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NCERT Solutions for Class 8 Maths - Practical Geometry - 3

Question 1. How will you construct a rectangle PQRS if you know only the lengths PQ and QR?

Solution: A rectangle can be constructed by taking PQ as the length. 

Making an ∠90° at Q and cutting off QR, the breadth the ray NCERT Solutions for Class 8 Maths - Practical Geometry - 3 Remaining two points R and S can be located by taking P and R as centres and radii as QR and PQ respectively draw arcs to intersect at S. Thus, PQRS is the required parallelogram.

NCERT Solutions for Class 8 Maths - Practical Geometry - 3

Question 2. Construct the kite EASY if AY = 8 cm, EY = 4 cm and SY = 6 cm. Which properties of the kite did you use in the process?

NCERT Solutions for Class 8 Maths - Practical Geometry - 3

Solution: Following properties have been used in constructing the KITE:

(i) Diagonals are at right angles.

(ii) One of the diagonal bisects the other.

(iii) Pairs of consecutive sides are equal.

Steps of construction:

I. Draw a line segment AY = 8 cm. 
II. Draw NCERT Solutions for Class 8 Maths - Practical Geometry - 3 , the perpendicular bisector of AY such that it meets AY at O.
III. We cannot locate a point E on PQ at 4 cm from Y and A, i.e. EY = 4 cm = EA is not possible. It is possible only when E and O coincide. In that case the kite does not exist.

NCERT Solutions for Class 8 Maths - Practical Geometry - 3

EXERCISE 4.5
Question: Draw the following.
1. The square READ with RE = 5.1 cm.
2. A rhombus whose diagonals are 5.2 cm and 6.4 cm long.
3. A rectangle with adjacent sides of lengths 5 cm and 4 cm.
4. A parallelogram OKAY where OK = 5.5 cm and KA = 4.2 cm.

NCERT Solutions for Class 8 Maths - Practical Geometry - 3

Solution:1. Steps of construction:

I. Draw a line segment RE = 5.1 cm.

II. At E, draw NCERT Solutions for Class 8 Maths - Practical Geometry - 3, such that ∠REX = 90°. 

III. From NCERT Solutions for Class 8 Maths - Practical Geometry - 3, cut-off NCERT Solutions for Class 8 Maths - Practical Geometry - 3 = 5.1 cm.

IV. With centre at A, draw an arc above RE of radius = 5.1 cm.

V. With centre at R, and radius = 5.1 cm, draw another arc to intersect the previous arc at D.

VI. Join DA and DR.

Thus, READ is the required square.

2. Steps of construction:

NCERT Solutions for Class 8 Maths - Practical Geometry - 3

Note: The diagonals of a rhombus bisect each other at right angles.

I. Draw a line segment AC = 5.2 cm.

II. Draw NCERT Solutions for Class 8 Maths - Practical Geometry - 3  , the perpendicular bisector of AC. 

III. From XY, cut-off OD NCERT Solutions for Class 8 Maths - Practical Geometry - 3

IV. Similarly, cut-off OB NCERT Solutions for Class 8 Maths - Practical Geometry - 3

V. Join AD, DC, CB and BA.

Thus, ABCD is the required rhombus.

3. Steps of construction:

NCERT Solutions for Class 8 Maths - Practical Geometry - 3

I. Draw a line segment PQ = 5 cm.

II. At P, draw NCERT Solutions for Class 8 Maths - Practical Geometry - 3 , such that ∠QPX = 90° 

III. FromNCERT Solutions for Class 8 Maths - Practical Geometry - 3 , cut-off PS = 4 cm.

IV. With centre at 5 and radius = 5 cm, mark an arc towards Q.

V. With centre Q and radius = 4 cm, mark an arc to intersect the previous arc at R.

VI. Join RQ and RS

Thus, PQRS is the required rectangle.

 

4. Steps of construction:

NCERT Solutions for Class 8 Maths - Practical Geometry - 3

I. Draw a line segment OK = 5.5 cm.

II. At K, draw a ray NCERT Solutions for Class 8 Maths - Practical Geometry - 3 .

III. From NCERT Solutions for Class 8 Maths - Practical Geometry - 3 , cut-off KA = 4.2 cm.

IV. With centre at A and radius = 5.5 cm, draw an arc above OK.

V. With centre O and radius = 4.2 cm, draw another arc to intersect the previous arc at Y.

VI. Join YO and YA.

Thus, OKAY is the required parallelogram.

The document NCERT Solutions for Class 8 Maths - Practical Geometry - 3 is a part of the Class 8 Course NCERT Textbooks & Solutions for Class 8.
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FAQs on NCERT Solutions for Class 8 Maths - Practical Geometry - 3

1. What are the different types of triangles based on their sides?
Ans. There are three types of triangles based on their sides: 1. Equilateral Triangle: All three sides of an equilateral triangle are equal in length. 2. Isosceles Triangle: In an isosceles triangle, two sides are equal in length. 3. Scalene Triangle: A scalene triangle has all three sides of different lengths.
2. What are the different types of triangles based on their angles?
Ans. There are three types of triangles based on their angles: 1. Acute Triangle: An acute triangle has all three angles less than 90 degrees. 2. Obtuse Triangle: In an obtuse triangle, one angle is greater than 90 degrees. 3. Right Triangle: A right triangle has one angle equal to 90 degrees.
3. How can we construct an equilateral triangle using a compass and a ruler?
Ans. To construct an equilateral triangle using a compass and a ruler, follow these steps: 1. Draw a line segment and mark its midpoint. 2. Using the compass, open it to the length of the line segment. 3. Place the compass on the midpoint and draw arcs on both sides of the line segment. 4. Without changing the compass width, place the compass on one of the arc intersections and draw another arc. 5. Repeat the previous step on the other arc intersection. 6. Connect the three arc intersections to form an equilateral triangle.
4. How can we 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. 2. Using the compass, open it to a width greater than half of the line segment. 3. Place the compass on one end of the line segment and draw an arc above and below the line. 4. Without changing the compass width, place the compass on the other end of the line segment and draw arcs intersecting the previous arcs. 5. Connect the intersection points of the arcs to form the perpendicular bisector.
5. How can we construct an angle bisector of an angle?
Ans. To construct an angle bisector of an angle, follow these steps: 1. Draw the given angle. 2. Place the compass on the vertex of the angle and draw an arc that intersects both sides of the angle. 3. Without changing the compass width, place the compass on each intersection point and draw arcs that intersect each other. 4. Connect the vertex of the angle with the intersection point of the arcs to form the angle bisector.
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