Viii practical geometry Rujul vyas

Chapter 4- Practical Geometry
Std : 8 by Rujul Vyas
Std : VIII Subject : Maths Topic : Practical Geometry

Construct a quadrilateral PQRS where PQ = 4 cm,QR = 6 cm, RS = 5
cm, PS = 5.5 cm and PR = 7 cm.

Given : Sides PQ = 4cm , QR = 6cm , RS = 5cm , PS = 5.5cm ,
Diagonal : PR = 7cm

Given : Sides PQ = 4cm , QR = 6cm , RS = 5cm , PS = 5.5cm ,
Diagonal : PR = 7cm
Rough Sketch

Construct a parallelogram PQRS where PQ = 6cm , QR = 4cm
QS = 8cm.

QS = 8cm.
Given : sides PQ = 6 cm = RS , QR = 4 cm = PS
Diagonal : QS = 8cm

QS = 8cm.
Diagonal : QS = 8cm
P
S
Q
R

QS = 8cm.
Diagonal : QS = 8cm
S R

QS = 8cm.
Diagonal : QS = 8cm
S R
Step 1 : Draw RS = 6 cm

QS = 8cm.
Diagonal : QS = 8cm
S R
Step 2 : For drawing diagonal QS take radius as QS = 8cm and to locate point Q,
draw an arc above RS(or SR) by taking centre as S.

QS = 8cm.
Diagonal : QS = 8cm
S R
Step 2 : For drawing diagonal QS take radius as QS = 8cm and to locate point Q,
draw an arc above RS(or SR) by taking centre as S.
Step 3: Now, take radius QR= 4cm and centre as R draw an arc on previous arc.
Intersection of both arcs is our point Q

QS = 8cm.
Diagonal : QS = 8cm
S
Q
R
Step 4 : Join SQ and QR.

QS = 8cm.
Diagonal : QS = 8cm
S
Q
R
Step 4 : Join SQ and QR.
Step 5 : For locating point P, we take radius SP= 4cm and S as a centre draw an arc.
Step 6 : Now taking PQ = 6cm as radius and draw another arcon previous arc.
Intersection of both arcs is our point P.

QS = 8cm.
Given : sides PQ = 6 cm = RS , QR = 4 cm = PR
Diagonal : QS = 8cm
P
S
Q
R
Step 7 : Join SP and PQ. PQRS is our required parallelogram

Given : Sides WX = 5 cm = XY = YZ = ZX (it’s a rhombus)
Diagonal : XZ = 7 cm
Construct a Rhombus WXYZ where WX = 5 cm and XZ = 7 cm
W
Z
Y
X
7 cm

Given : Sides WX = 5 cm = XY = YZ = ZX (it’s a rhombus)
Diagonal : XZ = 7 cm
Construct a Rhombus WXYZ where WX = 5 cm and XZ = 7 cm
Z X
• Steps:
1)Draw XZ = 7 cm.
2)For locating point W, take radius WZ = 5 cm and
centre as point X draw an arc above XZ.
3)Similarly, take radius WX = 5 cm and Z as a centre
draw another arc on previous arc.Intersection of
both arcs is point W.
W
4)Join WX and WZ.
5)Similarly, for locating Y, take radius take radius XY=
5 cm and centre as point X draw an arc below XZ.
6)Then, take radius YZ= 5 cm and Z as a centre draw
another arc on previous arc.Intersection of both arcs
is point Y.
Y
7)Join XY and YZ.Hence, WXYZ is required Rhombus.

Given : Sides PQ = 4 cm ,QR = 4.5 cm,RS = 5 cm.
Diagonal : PR = 5 cm, QS = 7 cm
Construct a Quadrilateral PQRS where PR = 5 cm, QS = 7 cm,PQ = 4 cm
,QR = 4.5 cm,RS = 5 cm.
P
S Q
7 cm
R

Given : Sides PQ = 4 cm ,QR = 4.5 cm, RS = 5 cm.
Diagonal : PR = 5 cm, QS = 7 cm
Construct a Quadrilateral PQRS where PR = 5 cm, QS = 7 cm,PQ = 4 cm
,QR = 4.5 cm,RS = 5 cm.
• Steps:
1)Draw QS = 7 cm.
2)To locate point R, take QR= 4.5 cm radius, and Q
as a centre, draw arc below QS.
3)Now, take RS = 5 cm radius and S as a centre, draw
another arc on previous arc.Intersection of both arcs
is our point R.
4)Join QR and SR.
5)Similarly, for locating P, take radius take radius PQ
= 4 cm and centre as point Q draw an arc above QS.
6)Then, take radius PR = 5 cm and R as a centre
draw another arc on previous arc.Intersection of
both arcs is point P.
7)Join PS , PQ and PR.Hence ,PQRS is required
Rhombus.
S Q
R
P

Construct a Rhombus ABCD where AC = 5 cm, BD= 7 cm.
A
D
C
B
7 cm
Given :Diagonal AC= 5 cm, BD = 7 cm

Construct a Rhombus ABCD where AC = 5 cm, BD= 7 cm.
Given :Diagonal AC= 5 cm, BD = 7 cm
D B
• Steps:
1)Draw BD= 7 cm.
2)To draw diagonal AC , draw an arc above BD and
below BD by taking B as a centre and radius equal to
half or more than half of DB
3)Now,take D as a centre and radius half or more than
half of BD draw another arc above and below BD.
4)Intersection of both arcs above BD is point P and
below BD is point Q. Join PQ.Its perpendicular bisector
of BD.
5)Perpendicular bisector PQ intersects BD at poInt R.
6)Now , to locate point A and C , take R as a centre
and half of AC = 2.5 cm radius draw arcs on
perpendicular bisector PQ above and below BD.
7)Intersections are point A above and C below BD.
8)Join AB,AD,CD and BC.Hence, ABCD is required
rhombus.
P
Q
R
A
C

Completed till now :
• 4 sides and 1 diagonals (quadrilateral,parallelogram,rhombus)
• 3 sides and 2 diagonals (quadrilateral,rhombus)
Let’s go with 2 sides and 3 angles

o0
degree
180
degree
60 degree
90 degree
120 degree
30 degree 150 degree
165 degree15 degree

Construct a quadrilateral ABCD where AB = 5 cm and BC = 7 cm , angle B =
60 degree , angle A = 90 degree , angle C = 120 degree.
Given : AB = 5 cm , BC = 7 cm
angle B = 60 degree , angle A = 90 degree , angle C = 120 degree
B C
• Steps:
1)Draw BC= 7 cm.
2)Make 60 degree angle on B
3) To locate A,Take B as a centre and
radius AB = 5 cm , draw an arc on ray.
Intersection on ray is point A.
A
4) Make 120 degree on point C.
5) Make an angle A =90 degree
6) A ray from point A intersects
another ray from C point.
7) Intersection of both ray is point D
D
8) ABCD is required quadrilateral

Construct a parallelogram ABCD where AB = 5 cm and BC = 7 cm , ∠B = 85°
Given : AB = 5 cm , BC = 7 cm , ∠B = 85°
B C
A D
85°
85°
360 - (85 + 85)
360 - (170)
190
190 / 2 = 95
95°
95°
7 cm
5cm

Construct a parallelogram ABCD where AB = 5 cm and BC = 7 cm , ∠B = 85°
Given : AB = 5 cm , BC = 7 cm , ∠B = 85°
• Steps:
1)Draw BC = 7 cm.
B C
2) Make 85 degree angle on point B.
3) To locate point A take AB = 5 cm as radius and
B as a centre draw an arc on ray from point
B.Intersection is our point A.
A
4) Make 95 degree angle on point C.
{(360-(85+85))/2}
5) Make 95 degree angle on point A. intersection
of both ray from point A and C gives point D.
D
6) ABCD is required parallelogram

Construct a Rectangle ABCD where AB = 5 cm and BC = 7 cm.
Given : AB = 5 cm , BC = 7 cm.
A
B
D
C
7 cm
5 cm
7 cm
5 cm

Given : AB = 5 cm , BC = 7 cm.
• Steps:
1)Draw BC = 7 cm.
B C
2) Make 90° angle on point B.
3) To locate point A, take point B as a
centre and radius AB = 5 cm , draw an arc
on ray from point B.
4) Intersection is point A
A
6) To locate point D, take point C as a
centre and radius CD = 5 cm , draw an arc
on ray from point C.
5) Make 90° angle on point B.
7) Intersection is point D.
D
8) Join AD.ABCD is required rectangle.

Construct a Quadrilateral ABCD where AB = 5 cm , BC = 7 cm and CD = 5.5
cm .Included angles ∠𝐵 = 60° , ∠𝐶 = 45 ° .
Given : AB = 5 cm , BC = 7 cm , CD = 5.5 cm .
angles ∠𝐵 = 60° , ∠𝐶 = 45 ° .
A
CB
D
7 cm
5.5 cm5 cm
60° 45 °

• Steps:
1)Draw BC = 7 cm.
Given : AB = 5 cm , BC = 7 cm , CD = 5.5 cm .
angles ∠𝐵 = 60° , ∠𝐶 = 45 ° .
B C
5) Make 45° angle on point c.
4) Intersection is point A.
3) To locate point A, take point B as a
centre and radius AB = 5 cm , draw an arc
on ray from point B.A
6) To locate point D, take point C as a
centre and radius CD = 5.5 cm , draw an
arc on ray from point C.
Construct a Quadrilateral ABCD where AB = 5 cm , BC = 7 cm and CD =
5.5 cm .Included angles ∠𝐵 = 60° , ∠𝐶 = 45 ° .
7) Intersection is point D.
D
8) Join AD. ABCD is required
quadrilateral.

By Some special cases
Diagonals Equal Bisects 90°

By Some special cases
Sides Opposite
Equal
Opposite
Parallel
Adjacent
Equal

Construct a Kite ABCD where AB = 5 cm , BC = 7 cm and BD= 6.5 cm.
Given : AB = 5 cm , BC = 7 cm , BD = 6.5 cm .
A
B
C
D

Given : AB = 5 cm , BC = 7 cm , BD = 6.5 cm .
Construct a Kite ABCD where AB = 5 cm , BC = 7 cm and BD= 6.5 cm.
D B
• Steps:
1)Draw BD = 6.5cm.
2)Take more than half of BD as radius and
B and D as centres draw perpendicular
bisector of BD.
3) Now take B as a centre and AB = 5 cm,
draw arc on bisector above
BD.Intersection is point A.
A
4)Now take B as a centre and BC= 7 cm,
draw arc on bisector below BD
.Intersection is point C.
C
5) Join AD,AB,BC,CD.ABCD is required
kite.

Viii practical geometry Rujul vyas

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