Question

\(\triangle\)ABC \(\sim\) \(\triangle\)PQR. In \(\triangle\)ABC, AB = 3.6 cm, BC = 4 cm and AC = 4.2 cm. The corresponding sides of \(\triangle\)ABC and \(\triangle\)PQR are in the ratio 2 : 3, construct \(\triangle\)ABC and \(\triangle\)PQR.

Hint:

When constructing similar triangles with a scale factor \(m/n\), draw a ray and mark \(max(m,n)\) points. Join the \(n\)-th point to the vertex if the given triangle's ratio is \(n\). Then draw a parallel line from the \(m\)-th point. This works for both enlargement (\(m > n\)) and reduction (\(m < n\)).

Answer 1

Step 1: Understanding the Concept: 
We first need to construct \(\triangle\)ABC from its given side lengths. Then, we construct \(\triangle\)PQR, which is similar to \(\triangle\)ABC, using the given ratio of corresponding sides. Since the ratio is 2:3, \(\triangle\)PQR will be larger than \(\triangle\)ABC. The scale factor for enlargement is \(\frac{3}{2}\). 
 

Step 2: Key Formula or Approach: 
The ratio of corresponding sides is given by: \[ \frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR} = \frac{2}{3} \] We can calculate the side lengths of \(\triangle\)PQR: \[\begin{array}{rl} \bullet & \text{\(PQ = \frac{3}{2} \times AB = \frac{3}{2} \times 3.6 = 5.4\) cm} \\ \bullet & \text{\(QR = \frac{3}{2} \times BC = \frac{3}{2} \times 4 = 6\) cm} \\ \bullet & \text{\(PR = \frac{3}{2} \times AC = \frac{3}{2} \times 4.2 = 6.3\) cm} \\ \end{array}\] We can construct both triangles separately, or use a combined construction method. The combined method is more elegant. 
 

Step 3: Detailed Explanation (Construction Steps): 
Part I: Constructing \(\triangle\)ABC \[\begin{array}{rl} 1. & \text{Draw a line segment BC of length 4 cm.} \\ 2. & \text{With B as the centre and radius 3.6 cm, draw an arc.} \\ 3. & \text{With C as the centre and radius 4.2 cm, draw another arc intersecting the first arc at point A.} \\ 4. & \text{Join AB and AC. \(\triangle\)ABC is the required triangle.} \\ \end{array}\]

Part II: Constructing \(\triangle\)PQR (similar to \(\triangle\)ABC) \[\begin{array}{rl} 1. & \text{From point B of \(\triangle\)ABC, draw a ray BX making an acute angle with the side BC, on the side opposite to vertex A.} \\ 2. & \text{On ray BX, mark 3 points (the larger number in the ratio 2:3) \(B_1, B_2, B_3\) such that \(BB_1 = B_1B_2 = B_2B_3\).} \\ 3. & \text{Join \(B_2\) (the smaller number in the ratio) to point C.} \\ 4. & \text{From point \(B_3\), draw a line parallel to \(B_2C\), which intersects the extended line segment BC at point R.} \\ 5. & \text{From point R, draw a line parallel to side AC, which intersects the extended line segment BA at point P.} \\ 6. & \text{\(\triangle\)PBR is the required triangle \(\triangle\)PQR (with B corresponding to Q).} \\ \end{array}\] This construction creates \(\triangle\)PBR which is similar to \(\triangle\)ABC and its sides are \(\frac{3}{2}\) times the sides of \(\triangle\)ABC. 
 

Step 4: Final Answer: 
The triangles \(\triangle\)ABC and \(\triangle\)PQR are constructed as per the steps above.