Question

In a Fine Arts class, students were asked to design triangular tiles in geometric pattern.

Neelima made a circular design inside an equilateral triangle ABC. The radius of the circle is 4 cm. Observe the diagram and answer the following questions:

(i) Determine the length OB.

(ii) Is $D E ∥ C A$ ? Give reason for your answer.

(iii) (a) Write all angles of quadrilateral OEBD and show that it is a cyclic quadrilateral.

(iii) (b) Find the perimeter of $△ A BC$ . (Use $3 = 1.73$ )

Approach

In equilateral $△ A BC$ with inscribed circle radius 4 cm, the perpendicular from center O to side gives $sin 30° = \frac{4}{OB}$ , so $OB = 8$ cm. Since AD is median and altitude, D is midpoint of BC. By symmetry, E is midpoint of AB, so DE || CA. For cyclic quadrilateral OEBD: $∠ OEB = ∠ O D B = 90°$ (radius perpendicular to tangent). Sum $= 180°$ , hence cyclic. Alternatively, in $△ OB D$ , $cos 30° = \frac{B D}{8}$ gives $B D = 43 \approx 6.92$ cm, so $BC = 13.84$ cm. Perimeter of equilateral $△ A BC = 3 \times 13.84 \approx 41.52$ cm.

Step-by-step working

1
(i) In right $△ O D B$ , $sin 30° = \frac{O D}{OB} = \frac{4}{OB}$ , so $OB = 8$ cm
1 mark
2
(ii) $△ A BC$ equilateral, AD is median and altitude, so D is midpoint of BC. Similarly E is midpoint of AB. By midpoint theorem, $D E ∥ C A$
1 mark
3
(iii)(a) $∠ EB D = 60°$ , so $∠ EO D = 120°$ . $∠ OEB = ∠ O D B = 90°$ (radius $⊥$ tangent). Sum $= 180°$ , hence OEBD is cyclic
1.5 marks
4
(iii)(b) OR: In $△ OB D$ , $cos 30° = \frac{B D}{8} \Rightarrow B D = 8 \times \frac{3}{2} = 43 \approx 6.92$ cm. $BC = 2 \times 6.92 = 13.84$ cm. Perimeter $= 3 \times 13.84 = 41.52$ cm
1.5 marks

Concepts used

equilateral trianglecirclecyclic quadrilateralinscribed circlefind lengthfind perimeterprove

Approach

Step-by-step working

Concepts used

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