Construct a triangle in which AB= 4 cms . BC = 5 cms. AC = 6 cms. Then construct another
triangle whose sides are 2/3 times the corresponding sides of triangle ABC.
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Answer:
Steps:
1. draw base AB=4cm
2. with A as center, with a scale of 5cm as radius draw an arc and with B as center, with a scale of 6cm as radius draw an arc.
3. let C be the point of intersection of the above 2 lines.
4. draw a ray AX making an acute angle with line AB on the opposite side if the vertex C.
5. mark 3 points A1,A2,A3 such tha A1A2=A2A3=A3A1
6. join A3B and draw a parallel line B
′
A2
7. draw a line parallel to BC to intersect AC at C
′
Thus AB
′
C
′
is a required triangle.
Justification;
By construction,
AB
AB
′
=
AA3
AA2
=
3
2
also B
′
C
′
∥BC
∴∠AB
′
C
′
=∠ABC
In △AB
′
C
′
,△ABC
∠A=∠A
∠AB
′
C
′
=∠ABC
△AB
′
C
′
∼△ABC
⇒
AB
AB
′
=
AC
AC
′
=
BC
B
′
C
′
∴
AB
AB
′
=
AC
AC
′
=
BC
B
′
C
′
=
3
2
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