Parallelogram Maths Tuition: Solution

parallelogram-maths-tuition

Solution:

(a) We have \angle APQ=\angle ARQ (opp. angles of parallelogram)

AP=RQ (opp. sides of parallelogram)

AR=PQ (opp. sides of parallelogram)

Thus, \triangle APQ\equiv\triangle QRA (SAS)

Similarly, \triangle ABC\equiv\triangle CDA (SAS)

\triangle CHQ\equiv\triangle QKC (SAS)

Thus, \begin{array}{rcl}\text{area of BPHC}&=&\triangle APQ-\triangle ABC-\triangle CHQ\\    &=&\triangle QRA-\triangle CDA-\triangle QKC\\    &=& \text{area of DCKR}    \end{array}

(proved)

(b)

\angle ACD=\angle HCQ (vert. opp. angles)

\angle ADC=\angle CHQ (alt. angles)

\angle DAC=\angle CQH (alt. angles)

Thus, \triangle ADC is similar to \triangle QHC (AAA)

Hence, \displaystyle\frac{AC}{DC}=\frac{QC}{HC}

Thus, AC\cdot HC=DC\cdot QC

(proved)

Congruent Triangles Maths Tuition: Solution

congruent-maths-tuition

Solution:

BS=BE+ES=ST+ES=ET

\angle DES=\angle ESA=90^\circ

BA=DT (given)

Thus, \triangle ASB is congruent to \triangle DET (RHS)

Hence \angle DTE=\angle SBA

Thus DT//BA (alt. angles)

(Proved)

By Pythagoras’ Theorem, we have

\begin{array}{rcl}DB&=&\sqrt{DE^2+BE^2}\\    &=&\sqrt{SA^2+ST^2}\\    &=&TA    \end{array}

Hence \triangle DEB and \triangle AST are congruent (SSS).

Hence \angle DBE=\angle STA

Thus DB//TA (alt. angles)

Therefore, ABDT is a parallelogram since it has two pairs of parallel sides.

(shown)