Suppose the medians $BD$ and $CE$ of a triangle $ABC$ intersect at a point $O$. If area of triangle $ABC$ is $108$ sq. cm., then, the area of the triangle $EOD$, in sq. cm., is

A triangle $ABC$ has area $32$ sq units and its side $BC$ , of length $8$ units, lies on the line $\mathrm{x}=4$. Then the shortest possible distance between $A$ and the point $(0,0)$ is

In a triangle ABC, $\angle BCA = 50^\circ$. D and E are points on AB and AC, respectively, such that AD = DE. If F is a point on BC such that BD = DF, then $\angle FDE$, in degrees, is equal to

In triangle ABC, altitudes AD and BE are drawn to the corresponding bases. If $\angle BAC = 45^\circ$ and $\angle ABC = \theta$, then $\frac{AD}{BE}$ equals

Two ships are approaching a port along straight routes at constant speeds. Initially, the two ships and the port formed an equilateral triangle with sides of length $24$ km. When the slower ship travelled $8$ km, the triangle formed by the new positions of the two ships and the port became right-angled. When the faster ship reaches the port, the distance, in km, between the other ship and the port will be

A circle of diameter $8$ inches is inscribed in a triangle ABC where $\angle A B C=90^{\circ}$. If $B C=10$ inches then the area of the triangle in square inches is

Let $\triangle ABC$ be an isosceles triangle such that $AB$ and $AC$ are of equal length. $AD$ is the altitude from $A$ on $BC$ and $BE$ is the altitude from $B$ on $AC$. If $AD$ and $BE$ intersect at $O$ such that $\angle AOB = 105^\circ$, then $AD/BE$ equals