HKDSE-MATH - 2018 - Paper 2 -

Ans: B
Consider $\Delta BDE$.

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BD �

Ans: B
Join $AC$ and $AD$.

Consider $\Delta ABC$.

$\begin{a

Ans: (a) $25^\circ$ (b) No

1. Consider quadrilateral $OAED$,

Ans: B
In $\Delta OBD$ and $\Delta OAC$,

$\begin{array}{rc

Ans: B
Consider cyclic quadrilateral $ACDE$.

$\begin{arr

Ans: (b) (i) $1547\pi\text{ cm}$ (ii) Yes

1. In $\Delta UTV$ an

Ans: C
Join $AC$ and $AD$.

Consider $\Delta ABC$. Since $\an

Ans: A
Add a point $F$ on the circle $CDE$. Join $DF$ and $EF$.

Ans: A

$\begin{array}{rcll}
\angle ACD & = & 180

Ans: D

Join $AD$.

In $\Delta CDE$,

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\a

Ans: B

Join $OA$.

$\begin{array}{rcll}
\angle CAB & =

Ans: C

Join $AE$. Since $\angle ABE = 90^\circ$ and $ABED$ is

Ans: D

Join $AC$ and $BD$.

$\begin{array}{rcll}
\angle BDC

Ans: $\angle RQS=27^\circ$, $\angle PQS=63^\circ$

In $\D

Ans: B

Let $\angle TSU =x$. In $\Delta STU$, we have

$\begin{

Ans: C

Join $P$ and $R$.

Since $SU$ and $ST$ are tangent to the