is the required quadrilateral.

Chapter 18 - Practical Geometry (Constructions) (Part - 3), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics


Question 4:

Construct a parallelogram PQRS such that PQ  = 5.2 cm, PR  = 6.8 cm and QS  = 8.2 cm.

ANSWER:

Chapter 18 - Practical Geometry (Constructions) (Part - 3), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics

 

In a parallelogram opposite sides are equal.

Thus, we have to construct a quadrilateral PQRS in which

PQ = 5.2 cm,

PR = 6.8 cm and QS = 8.2 cm.

Steps of construction:

Step I: Draw QS = 8.2 cm

Step II: With Q as the centre and radius 5.2 cm, draw an arc.

Step III: With S as the centre and radius 5.2 cm, draw an arc to intersect the arc drawn in Step II at  C.

Step IV: With P as the centre and radius 6.8 cm.

Step V: With Q as the centre and radius 5.2 cm, draw an arc to intersect the arc drawn in Step IV at  A.

Step VI: Join QR, QP, PS and SR.

The quadrilateral PQRS so obtained is the required quadrilateral.


Question 5:

Construct a rhombus with side 6 cm and one diagonal 8 cm. Measure the other diagonal.

ANSWER:

Chapter 18 - Practical Geometry (Constructions) (Part - 3), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics

 

Steps of construction:

Step 1: Draw AC = 8 cm.

Step 2: With A as the centre and radius = 6 cm, draw arcs on both sides.

Step 3: With C as the centre and radius = 6 cm, 

draw arcs on both sides, intersecting the previous arcs at points B and D.

Step 4: Join BD  = 8.9cm.

Thus, ABCD is the required rhombus.


Question 6:

Construct a kite ABCD in which AB  = 4 cm, BC  = 4.9 cm and AC = 7.2 cm.

ANSWER:

Steps of construction:

Step I: Draw AC = 7.2 cm.

Step II: With A as the centre and radius 4cm, draw arcs on both sides of the line segment AC.

Step III : With C as the centre and radius 4.9 cm, draw arcs on both sides of AC intersecting the previous arcs of step II at B and D.

Step IV: Join BA, DA, BC and CD.

 Thus, the quadrilateral ABCD so obtained is the required kite.

Chapter 18 - Practical Geometry (Constructions) (Part - 3), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics


Question 7:

Construct, if possible, a quadrilateral ABCD given AB  = 6 cm, BC  = 3.7 cm, CD  = 5.7 cm, AD  = 5.5 cm and BD  = 6.1 cm. Give reasons for not being able to construct it, if you cannot.

ANSWER:

Steps of construction:

Step I: Draw AB = 6 cm.

Step II: With A as the centre and radius 5.5 cm, draw an arc.

Step III : With B as the centre and radius 6.1 cm, draw an arc to intersect the arc drawn in StepII at D.

Step IV: With B as the centre and radius 3.7 cm, draw an arc on the side.

Step V :With D as the centre and radius 5.7 cm, draw an arc to intersect the arc drawn in StepIV at C.

Step VI: Join BD, DA, BC and CD.

 The quadrilateral ABCD so obtained is the required quadrilateral.

Chapter 18 - Practical Geometry (Constructions) (Part - 3), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics


Question 8:

Construct, if possible, a quadrilateral ABCD in which AB  = 6 cm, BC  = 7 cm, CD  = 3 cm, AD  = 5.5 cm and AC  = 11 cm. Give reasons for not being able to construct, if you cannot. (Not possible, because in triangle ACD, AD + CD < AC).

ANSWER:

Such a quadrilateral cannot be constructed because in a triangle, the sum of the length of its two sides must be greater than the that of the third side
But here in triangle ACD,
AD + CD = 5.5 + 3 = 8.5 cm
and AC = 11 cm
i.e., AD + CD <  AC, which is not possible.
So, the construction is not possible.

The document Chapter 18 - Practical Geometry (Constructions) (Part - 3), 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 - 3), Class 8, Maths RD Sharma Solutions - RD Sharma Solutions for Class 8 Mathematics

1. What are the different types of constructions discussed in Chapter 18 of RD Sharma Solutions for Class 8 Maths?
Ans. In Chapter 18 of RD Sharma Solutions for Class 8 Maths, the different types of constructions discussed are: 1. Construction of perpendicular bisector of a line segment 2. Construction of angle bisector 3. Construction of perpendicular from a point to a line 4. Construction of triangle when its base, sum/difference of the other two sides, and one base angle are given 5. Construction of a triangle when its base, base angles, and one side are given
2. How do you construct the perpendicular bisector of a line segment using the steps given in RD Sharma Solutions for Class 8 Maths?
Ans. To construct the perpendicular bisector of a line segment, follow these steps: 1. Draw a 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 of the same radius intersecting the previous arc. 4. Let the intersection point be P. Join AP and BP. 5. The line passing through P and perpendicular to AB is the perpendicular bisector of the line segment AB.
3. How can I construct an angle bisector using the steps provided in RD Sharma Solutions for Class 8 Maths?
Ans. To construct an angle bisector, follow these steps: 1. Draw an angle XYZ. 2. With X as the center, draw an arc that intersects both rays of the angle. 3. With Y as the center, draw an arc of the same radius intersecting the previous arc. 4. Let the intersection point be P. Join XP. 5. The line XP is the bisector of angle XYZ.
4. Can you explain how to construct a perpendicular from a point to a line using the steps given in RD Sharma Solutions for Class 8 Maths?
Ans. Yes, here are the steps to construct a perpendicular from a point to a line: 1. Draw a line and a point not on the line. 2. From the given point, draw a line segment that intersects the given line at a right angle. 3. The line segment drawn is the perpendicular from the given point to the given line.
5. How do you construct a triangle when its base, sum/difference of the other two sides, and one base angle are given, as explained in RD Sharma Solutions for Class 8 Maths?
Ans. To construct a triangle when its base, sum/difference of the other two sides, and one base angle are given, follow these steps: 1. Draw a line segment AB as the base of the triangle. 2. Using the given sum/difference of the other two sides, mark points C and D on AB. 3. Construct an angle at B, which is the given base angle. 4. With C and D as centers, construct arcs that intersect at a point E. 5. Join AE. The triangle ABE is the required triangle.
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