Rectangle Maths Tuition

rectangle-maths-tuition

Solution:

(a) \angle SPC=\angle SRQ=90^\circ

\angle PCS=\angle SQR (given)

Thus \triangle PCS is similar to \triangle RQS (AAA)

(b)(I)

\angle KRQ=90^\circ /2=45^\circ (KR bisects \angle QRS)

Thus \angle QKR=180^\circ-90^\circ-45^\circ=45^\circ

Hence \triangle KQR is an isosceles triangle.

So KQ=QR=PS=\frac{1}{2}PQ

Hence,

\begin{array}{rcl}PK&=&PQ-KQ\\    &=&PQ-\frac{1}{2}PQ\\    &=&\frac{1}{2}PQ\\    &=&\frac{1}{2}(2PS)\\    &=&PS    \end{array}

(shown)

(ii) By similar triangles,

\displaystyle\frac{PC}{PS}=\frac{RQ}{RS}=\frac{PS}{PQ}=\frac{PS}{2PS}=\frac{1}{2}

Thus PC=\frac{1}{2}PS=\frac{1}{2} PK

Hence

\begin{array}{rcl}CK&=&PK-PC\\    &=&PK-\frac{1}{2}PK\\    &=&\frac{1}{2}PK\\    &=&PC    \end{array}

(shown)

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