HKCEE - 2006 - Paper 1 -

Ans: (a) $\dfrac{305}{9}\text{ cm}$ (b) (i) (1) $63.4\tex

Ans: B
Note that the projection of $BD$ on the plane $CDEF$ is

Ans: C

Note that each faces of the tetrahedron are equilater

Ans: (a) $10\sqrt{2}\text{ cm}^2$, $210\sqrt{2}\text{ c

Ans: D
Note that the line of intersection of planes $ABC$ and

Ans: (a) $148\text{ m}$ (b) (i) $37.8^\circ$ (ii) $\angle A

Ans: D
Note that $CF=DE$. Therefore, the angle of inclinati

Ans: (a) (i) $22.3\text{ m}$ (ii) $48.5^\circ$ (iii) $337\

Ans: B

Let $N$ be the mid-point of $HG$. Then the angle betwee

Ans: (a) $18.7\text{ cm}$ (b) (i) $26.8\text{ cm}$ (ii) $97

Ans: D
Note that $BC$ is the line of intersection of planes $A

Ans: D

Join $NC$. Let $2x$ be the length of the cube, then $MB=

Ans: (a) (i) $25.1\text{ cm}$ (ii) $41.4^\circ$ (iii) $15.

Ans: C
Note that the projection of $EB$ on the plane $ABCD$ is

Ans: B
Consider $\Delta BFG$,

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\tan y &