Question
Draw a triangle $A B C$ in which $AB =5 cm$, $B C=6 cm$ and $\angle A B C=60^{\circ}$. Then construct a triangle whose sides are $\frac{5}{7}$ times the corresponding sides of $\triangle A B C$.
✓
Answer
1. Steps of construction for triangle $A B C$ :
(1) Draw a line segment $B C=6 cm$
(2) From point $B$, draw a ray making an angle of $60^{\circ}$ with $B C$.
(3) Now with $B$ as center and radius 5 cm draw an are cutting the ray at point $A$.
(4) Join $A C$, to form $\triangle A B C$.
Now steps of construction for similar triangle,
(1) Draw a ray $B X$ making an acute angle opposite to vertex $A$.
(2) Mark 7 points $B_1, B_2, B_3, B_4, B_5, B_6, B_7$ at equal distance from each other on $B X$ such that
$
B B_1=B_1 B_2=B_2 B_3=B_3 B_4=B_4 B_5=B_5 B_6=B_6 B_7
$
(3) Join the point $C$ and $B_7$ and draw $B_5 C$ 'parallel to $B_7 C$.
(4) Draw $C$ 'A' parallel to $C A$.
Hence $A^{\prime} B^{\prime} C$ is the required triangle as shown below.
(1) Draw a line segment $B C=6 cm$
(2) From point $B$, draw a ray making an angle of $60^{\circ}$ with $B C$.
(3) Now with $B$ as center and radius 5 cm draw an are cutting the ray at point $A$.
(4) Join $A C$, to form $\triangle A B C$.
Now steps of construction for similar triangle,
(1) Draw a ray $B X$ making an acute angle opposite to vertex $A$.
(2) Mark 7 points $B_1, B_2, B_3, B_4, B_5, B_6, B_7$ at equal distance from each other on $B X$ such that
$
B B_1=B_1 B_2=B_2 B_3=B_3 B_4=B_4 B_5=B_5 B_6=B_6 B_7
$
(3) Join the point $C$ and $B_7$ and draw $B_5 C$ 'parallel to $B_7 C$.
(4) Draw $C$ 'A' parallel to $C A$.
Hence $A^{\prime} B^{\prime} C$ is the required triangle as shown below.
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