In the given figure, AB is diameter of the circle and points C and D are on the circumference such that $\angle CAD = 30^\circ$ and $\angle CBA = 70^\circ$. What is the measure of $\angle ACD$?
In the adjoining figure, AC + AB = 5AD and AC – AD = 8. Then the area of the rectangle ABCD is
PQRS is a square. SR is a tangent (at point S) to the circle with centre O and TR = OS. Then the ratio of area of the circle to the area of the square is:
AB $\perp$ BC, BD $\perp$ AC and CE bisects $\angle C$, $\angle A = 30^\circ$. Then what is $\angle CED$?
ABCD is a square of area $4$, which is divided into four non-overlapping triangles as shown in the figure. Then the sum of the perimeters of the triangles is:
$le(x, y) = \text{Least of} \ (x, y)$
$mo(x) = |x|$
$me(x, y) = \text{Maximum of} \ (x, y)$