construct a triangle shadow similar to the given triangle ABC with its equal to 5by3 of the corresponding sides of the triangle ABC
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Solution
Step I : Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.
Step II : From B cut off 5 ares
B
1
,B
2
,B
3
B
4
andB−5 on BX so that
BB
1
=B
1
B
2
=B
2
B
3
=B
3
B
4
=B
4
B
5
Step III : Join B
3
to C and dra a line through B
5
parallel to B
3
C. intersecting the extended line segment BC at C.
Step IV : Draw a line through C parallel to CA intersecting the extended line segment BA at A
′
Then A
′
BC
′
is the required triangle.
justification :
Note that ΔABC∼Δ
′
BC
′
( Since AC∣∣A
′
C)
therefore ,
A
′
B
AB
=
A
′
C
′
AC
=
BC
′
BC
But,
BC
′
BC
=
BB
5
BB
3
=
5
3
,
Therefore .
AB
A
′
B
=
AC
A
′
C
′
=
BC
BC
′
=
3
5
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