The area of a $\Delta ABC$ is equal to

The ratios of the sides of a triangle $ABC$ are $5:12:13$ and its area is $270$. Then the sides of the triangle are:

If $7$ and $8$ are the lengths of two sides of a triangle and '$a$' is the length of its smallest side. The angles of the triangle are in $AP$ and '$a$' has two values $a_1$ and $a_2$ satisfying this condition. If $a_1 < a_2$ then $2 a_1 + 3 a_2 =$

If $p_1, p_2, p_3$ are altitudes of a triangle $ABC$ from the vertices $A, B, C$ respectively and if $\Delta$ is the area of the triangle,$S$ is the semi-perimeter of the triangle,then find the value of $\frac{\cos A}{p_1} + \frac{\cos B}{p_2} + \frac{\cos C}{p_3}$.

In a $\triangle ABC$,if $2 \Delta^2 = \frac{a^2 b^2 c^2}{a^2+b^2+c^2}$,then the triangle is

In a $\triangle ABC$,$\angle C = 60^{\circ}$ and $\angle A = 75^{\circ}$. If $D$ is a point on $AC$ such that the area of $\triangle BAD$ is $\sqrt{3}$ times the area of $\triangle BCD$,then the measure of $\angle ABD$ is (in $^{\circ}$)