Construct a triangle with sides $ 5 \mathrm{~cm}, 6 \mathrm{~cm} $ and $ 7 \mathrm{~cm} $ and then another triangle whose sides are $ \frac{5}{7} $ of the corresponding sides of the first triangle.
Given:
A triangle with sides \( 5 \mathrm{~cm}, 6 \mathrm{~cm} \) and \( 7 \mathrm{~cm} \).
To do:
We have to construct a triangle with sides \( 5 \mathrm{~cm}, 6 \mathrm{~cm} \) and \( 7 \mathrm{~cm} \) and then another triangle whose sides are \( \frac{5}{7} \) of the corresponding sides of the first triangle.
Solution:
Steps of construction:
(i) Draw a line segment $BC = 5\ cm$.
(ii) With centre $B$ and radius $6\ cm$ and with centre $C$ and radius $7\ cm$, draw arcs intersecting each other at $A$.
(iii) Join $AB$ and $AC$.
$ABC$ is the required triangle.
(iv) Draw a ray $BX$ making an acute angle with $BC$ and cut off seven equal parts making $BB_1 = B_1B_2 = B_2B_3 = B_3B_4 = B_4B_5 = B_5B_6 = B_6B_7$.
(v) Join $B_7$ and $C$.
(vi) From $B_5$, draw $B_5C'$ parallel to $B_7C$ and $C’A’$ parallel to $CA$.
$A’BC’$ is the required triangle.
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