HKDSE-MATH - 2015 - Paper 1 -

Ans: (a) $75^\circ$ (b) $(5\pi + 24) \text{ cm}$

1. Let $\theta

Ans: C
By applying the Pythagorus Theorem to $\Delta CDN$, w

Ans: B

Let $x\text{ cm}$ be the radius of the upper base of the

Ans: D
Note that $ABCD$ is a parallelogram, then $AD//BC$. H

Ans: (a) $768\pi \text{ cm}^3$ (b) No

1. Let $x\text{ cm}^3$ be

Ans: C
Add a point $E$ on $CD$ such that $BE\perp CD$.

Since $A

Ans: A

I is true. Let $\theta$ be the angle of the sector $OAB$

Ans: C

Since $ABCD$, $CDEF$ and $EFGH$ are squares, then we h

Ans: (a) $1296\text{ cm}^3$ (b) $384\text{ cm}^2$

1. The tota

Ans: B
Let $x \text{ cm}$ be the length of $AD$. Then we have

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Ans: C
Let $r\text{ cm}$ and $h\text{ cm}$ be the base radius

Ans: D

Since $\Delta CEH \sim \Delta CBG$, then we have

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Ans: (a) $4096\pi\text{ cm}^3$ (b) $48\text{ cm}$ (c) No

1. Th

Ans: D
The base of the triangle base

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Ans: A
Note that $ABCD$ is a parallelogram, then $AD = BC$. He

Ans: B
Since $OC \perp AD$, then $AE = DE$. Hence, we have

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Ans: (a) $288\pi\text{ cm}^3$ (b) $180\pi\text{ cm}^2$

1. Le

Ans: D

The height of the lateral face

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Ans: D
Since $ABCD$ is a parallelogram, $BC = AD$ (opp. sides

Ans: A
By applying the Pythagoras Theorem to $\Delta ABD$, w