A triangle ABC is formed with AB = AC = 50 cm and BC = 80 cm. Then, the sum of the lengths, in cm, of all three altitudes of the triangle ABC is

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

Let $D$ and $E$ be points on sides $A B$ and $A C$, respectively, of a triangle $A B C$, such that $A D: B D=2: 1$ and $A E: C E=2: 3$. If the area of the triangle $A D E$ is $8 \mathrm{ sq } \mathrm{cm}$, then the area of the triangle $ABC$ , in sq cm , is

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 $T$ be the triangle formed by the straight line $3x + 5y - 45 = 0$ and the coordinate axes. Let the circumcircle of $T$ have radius of length $L,$ measured in the same unit as the coordinate axes. Then, the integer closest to $L$ is

The midpoints of sides $AB$, $BC$, and $AC$ in $\triangle ABC$ are $M$, $N$, and $P$, respectively. The medians drawn from $A$, $B$, and $C$ intersect the line segments $MP$, $MN$ and $NP$ at $X$, $Y$, and $Z$, respectively. If the area of $\triangle ABC$ is $1440$ sq cm, then the area, in sq cm, of $\triangle XYZ$ 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