Transcript Power Point Prepared by - e-CTLT

Power Point Prepared
By
N K Srivastava
KV NTPC Shaktinagar
Class : VII
Topic : Practical Geometry
Construction of a line parallel to a
given line , through a point not on
the line.
Step 1 : Take a line ‘l’ and a point ‘A’
not on ‘l’
.A
l
Step 2: Take any point B on ‘l’ and join
B to A .
A
B
l
Step 3 : With B as centre and a
convenient radius , draw an arc cutting l
at C and BA at D.
A
D
B
C
l
Step 4 : Now with A as centre and the same
radius as in Step 3,draw an arc EF cutting AB at
G. Place the pointed tip of the compass at C and
adjust the opening so that the pencil tip is at D
F
A
E
G
D
B
C
l
Step 5 : With the same opening as in step 4 and
with G as centre draw an arc cutting the arc EF
at H.
F
A
H
E
G
D
B
C
l
Step 6 : Now , join AH to draw a line m . This line
m is the required parallel line parallel to line ‘l’
and passing through A .
F
A
H
E
m
G
D
B
C
l
Construction Of Triangles
1.Construction of a triangle when the
measurements of all the three sides are given.
( SSS Criterion )
Example : Construct a triangle ABC , given that
AB = 5 cm , BC = 6 cm and AC = 7 cm
Solution
Step 1 : First , we draw a rough sketch
with given measures.
A
B
6 cm
C
Step 2 : Draw a line segment BC of
length 6 cm .
B
6 cm
C
Step 3 : From point B , point A is at
a distance of 5 cm . So , with B as
centre , draw an arc of radius 5 cm.
B
6 cm
C
Step 4 : From point C , point A is at
a distance of 7 cm . So , with C as
centre , draw an arc of radius 7 cm.
B
6 cm
C
Step 5 : Mark the point of intersection
of arcs as A . Join AB and AC .Δ ABC is
the required triangle.
A
B
6 cm
C
2.Construction of a triangle when the lengths
of two sides and the measurement of angle
between them are given.
( SAS Criterion )
Example : Construct a triangle PQR , given that
PQ = 3 cm , QR = 5.5 cm and ˂ PQR = 60⁰
Solution
Step 1 : First , we draw a rough sketch
with given measures.
P
60⁰
Q
5.5 cm
R
Step 2 : Draw a line segment QR of
length 5.5 cm .
Q
5.5 cm
R
Step 3 : AT point Q , Draw QX
making 60⁰ with QR.
X
60⁰
Q
5.5 cm
R
Step 4 : With Q as centre draw an
arc of radius 3 cm. IT cuts QX at
point P.
X
P
60⁰
Q
5.5 cm
R
Step 5 : Join PR .ΔPQR is the required
triangle.
X
P
60⁰
Q
5.5 cm
R
3.Construction of a triangle when the
measures of two of its angles and the length
of side between them is given.
( ASA Criterion )
Example : Construct a triangle PQR , given that
XY = 6 cm ,m YXZ = 30⁰ and m XYZ = 100⁰
Solution
Step 1 : First , we draw a rough sketch
with given measures.Z
X
100⁰
30⁰
6 cm
Y
Step 2 : Draw a line segment XY of
length 6 cm .
X
6 cm
Y
Step 3 : AT point X , Draw XP making
30⁰ with XY and at point Y draw YQ
making an angle 100⁰ with YX.
P
Q
X
100⁰
30⁰
6 cm
Y
Step 4 : Z has to lie on both the rays XP and YQ.
So the point of Intersection of these two rays is
Z. Δ XYZ is the required triangle.
Q
Z
X
100⁰
30⁰
6 cm
Y
P
4.Construction of a Right -angled triangle
when the length of its one leg and its
hypotenuse are given.
( RHS Criterion )
Example : Construct a triangle LMN , right –
angled at M , given that LN = 5 cm and MN = 3
cm.
Solution
Step 1 : First , we draw a rough sketch
with given measures.
L
90⁰
M
3 cm
N
Step 2 : Draw a line segment MN of
length 3 cm .
M
3 cm
N
Step 3 : AT point M , Draw MX
perpendicular to MN.
X
90⁰
M
3 cm
N
Step 4 : With N as centre , draw an
arc of radius 5 cm.
X
L
90⁰
M
3 cm
N
Step 5 : Now join LN.Δ MNL is the
required right – angled triangle.
X
L
90⁰
M
3 cm
N
HOME WORK
1.
2.
3.
4.
Take any line m and draw a line perpendicular to passing through it and not on it.
Construct a triangle ABC in which AB = 6 cm , BC = 7 cm and AC = 4 cm
Construct a triangle PQR in which PQ = 8 cm Q = 60⁰ and QR = 6 cm
Construct a triangle XYZ in which YZ = 7 cm ,
Y = 30⁰ and
Z = 100⁰ .