Question

Mark (✓) against the correct answer:
The area of a rhombus is 120 cm² and one of its diagonals is 24 cm. Each side of the rhombus is
(a) 10 cm
(b) 13 cm
(c) 12 cm
(d) 15 cm

Answer 1

(b) 13 cm



$$\begin{gathered} \mathrm{Let}ABCD\mathrm{be}\mathrm{a}\mathrm{rhombus}\mathrm{whose}\mathrm{diagonals}AC\mathrm{and}BD\mathrm{intersect}\mathrm{at}\mathrm{a}\mathrm{point}\mathrm{O}. \\ \mathrm{Let}\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{diagonal}ACbe24\mathrm{cm}. \\ A\mathrm{rea}\mathrm{of}\mathrm{the}\mathrm{rhombus}=\left(\frac{1}{2}\times AC\times BD\right){\mathrm{cm}}^2 \\ \mathrm{But}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{rohmbus}is120{\mathrm{cm}}^2\left(\mathrm{given}\right) \\ \therefore \frac{1}{2}\times AC\mathit{\times }BD\mathit{}=120 \\ \mathrm{or}\frac{1}{2}\times 24\times BD=120 \\ \mathrm{or}12\times BD=120 \\ \mathrm{or}BD=\frac{120}{12}=10\mathrm{cm} \\ \mathrm{OB}=\frac{\mathrm{BD}}{2}=\frac{10}{2}=5\mathrm{cm} \end{gathered}$$

$$\begin{gathered} \mathrm{And}OA=\frac{AC}{2}=\frac{24}{2}=12\mathrm{cm} \\ \mathrm{Now},\mathrm{in}\mathrm{right}\mathrm{triangle}AOB: \\ {\left(AB\right)}^2={\left(OA\right)}^2+{\left(OB\right)}^2 \\ \mathrm{or}{\left(AB\right)}^2={12}^2+5^2 \\ =144+25 \\ =169 \\ \mathrm{or}AB=\sqrt{169}=13\mathrm{cm} \\ \mathrm{Therefore},\mathrm{each}\mathrm{side}\mathrm{of}\mathrm{the}\mathrm{rhombus}\mathrm{is}13\mathrm{cm}. \end{gathered}$$