Chapter: Similarity· Mathematics Part 2 (Geometry)
Answer: MN = 15
\(9 \times A(\Delta PQR) = 16 \times A(\Delta LMN)\) \(\therefore \dfrac{A(\Delta PQR)}{A(\Delta LMN)} = \dfrac{16}{9}\).
\(\Delta PQR \sim \Delta LMN\), so by the theorem of areas of similar triangles, \(\dfrac{A(\Delta PQR)}{A(\Delta LMN)} = \dfrac{QR^2}{MN^2}\).
\(\therefore \dfrac{QR^2}{MN^2} = \dfrac{16}{9}\) \(\therefore \dfrac{QR}{MN} = \dfrac{4}{3}\) (taking square roots)
\(\therefore \dfrac{20}{MN} = \dfrac{4}{3}\) \(\therefore MN = \dfrac{20 \times 3}{4} = 15\).