A cyclic quadrilateral has sides 5, 5, 12, 12 units. Find its area.
Step-by-step solution
To find: Area of ABCD
Idea: Use the fact that opposite angles of a cyclic quadrilateral add up to 180°. Equal sides give congruent triangles and hence two equal opposite angles, which must then each be 90°.
- Sides in the order 5, 5, 12, 12 (AB = AD = 5, CB = CD = 12): in ΔABC and ΔADC, AB = AD, CB = CD and AC is common, so ΔABC ≅ ΔADC (SSS) and ∠B = ∠D.1 mark
- ABCD is cyclic, so ∠B + ∠D = 180°. With ∠B = ∠D, each is 90°.1 mark
- Area = area ΔABC + area ΔADC = ½ × 5 × 12 + ½ × 5 × 12 = 30 + 30 = 60 square units.1 mark
- Sides in the order 5, 12, 5, 12: opposite sides are equal, so ABCD is a parallelogram; its opposite angles are equal and also add to 180°, so each angle is 90°. It is a 5 × 12 rectangle: area = 60 square units again.
Check: Brahmagupta’s formula for a cyclic quadrilateral, √((s − a)(s − b)(s − c)(s − d)) with s = 17, gives √(12 × 12 × 5 × 5) = 60 ✓. Also AC = √(52 + 122) = 13 is a diameter.
Answer to write in the exam
Let AB = AD = 5, CB = CD = 12.
ΔABC ≅ ΔADC (SSS: AB = AD, CB = CD, AC common) ⇒ ∠B = ∠D (CPCT)
∠B + ∠D = 180° (opposite angles of a cyclic quadrilateral) ⇒ ∠B = ∠D = 90°
Area = ½ × 5 × 12 + ½ × 5 × 12 = 30 + 30
∴ Area = 60 square units
(Order 5, 12, 5, 12: a parallelogram that is cyclic is a rectangle; area = 5 × 12 = 60 square units.)
Common mistakes that cost marks
- Multiplying 5 × 12 × 2 = 120 by treating it as a rectangle of sides 5 + 5 and 12 (wrong shape).
- Assuming the right angles without proof. They come from ∠B = ∠D together with ∠B + ∠D = 180°.
- Using ½ × 5 × 12 only once (30), forgetting the second triangle.
How this can come in the exam
A cyclic quadrilateral has sides 6, 6, 8, 8 units, with the equal sides next to each other. Its area is
- 24 sq units
- 48 sq units
- 96 sq units
- 28 sq units
Show answer
(B) 48 sq units
The angles between the 6 and 8 sides are 90°, so area = 2 × ½ × 6 × 8 = 48 square units.
Try one yourself
A cyclic quadrilateral PQRS has PQ = PS = 9 cm and RQ = RS = 12 cm. Find its area and the length of PR.
Show answer
∠Q = ∠S = 90°, so area = 2 × ½ × 9 × 12 = 108 cm2 and PR = √(81 + 144) = 15 cm (a diameter).
More questions like this
- Consider a cyclic quadrilateral. Without drawing its circumcircle, how can we find out whether the centre of the circumcircle lies inside the quadrilateral or outside? What is the best way of finding out?
- When two chords intersect, each of them is divided into two line segments. Show that if the intersecting chords are of equal length, then the line segments of one chord are equal to the corresponding line segments of the other chord.
- Draw a circle in which a chord of 6 cm length stands at a distance of 3 cm from the centre.
(Hint: Is it a circumcircle of a suitable triangle?) - Show that rectangle is the only parallelogram that can be inscribed in a circle.
- Show that if a rectangle is inscribed in a circle, then the point of intersection of its diagonals must lie at the centre of the circle.