Practical Geometry Class 8 Extra Questions with Answers

These Class 8 Maths Extra Questions Chapter 4 Practical Geometry will help students prepare well for the exams.

Class 8 Maths Chapter 4 Extra Questions Practical Geometry

Extra Questions of Practical Geometry Class 8

Question 1.
Construct a quadrilateral ABCD given AB = 5.1 cm, AD = 4 cm, BC = 2.5 cm, ∠A = 60°, and ∠B = 85°
Solution:
Given AB = 5.1 cm, AD = 4 cm,
BC = 2.5 cm, ∠A = 60°, ∠B = 85°

Steps of construction :

1. Draw a rough sketch with given measurements.
2. Draw a line segment AB = 5.1 cm and draw ∠BAX = 60° at A
3. With ‘L’ as centre. Draw an arc with radius 4 cm, so as to intersect AX at ‘D’.
4. Draw ∠ABY = 85° at B and draw an arc with B as centre 2.5 cm as radius to intersect AY at C.
5. Join ‘D’ and ‘C’. So, ABCD is the required quadrilateral.

Question 2.
Construct a parallelogram ABCD with AB = 3.5 cm., BC = 4.5 cm. and diagonal BD = 5.5 cm.
Solution:
Given parallelogram ABCD mesaures
AB = 3.5 cm
BC = 4.5 cm
and Diagonal BD = 5.5 cm

Steps of construction :

1. Draw a rough sketch of a parallelogram ABCD and mark the given measurements.
2. Analysis : As ABCD is parallelogram, opposite sides are equal
   ∴ AB = CD = 3.5 cm BC = AD = 4.5 cm
3. Draw a line segment BC = 4.5 cm
4. C as a centre draw an arc with 3.5 cm as a radius and draw another arc with B as centre 5.5 cm as radius to cut previous arc and name as ‘D’, Join CD and BD.
5. Draw ‘B’ as centre with 3.5 cm as radius and draw another arc ‘D’ as centre with 4.5 cm as radius to cut the arc, name as ‘A’. Join A and D. This is our required parallelogram ABCD.

Question 3.
Construct a quadrilateral WXYZ in which ZW = 4 cm, WX = 5.5 cm, YZ = 7 cm, YW = 8 cm and XZ – 6.5 cm.
Solution:

Steps of construction :

1. Draw a rough sketch with marking the given measurements.
2. Draw the diagonal WY equals 8 cm using the ruler.
3. Draw the triangle WZY using the SSS construction with the measurements of WZ = 4 cm and YZ = 8 cm.
4. Draw an arc of radius 6.5 cm with Z as centre.
5. Draw another arc of radius 5.5 cm with W as centre.
6. Name the point of intersection as ‘X’.
7. Joins the points Z to X, W to X and Y to X to complete the quadrilateral WXYZ.

Practical Geometry Class 8 Extra Questions

Question 1.
Construct a quadrilateral PQRS where PQ = 4 cm, QR = 6 cm, RS = 5 cm, PS = 5.5 cm and PR = 7 cm.
Solution:
[A rough sketch will help us in visualising the quadrilateral. We draw this first and mark the measurements.]

Step 1 : From the rough sketch, it is easy to see that ΔPQR can be constructed using SSS construction condition.
Draw ΔPQR.

Step 2 : Now, we have to locate the fourth point S. This ‘S’ would be on the side opposite to Q with reference to PR. For that, we have two measurements.

S is 5.5 cm away from P. So, with P as centre, draw an arc of radius 5.5 cm. (The point S is somewhere on this arc!)

Step 3 : S is 5 cm away from R. So with R as centre, draw an arc of radius 5 cm (The point S is somewhere on this arc also.)

Step 4 : S should lie on both the arcs drawn. So it is the point of intersection of the two arcs. Mark S and complete PQRS. PQRS is the required quadrilateral.

Question 2.
Construct a quadrilateral ABCD, given that BC = 4.5 cm, AD = 5.5 cm, CD = 5 cm the diagonal AC = 5.5 cm and diagonal BD = 7 cm.
Solution:
Here is the rough sketch of the quadrilateral ABCD.

Studying this sketch, we can easily see that it is possible to draw ΔACD first. Step 1: Draw ΔACD using SSS construction.
(We now need to find B at a distance of 4.5 cm from C and 7 cm from D).

Step 2 : With D as centre, draw an arc of radius 7 cm.(B is somewhere on this arc).

Step 3 : With C as centre, draw an arc of radius 4.5 cm (B is somewhere on this arc also).

Step 4 : Since B lies on both the arcs, B is the point intersection of the two arcs. Mark B and complete ABCD. ABCD is the required quadrilateral.

Question 3.
Construct a quadrilateral MIST where MI = 3.5 cm, IS = 6.5 cm, ∠M = 75°, ∠I = 105° and ∠S = 120°.
Solution:
Here is a rough sketch that would help us in deciding our steps of construction. We give only hints for various steps.

Step 1 : How do you locate the points? What choice do you make for the base and what is the first step ?

Step 2 : Make ∠ISY = 120° at S.

Step 3 : Make ∠IMZ = 75° at M. (where will SY and MZ meet?) Mark that point as T. We get the required quadrilateral MIST.

Question 4.
Construct a quadrilateral ABCD, where AB = 4 cm, BC = 5 cm, CD = 6.5 cm and ∠B = 105° and ∠C = 80°.
Solution:
We draw a rough sketch, as usual, to get an idea of how we can start off. Then we can devise a plan to locate the four points.

Step 1 : Start with taking BC = 5 cm on B. Draw an angle of 105° along BX. Locate A 4 cm away on this. We now have B, C and A.

Step 2 : The fourth point D is on CY which is inclined at 80° to BC. So make ∠BCY = 80° at C on BC.

Step 3 : D is at a distance of 6.5 cm on CY. With C as centre, draw an arc of length 6.5 cm. It cuts CY at D.

Step 4 : Complete the quadrilateral ABCD. ABCD is the required quadrilateral.

Question 5.
Draw a square of side 4.5 cm.
Solution:
Initially it appears that only one measurement has been given. Actually we have many more details with us, because the figure is a special quadrilateral, namely a square. We now know that each of its angles is a right angle. (See the rough figure)

This enables us to draw AABC using SAS condition. Then D can be easily located.

Question 6.
Is it possible to construct a rhombus ABCD where AC = 6 cm and BD = 7 cm ? Justify your answer.
Solution:
Only two (diagonal) measurements of the rhombus are given. However, since it is a rhombus, we can find more help from its properties.

The diagonals of a rhombus are perpendicular bisectors of one another.

So, first draw AC = 7 cm and then construct its perpendicular bisector. Let them meet at 0. Cut off 3 cm lengths on either side of the drawn bisector. You now get B and D.

Draw the rhombus now, based on the method described above.