Qpqr is a rhombus ,3 of whose vertices lie on the circle with centre o .if the area of the rhombus is 32 root 3 find the radius of the radius
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To find the radius of the circle, we need to understand the properties of a rhombus and use the formula for its area.
A rhombus is a quadrilateral with all four sides of equal length. In this case, we have a rhombus QPQR, where Q, P, and R are vertices lying on the circle with center O.
The diagonal of a rhombus divides it into two congruent right-angled triangles. Let's consider one of these triangles, say QOP.
The area of a triangle can be found using the formula:
Area = (base * height) / 2
In triangle QOP, the base is equal to the length of side QP, and the height is equal to the radius of the circle (OQ). So, the area of triangle QOP can be written as:
Area(QOP) = (QP * OQ) / 2
Since the diagonals of a rhombus are perpendicular and they bisect each other, we know that QP is twice the length of OQ. Therefore, QP = 2 * OQ.
Substituting this value into the area formula for triangle QOP, we get:
Area(QOP) = (2 * OQ * OQ) / 2
Area(QOP) = OQ^2
Since triangle QOP is a right-angled triangle, we can use the Pythagorean theorem to relate the sides. We know that QP = 2 * OQ, so we have:
QP^2 = OQ^2 + OQ^2
QP^2 = 2 * OQ^2
Now we can calculate the area of the rhombus using the given value of 32√3:
Area(rhombus) = QP * OQ
32√3 = 2 * OQ * 2 * OQ
32√3 = 4 * OQ^2
8√3 = OQ^2
Taking the square root of both sides, we get:
OQ = √(8√3)
OQ = √(8) * √(√3)
OQ = 2√2 * √(√3)
Therefore, the radius of the circle, OQ, is equal to 2√2 * √(√3).