OABC is a rhombus whose three vertices lie on a circle with centre at O. If the area of the rhombus is 32√3 square cm, then what is the radius of the circle ?
8 cm
Solving the Rhombus and Circle Geometry Problem
The problem asks us to find the radius of a circle on which three vertices of a rhombus OABC lie, given that the center of the circle is O and the area of the rhombus is \(32\sqrt{3}\) square cm.
Understanding the Setup
We are given a rhombus OABC. The center of the circle is O, and three vertices of the rhombus lie on this circle. Since O is the center of the circle, a vertex can only lie on the circle if its distance from O is equal to the radius. If O itself were one of the three vertices on the circle, its distance from the center O would be 0, implying a radius of 0. A circle with radius 0 cannot enclose a rhombus with an area of \(32\sqrt{3}\) square cm. Therefore, the three vertices of the rhombus that lie on the circle must be the vertices A, B, and C.
So, we have:
- The center of the circle is O.
- Vertices A, B, and C of the rhombus lie on the circle.
This means the distance from the center O to each of these vertices is equal to the radius of the circle. Let 'r' be the radius of the circle.
Thus, \(OA = OB = OC = r\).
Relating Rhombus Properties to the Circle
OABC is a rhombus. By definition, all sides of a rhombus are equal in length. Let 's' be the side length of the rhombus.
The sides of the rhombus OABC are OA, AB, BC, and CO. So, \(OA = AB = BC = CO = s\).
From our understanding of the circle setup, we know \(OA = r\). Since OA is also a side of the rhombus, the side length of the rhombus is equal to the radius of the circle.
So, \(s = r\).
This means the rhombus has a side length equal to the radius 'r'. All sides are \(OA = AB = BC = CO = r\).
Now consider the diagonals of the rhombus OABC. The diagonals are AC and OB. We know from the circle properties that \(OB = r\) because vertex B lies on the circle centered at O.
So, one diagonal of the rhombus, OB, has a length equal to the side length, which is also equal to the radius 'r'.
Calculating the Length of the Other Diagonal
In a rhombus, the diagonals bisect each other at right angles. Let the lengths of the diagonals be \(d_1\) and \(d_2\). If 's' is the side length, the relationship between the side and diagonals is:
\(\qquad s^2 = \left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2\)
We have the side length \(s = r\), and one diagonal \(d_1 = OB = r\). Let the other diagonal be \(d_2 = AC\). Substituting these values:
\(\qquad r^2 = \left(\frac{r}{2}\right)^2 + \left(\frac{AC}{2}\right)^2\)
\(\qquad r^2 = \frac{r^2}{4} + \frac{AC^2}{4}\)
Multiply the equation by 4:
\(\qquad 4r^2 = r^2 + AC^2\)
\(\qquad AC^2 = 4r^2 - r^2\)
\(\qquad AC^2 = 3r^2\)
\(\qquad AC = \sqrt{3r^2} = \sqrt{3}r\) (Since length must be positive)
So, the lengths of the diagonals of the rhombus are \(r\) and \(\sqrt{3}r\).
Using the Area of the Rhombus
The area of a rhombus can be calculated using the formula:
\(\qquad \text{Area} = \frac{1}{2} \times d_1 \times d_2\)
We are given that the area is \(32\sqrt{3}\) square cm. Using our diagonal lengths \(d_1=r\) and \(d_2=\sqrt{3}r\):
\(\qquad 32\sqrt{3} = \frac{1}{2} \times r \times (\sqrt{3}r)\)
\(\qquad 32\sqrt{3} = \frac{\sqrt{3}}{2} r^2\)
Now, we solve for r:
Divide both sides by \(\sqrt{3}\):
\(\qquad 32 = \frac{1}{2} r^2\)
Multiply both sides by 2:
\(\qquad 64 = r^2\)
Take the square root of both sides (radius must be positive):
\(\qquad r = \sqrt{64} = 8\)
The radius of the circle is 8 cm.
Conclusion
The radius of the circle on which three vertices (A, B, C) of the rhombus OABC lie, with the center at O and an area of \(32\sqrt{3}\) square cm, is 8 cm.
| Property | Value / Relation |
|---|---|
| Center of Circle | O |
| Vertices on Circle | A, B, C |
| Radius of Circle (r) | OA = OB = OC = r |
| Rhombus OABC Sides (s) | OA = AB = BC = CO = s |
| Relation s and r | s = r |
| Diagonal \(d_1\) | OB = r |
| Diagonal \(d_2\) | AC = \(\sqrt{3}r\) |
| Area of Rhombus | \(\frac{1}{2} d_1 d_2 = \frac{\sqrt{3}}{2} r^2\) |
| Given Area | \(32\sqrt{3}\) cm² |
| Calculated Radius (r) | 8 cm |
Revision Table: Rhombus and Circle Properties
| Concept | Description | Key Formula / Property |
|---|---|---|
| Rhombus | A quadrilateral with all four sides of equal length. Opposite angles are equal, and consecutive angles are supplementary. Diagonals bisect each other at right angles. | Sides equal; Diagonals are perpendicular bisectors. |
| Area of Rhombus | The space enclosed by the sides of the rhombus. | Area = \(\frac{1}{2} \times d_1 \times d_2\) (where \(d_1, d_2\) are diagonal lengths) |
| Circle | A set of points in a plane equidistant from a fixed point (the center). | Distance from center to any point on circle = Radius (r). |
| Rhombus Side & Diagonals | Relationship between the side length (s) and half-diagonals (\(d_1/2\), \(d_2/2\)) forming right triangles. | \(s^2 = \left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2\) |
Additional Information: Rhombus Geometry
A rhombus is a special type of parallelogram where all sides are congruent. Because its diagonals are perpendicular bisectors, a rhombus has perpendicular diagonals. The diagonals also bisect the interior angles of the rhombus.
In the specific case encountered in this problem, where the side length of the rhombus is equal to the length of one of its diagonals, the rhombus is composed of two equilateral triangles joined along that diagonal. For a rhombus with side length 's':
- If one diagonal is 's', the angles corresponding to the vertices connected by this diagonal are \(120^\circ\). The other two angles are \(60^\circ\).
- The other diagonal in such a rhombus has length \(\sqrt{3}s\).
This geometric property was key to solving the problem, as it allowed us to determine the lengths of both diagonals in terms of the radius 'r' (since side length = r and one diagonal = r).
Similar Questions
If the quadrilateral has an inscribed circle, then the sum of a pair of opposite sides equals:
ABCD is a cyclic quadrilateral. The bisectors of the angles A, B, C and D cut the circle at P, Q, R and S respectively. What is ∠PQR + ∠RSP equal to?
What is the magnitude (in radian) of the interior angle of a regular pentagon?
If the sum of all interior angles of a regular polygon is twice the sum of all its exterior angles then the polygon is
ABCD is a square field with AB = x. A vertical pole OP of height 2x stands at the centre O of the square field. If ∠APO = θ, then what is cot θ equal to ?
A square is inscribed in a right angled triangle with legs p and q and has a common right angle with triangle. The diagonal of the square is given by
ABCDA is a con-cyclic quadrilateral of a circle ABCD with radius r and centre at O. If AB is the diameter and CD is parallel and half of AB and if the circle completes one rotation about the centre O, then the locus of the middle point of CD is a circle of radius:
The diagonals of a rhombus are of length 20 cm and 48 cm. What is the length of a side of the rhombus?
The area of a regular hexagon of side ‘a’ is equal to
Two parallel sides of a trapezium are 29 cm and 21 cm. Non-parallel sides are equal and each is of length 8.5 cm. What is the area of the trapezium?
Important Questions from Quadrilaterals
The ratio between the length and breadth of a rectangular park is 3 : 2. If a man cycling along the boundary at the speed of 12 km per hour completes one round in 8 minutes, then the area of the park in square meter will be
The side of a rhombus is 26 cm. The length of one of its diagonals is 20 cm. The sum of the lengths of the diagonals of this rhombus is equal to the perimeter of a rectangle. If the difference between the length and breadth of the rectangle is 6 cm, then what is the area of the rectangle?
PQRS is a cyclic quadrilateral. If ∠P is 4 times ∠R, and ∠S is 3 times ∠Q, then the average of ∠Q and ∠R is:
ABCD is a trapezium in which AB || DC and DC is perpendicular to BC. If ∠DAB = 110°, then ∠ABC - ∠ADC =_____.
If the quadrilateral has an inscribed circle, then the sum of a pair of opposite sides equals:
