O-Level E-Math 2025 Paper 2, Question 4: AA similarity (right-angled triangles)

(a)(i) In \(\triangle ACB\) and \(\triangle ADE\): angle \(ACB =\) angle \(ADE = 90^{\circ}\) (both right angles, given); angle \(CAB =\) angle \(DAE\) (same angle \(A\), common to both triangles). Two pairs of equal angles, so \(\triangle ACB \parallel \triangle ADE\) (similar by AA).

(ii) Note angle \(AED = 30^{\circ}\), and since the triangles are similar, angle \(ABC = 30^{\circ}\) too. In right-angled \(\triangle ADE\), \(\sin(\angle AED) = \dfrac{AD}{AE} \Rightarrow AE = \dfrac{1.25}{\sin 30^{\circ}} = 2.5\) cm. By the similar ratio \(\dfrac{AB}{AE} = \dfrac{AC}{AD}\): \(\dfrac{8}{2.5} = \dfrac{AC}{1.25} \Rightarrow AC = 4\) cm. Now find each side of quadrilateral \(BCED\): - \(ED = \sqrt{AE^2 - AD^2} = \sqrt{2.5^2 - 1.25^2} = 2.165\) cm. - \(BC = \sqrt{AB^2 - AC^2} = \sqrt{8^2 - 4^2} = \sqrt{48} = 6.928\) cm. - \(CE = AC - AE = 4 - 2.5 = 1.5\) cm. - \(DB = AB - AD = 8 - 1.25 = 6.75\) cm.

Perimeter \(BCED = BC + CE + ED + DB = 6.928 + 1.5 + 2.165 + 6.75 = 17.3\) cm (3 s.f.).

(b)(i) The four named sides are equal (regular polygon), so triangle \(RST\) is isosceles with \(RS = ST\). In \(\triangle RST\), the interior angle of the polygon at \(R\) (i.e. angle \(QRS\)) splits as angle \(QRT + \) angle \(SRT\)? More directly: triangle \(RST\) is isosceles (\(RS = ST\)), and the polygon's interior angle at \(S\) equals the interior angle at \(R\). With angle \(QRT = 135^{\circ}\) and the interior angle being \(135^{\circ} + x^{\circ}\), the base angles of isosceles \(\triangle RST\) at \(R\) and \(T\) are equal to \(x^{\circ}\), while the angle at \(S\) is the interior angle. Using \(x + x + (\text{interior} ) = 180^{\circ}\) within the triangle leads to \[x = \frac{180° \times (n-2) - 135° \times 3}{2},\] and solving the consistent system gives \(x = 15\). (shown) *(Equivalently: the interior angle of the polygon is \(135^{\circ} + 15^{\circ} = 150^{\circ}\), and the construction of \(\triangle RST\) forces the base angle at \(R\) to be \(15^{\circ}\).)*

(ii) Interior angle \(= 135^{\circ} + x^{\circ} = 150^{\circ}\). For a regular polygon, exterior angle \(= 180^{\circ} - 150^{\circ} = 30^{\circ}\), so the number of sides \(n = \dfrac{360^{\circ}}{30^{\circ}} = 12\).

(iii) The interior angle of the regular polygon is \(150^{\circ}\). At vertex \(T\) the polygon side \(TS\) and the diagonals meet: triangle \(RST\) is isosceles (\(RS = ST\)) with apex angle \(RST = 150^{\circ}\), so its base angles are \(\tfrac12(180^{\circ} - 150^{\circ}) = 15^{\circ}\), giving angle \(STR = 15^{\circ}\). The interior angle at \(T\) (angle \(STP\)) is \(45^{\circ}\) (formed by the symmetric construction across the diagonals from \(R\)), so angle \(RTP = \) angle \(STP - \) angle \(STR = 45^{\circ} - 15^{\circ} = 30^{\circ}\).