Question

Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are `5/7` of the corresponding sides of the first triangle.

Answer 1

\[ \begin{array}{r l} \textbf{Given:} & \text{A triangle } ABC \text{ with } BC = 7 \text{ cm, } AB = 5 \text{ cm and } AC = 6 \text{ cm, to be reduced by the scale factor } \dfrac{5}{7}. \\[4pt] \textbf{To Find:} & \text{The construction of a triangle similar to } ABC \text{ with each side } \dfrac{5}{7} \text{ of the corresponding side, and the lengths of its sides.} \\[4pt] \textbf{Solution:} & \text{Step 1: Draw } BC = 7 \text{ cm; with centre } B \text{ and radius } 5 \text{ cm and with centre } C \text{ and radius } 6 \text{ cm draw arcs meeting at } A, \text{ and join } AB \text{ and } AC. \\[4pt] & \text{Step 2: Draw a ray } BX \text{ making an acute angle with } BC \text{ on the side opposite to } A. \\[4pt] & \text{Step 3: Since the greater of } 5 \text{ and } 7 \text{ is } 7, \text{ mark seven points } P, Q, R, S, T, U, V \text{ on } BX \text{ such that } BP = PQ = QR = RS = ST = TU = UV. \\[4pt] & \text{Step 4: Join } VC \text{ and through } T, \text{ the fifth point, draw a line parallel to } VC, \text{ meeting } BC \text{ at } C'. \\[4pt] & \text{Step 5: Through } C' \text{ draw a line parallel to } CA, \text{ meeting } BA \text{ at } A'; \text{ then } A'BC' \text{ is the required triangle.} \\[4pt] & \text{Justification: in triangle } BVC, \text{ since } TC' \text{ is parallel to } VC, \text{ the Basic Proportionality Theorem gives } \dfrac{BC'}{BC} = \dfrac{BT}{BV} = \dfrac{5}{7}. \\[4pt] & \text{Since } C'A' \text{ is parallel to } CA, \text{ the triangles } A'BC' \text{ and } ABC \text{ are similar by the AA criterion, so their sides are proportional:} \\[4pt] & \begin{aligned} \frac{A'B}{AB} = \frac{BC'}{BC} = \frac{A'C'}{AC} &= \frac{5}{7} \\[4pt] BC' &= \frac{5}{7} \times 7 \\[4pt] &= 5 \\[4pt] &= 5.00 \\[4pt] A'B &= \frac{5}{7} \times 5 \\[4pt] &= \frac{25}{7} \\[4pt] &\approx 3.57 \\[4pt] A'C' &= \frac{5}{7} \times 6 \\[4pt] &= \frac{30}{7} \\[4pt] &\approx 4.29 \end{aligned} \\[4pt] \textbf{Answer:} & \text{The required triangle } A'BC' \text{ has } BC' = 5 \text{ cm} = 5.00 \text{ cm, } A'B = \dfrac{25}{7} \text{ cm} \approx 3.57 \text{ cm and } A'C' = \dfrac{30}{7} \text{ cm} \approx 4.29 \text{ cm.} \end{array} \]