one diagonal of a cyclic quadrilateral coincides with a diameter of a circle whose area is 36pi cm^2. If the other diagonal which measures 8cm meets the first diagonal at right angles, find the area of quadrilateral. Show your solutions
To find the area of the cyclic quadrilateral, we need to find the lengths of both diagonals. Let's denote the length of the first diagonal (which coincides with a diameter of the circle) as d1, and the length of the second diagonal as d2.
We know that the area of the circle is given by A = πr^2, where r is the radius of the circle. In this case, the area A is given as 36π cm^2. So we can set up the equation:
A = πr^2
36π = πr^2
To find the radius, we can simplify the equation by canceling out π from both sides:
36 = r^2
Taking the square root of both sides, we get:
r = 6
Since the radius is half of the diameter, the length of the first diagonal (d1) is 2r:
d1 = 2r = 2(6) = 12 cm
Now, we are given that the second diagonal (d2) measures 8 cm and intersects the first diagonal at right angles. This implies that the diagonals of the cyclic quadrilateral are perpendicular bisectors of each other.
From here, we can start calculating the area of the cyclic quadrilateral. Since the diagonals of a cyclic quadrilateral are perpendicular bisectors of each other, the area can be found by multiplying half the product of the diagonals:
Area of quadrilateral = (1/2) * d1 * d2
Substituting the values we found:
Area of quadrilateral = (1/2) * 12 cm * 8 cm
Area of quadrilateral = 96 cm^2
Therefore, the area of the given cyclic quadrilateral is 96 cm^2.