CONSTRUCTIONS
A′B
AB
=
BC′
BC
=
A′C′
AC
=
3
2
B
A
C
8 cm
4 cm
Rough Figure
J
K
R
S
T
U
V
W
Ex-13.1 (Q.4)
C
B
8 cm
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B1
B2
B3
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15
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A
P
Q
D
C'
A'
4 cm
M
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Draw seg BC = 8cm
Let us draw perpendicular bisector of BC
B as centre radius more than half of BC, draw acrs up and down
C as centre and with the same radius cut the previously drawn arcs
Draw line PQ
M as centre radius = 4cm,
draw an arc
Draw AB and AC
Considering any suitable radius, draw 3 arcs on ray BD
Draw ray BD
Draw B2C
B2 as centre and any suitable radius, draw an arc intersecting BD and B2C at points J & K respectively
B3 as centre and with the same radius, draw an arc intersecting BD at point R
Now, consider radius = JK
R as centre, cut an arc and mark that point S
Draw B3S intersecting BC at C'
C as centre and any suitable radius, draw an arc intersecting BC and AC at points T & U respectively
C' as centre and with the same radius, drawn an arc intersecting BC at point V
Now, consider radius = TU
V as centre, cut an arc and mark that point W
Draw C'W intersecting AB at A'
A′
C′
Any point on the line as centre and radius = 8 cm, draw an arc
C as the centre and same radius, draw another arc
M
Justification :
∴
A′C′ ║ AC
[By construction]
ΔA′BC′ ~ ΔABC
[AA Similarity]
A′B
AB
=
BC′
BC
=
A′C′
AC
[corresponding sides of
similar triangles]
But,
BC′
BC
=
BB2
=
3
2
∴
BC′
BC
=
3
2
∴
A′B
AB
=
BC′
BC
=
A′C′
AC
=
3
2
C
B
8 cm
B1
B2
B3
A
P
Q
D
C'
A'
4 cm
M
[since ΔBB2C ~ΔBB3Cl]