Detailed Construction Steps (Beginner Level)
- Draw the given triangle:
- First, draw a line segment \(BC = 5\,\text{cm}\).
- With \(B\) as centre and radius \(6\,\text{cm}\), draw an arc.
- With \(C\) as centre and radius \(6\,\text{cm}\), draw another arc cutting the first one at \(A\).
- Join \(AB\) and \(AC\). Now triangle \(ABC\) is an isosceles triangle with \(AB=AC=6\,\text{cm}\) and \(BC=5\,\text{cm}\).
- Understand the requirement:
We need another triangle \(PQR\) similar to \(ABC\), but bigger in size. In the similar triangle, side \(PQ\) must be \(8\,\text{cm}\), while in the original triangle, side \(AB = 6\,\text{cm}\).
- Find the scale factor:
- Scale factor \(k = \dfrac{PQ}{AB} = \dfrac{8}{6} = \dfrac{4}{3}\).
- This means every side of the new triangle will be \(\tfrac{4}{3}\) times the corresponding side of the old triangle.
- Calculate the new sides:
- \(PR = AC \times k = 6 \times \dfrac{4}{3} = 8\,\text{cm}\).
- \(QR = BC \times k = 5 \times \dfrac{4}{3} = \dfrac{20}{3}\,\text{cm} \approx 6.67\,\text{cm}\).
- Construct the similar triangle:
- On \(AB\), use the standard similarity method (divide a ray into equal parts, use parallel lines) to enlarge \(AB\) into \(PQ=8\,\text{cm}\).
- Draw parallels through \(Q\) and use the same scale factor to get \(R\).
- Join \(P, Q, R\) to form \(\triangle PQR\).
Justification
By construction, the ratio of corresponding sides is the same:
\(\dfrac{PQ}{AB} = \dfrac{PR}{AC} = \dfrac{QR}{BC} = \dfrac{4}{3}\).
Since two triangles have their corresponding sides in the same ratio and their included angles are equal, by the AA similarity rule, \(\triangle PQR \sim \triangle ABC\).
Thus the construction is correct.