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# 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.

Steps of construction by the given data are:
Step 1. Draw a $$\triangle$$ ABC with sides as AB=5 cm, BC= 7 cm, AC= 6cm.

Step 2. At point B, draw an acute angle CBX below BC.

Step 4. Join $$B_5$$ and C.

Step 5.Draw $$B_7C'$$ parallel to $$B_5C$$, where C` is a point on extended line BC.

Step 6. Draw A'C' parallel to AC, where A' is a point on extended line BA.

Hence,A’BC’ is the required triangle.

JUSTIFICATION:

In $$\triangle$$ ABC and $$\triangle$$ A’BC’,

AC || A’C’

$$\frac{AB}{A’B}=\frac{BC}{BC’}$$
(By the basic Proportionality Theorem)...(i)

In $$ΔBB_5C$$and $$ΔBB_7C’$$,

$$B_5C || B_7C’$$

$$\frac{BC}{BC’}=\frac{BB_5}{BB_7}=\frac{5}7$$ (By the basic Proportionality Theorem)...(ii)

Equating (i) and (ii) $$\frac{AB}{A’B}=\frac{BB_5}{BB_7} \Rightarrow \frac{AB}{A’B}=\frac{5}7$$

$$\therefore A’B=\frac{7}5AB$$

$$\therefore$$ Sides of new triangle is $$\frac{7}5$$times the corresponding sides of the first triangle.