187
Rahim and Nadeem are two friends whose plots are adjacent to each other. Rahim's son made a drawing of the plots with necessary details. It is decided that Rahim will fence the triangular plot ABC and Nadeem will fence along the sides AF, FE and BE. Observe the diagram carefully and answer the following questions : (Use $\sqrt{2} = 1.41$ and $\sqrt{3} = 1.73$)
(i) Find length BC.
(ii) Find length AG.
(iii) (a) Calculate perimeter of $\Delta ABC$.
OR
(iii) (b) Calculate length of $(AF + FE + EB)$.
(i) Find length BC.
(ii) Find length AG.
(iii) (a) Calculate perimeter of $\Delta ABC$.
OR
(iii) (b) Calculate length of $(AF + FE + EB)$.
Show SolutionHide Solution↓
(i) $\cos 45^\circ = \frac{50}{BC} \Rightarrow BC = 50 \times 1.41 = 70.5\text{ m}$ [$1$ mark]
(ii) $\tan 45^\circ = \frac{38}{AG} \Rightarrow AG = 38\text{ m}$ [$1$ mark]
(iii) (a) $\sin 60^\circ = \frac{78}{AC} \Rightarrow AC = 89.96\text{ m}$
Perimeter of $\Delta ABC = 70.5 + 89.96 + 38 + 50 + 40 = 288.46\text{ m}$ [$1 + 1$ mark]
OR
(iii) (b) $\cos 30^\circ = \frac{40}{BE} \Rightarrow BE = 46.13\text{ m}$
$AF + FE + EB = 53.58 + 53 + 46.13 = 152.71\text{ m}$ [$1 + 1$ mark]
(ii) $\tan 45^\circ = \frac{38}{AG} \Rightarrow AG = 38\text{ m}$ [$1$ mark]
(iii) (a) $\sin 60^\circ = \frac{78}{AC} \Rightarrow AC = 89.96\text{ m}$
Perimeter of $\Delta ABC = 70.5 + 89.96 + 38 + 50 + 40 = 288.46\text{ m}$ [$1 + 1$ mark]
OR
(iii) (b) $\cos 30^\circ = \frac{40}{BE} \Rightarrow BE = 46.13\text{ m}$
$AF + FE + EB = 53.58 + 53 + 46.13 = 152.71\text{ m}$ [$1 + 1$ mark]