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Geometry is a branch of mathematics concerned
with questions of shape, size, relative position of
figures, and the properties of space. The word
geometry came from the Ancient Greek word :
γεωμετρία (geometron) .Where geo- means
“earth” and -Metron means “measurement”.
We shall learn how to construct a unique
quadrilateral given the following
measurements:
• When four sides and one diagonal are given.
• When two diagonals and three sides are given.
• When two adjacent sides and three angles are
given.
• When three sides and two included angles are
given.
• When other special properties are known.
We shall explain this construction through an example.
 Construct a quadrilateral PQRS where PQ = 4cm ,QR = 6
cm, RS = 5 cm, PS = 5.5 cm and PR = 7 cm.
Draw a rough sketch of the quadrilateral.
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.
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 ”.
7 cm
Q
R
S
P
Rough Sketch
Real Figure
S
R
Q
P
We shall explain this construction through an example.
 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.
Draw a rough sketch of the quadrilateral.
Step 1 : Draw Δ ACD 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
7 cm
5.5cm
C
B
A
D
Rough Sketch
C
B
D
A
5.5 cm
Real Figure
We shall explain this construction through an example.
 Construct a quadrilateral MIST where MI = 3.5 cm, IS = 6.5
cm, ∠M = 75°, ∠I = 105° and ∠S = 120°.
Draw a rough sketch of the quadrilateral.
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.”
120°
75°
S
T I
M
Rough Sketch
120°
M I
S
T
Y Z
X
Real Figure
We shall explain this construction through an example.
 Construct a quadrilateral ABCD, where AB = 4 cm, BC = 5
cm, CD = 6.5 cm and ∠B = 105° and ∠C = 80°.
Draw a rough sketch of the quadrilateral.
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.
6.5 cm
4 cm
D C
BA
Rough Sketch
D
B
A
C
X
Y
Real Figure
Practical geometry for class 8th
Practical geometry for class 8th

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Practical geometry for class 8th

  • 1.
  • 2. Geometry is a branch of mathematics concerned with questions of shape, size, relative position of figures, and the properties of space. The word geometry came from the Ancient Greek word : γεωμετρία (geometron) .Where geo- means “earth” and -Metron means “measurement”.
  • 3. We shall learn how to construct a unique quadrilateral given the following measurements: • When four sides and one diagonal are given. • When two diagonals and three sides are given. • When two adjacent sides and three angles are given. • When three sides and two included angles are given. • When other special properties are known.
  • 4. We shall explain this construction through an example.  Construct a quadrilateral PQRS where PQ = 4cm ,QR = 6 cm, RS = 5 cm, PS = 5.5 cm and PR = 7 cm. Draw a rough sketch of the quadrilateral. 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. 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 ”.
  • 6. We shall explain this construction through an example.  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. Draw a rough sketch of the quadrilateral. Step 1 : Draw Δ ACD 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
  • 8. We shall explain this construction through an example.  Construct a quadrilateral MIST where MI = 3.5 cm, IS = 6.5 cm, ∠M = 75°, ∠I = 105° and ∠S = 120°. Draw a rough sketch of the quadrilateral. 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.”
  • 9. 120° 75° S T I M Rough Sketch 120° M I S T Y Z X Real Figure
  • 10. We shall explain this construction through an example.  Construct a quadrilateral ABCD, where AB = 4 cm, BC = 5 cm, CD = 6.5 cm and ∠B = 105° and ∠C = 80°. Draw a rough sketch of the quadrilateral. 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.
  • 11. 6.5 cm 4 cm D C BA Rough Sketch D B A C X Y Real Figure