Question

Draw a line segment PQ = 7.5 cm. Divide it in the ratio 3: 1.

Answer 1

\[ \begin{array}{r l} \textbf{Given:} & \text{A line segment } PQ = 7.5 \text{ cm.} \\[4pt] \textbf{To Find:} & \text{The point } R \text{ on } PQ \text{ that divides it in the ratio } 3 : 1, \text{ and the two parts so obtained.} \\[4pt] \textbf{Solution:} & \text{Step 1: Draw the line segment } PQ = 7.5 \text{ cm.} \\[4pt] & \text{Step 2: Draw a ray } PX \text{ making an acute angle with } PQ. \\[4pt] & \text{Step 3: Since } 3 + 1 = 4, \text{ mark four points } A, B, C, D \text{ on } PX \text{ such that } PA = AB = BC = CD. \\[4pt] & \text{Step 4: Join } DQ. \\[4pt] & \text{Step 5: Through } C \text{ draw a line parallel to } DQ, \text{ meeting } PQ \text{ at } R; \text{ then } R \text{ is the required point.} \\[4pt] & \text{Justification: in triangle } PDQ, \text{ since } CR \text{ is parallel to } DQ, \text{ the Basic Proportionality Theorem gives:} \\[4pt] & \begin{aligned} \frac{PR}{RQ} &= \frac{PC}{CD} \\[4pt] &= \frac{3}{1} \\[4pt] PR &= \frac{3}{4} \times 7.5 \\[4pt] &= \frac{45}{8} \\[4pt] &= 5.625 \\[4pt] &\approx 5.63 \\[4pt] RQ &= \frac{1}{4} \times 7.5 \\[4pt] &= \frac{15}{8} \\[4pt] &= 1.875 \\[4pt] &\approx 1.88 \end{aligned} \\[4pt] \textbf{Answer:} & \text{The point } R \text{ divides } PQ \text{ with } PR = \dfrac{45}{8} \text{ cm} \approx 5.63 \text{ cm and } RQ = \dfrac{15}{8} \text{ cm} \approx 1.88 \text{ cm.} \end{array} \]