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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.

Answer :

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.


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.

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