3 Tutor System
Starting just at 265/hour

# Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are $$\frac{5}3$$times the corresponding sides of the given triangle.

Steps of Construction the figure by the given data:
Step 1. Construct a $$\triangle$$ ABC, with BC = 4 cm, CA = 3 cm and ∠BCA = 90°.

Step 2. By making an acute angle with BC, draw a ray BX .

Step 3. Mark five points $$B_1, B_2, B_3, B_4$$and $$B_5$$on BX, such that $$BB_1 = B_1B_2 = B_2B_3 = B_3B_4 = B_4B_5$$ and join $$B_3C$$.

Step 6. Through $$B_5$$, draw $$B_5C’$$ parallel to $$B_3C$$ intersecting BC produced at C’.

Step 7. Through C’, draw C’A’ parallel to CA intersecting AB produced at A’.

Thus, $$\triangle$$ A’BC’ is the required right triangle.

JUSTIFICATION:

In $$ΔA’BC’ and ΔABC$$,

$$∠ABC=∠ A’BC’$$(common)

$$∠ACB=∠ A’C’B$$ (corresponding angles)

$$∴ ΔABC∼ΔA’BC’$$ (By AA similarity)

$$∴ \frac{AB}{A’B}=\frac{AC}{A’C’}=\frac{BC}{BC’}$$

But, $$\frac{BC}{BC’}=\frac{BB_3}{BB_5}=\frac{3}5 ∴ \frac{BC’}{BC}=\frac{5}3$$

$$⇒ \frac{A’B}{AB}=\frac{A’C’}{AC}=\frac{BC’}{BC}=\frac{5}3 .$$

$$\therefore$$ Sides of the new triangle formed are $$\frac{5}{3}$$ times the corresponding sides of the first triangle.