O Level 2007 E Maths Paper 2 Q3 Solution (Geometry Question)

(a)
\angle DAB=\angle DCB
(opposite angles of parallelogram)

\displaystyle\angle PAD=\frac{1}{2}\angle DAB=\frac{1}{2}\angle DCB=\angle RCB
(shown)

(b)
AD=BC

\angle PAD=\angle RCB
(from part a)

\angle ADC=\angle ABC
(opposite angles of parallelogram)

\displaystyle\angle ADP=\frac{1}{2}\angle ADC=\frac{1}{2} ABC=\angle RBC

Thus, triangles ADP and CBR are congruent (ASA).

(c)(i)
\angle ADC+\angle DAB=180^\circ
\angle DAP+\angle ADP=90^\circ

Considering the triangle ADP,
\angle DPA=180^\circ-90^\circ=90^\circ
(shown)

(ii)
\angle DAB+\angle ABC=180^\circ

\angle BAQ+\angle ABQ=90^\circ

Considering the triangle ABQ,
\angle PQR=180^\circ-90^\circ=90^\circ
(shown)

Advertisements

About mathtuition88

http://mathtuition88.com
This entry was posted in geometry and tagged , , , . Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.