(a) −(r)
Let the man starts from A and goes 10 m due east at B and then 20 m due north at C.
Then, in right-angled triangle ABC, we have:
AB2 + BC2 = AC2
⟹ = \(\sqrt{10^2+20^2}\) = \(\sqrt{100+200}\) = 10√3
Hence, the man is 10 √3m away from the starting points
(b) −(q)
Let the triangle be ABC with altitude AD.
In right-angled triangle ABC, we have:
(c) – (p)
Area of an equilateral triangle with side a
(d) – (s)
Let the rectangle be ABCD with diagonals AC and BD.
In right-angled triangle ABC, we have:
AC2 = AB2 + BC2 = 82 + 62 = 64 + 36
⇒ AC = \(\sqrt{100}
\) = 10 m