Question

Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are $\frac{7}{5}$ of the corresponding sides of the first triangle.

Answer 1

Steps of Construction:
Step I: A triangle ABC with sides 5 cm, 6 cm and 7 cm is drawn.
Step II: A ray BX making an acute angle with BC is drawn opposite to vertex A.
Step III: 7 points, $B_1,B_2,B_3,B_4,B_5,B_{6}and{B}_7$ are marked on BX.
Step IV; Point $B_5$ is joined with C to draw $B_{5}C$.
Step V: $B_{7}C$' is drawn parallel to B5C and C'A' is parallel to CA.
Thus A'BC' is the required triangle.
Justification
$\Delta A{B}^{\prime }{C}^{\prime }\Delta ABC$ by AAA similarity condition
$\therefore \frac{AB}{A^{\prime }B}=\frac{AC}{A^{\prime }{C}^{\prime }}=\frac{BC}{B{C}^{\prime }}$
$and\frac{BC}{B{C}^{\prime }}=\frac{BB5}{BB7}=\frac{5}{7}$
$\therefore \frac{A^{\prime }B}{AB}=\frac{A^{\prime }{C}^{\prime }}{AC}=\frac{B{C}^{\prime }}{BC}=\frac{7}{5}$