Let `h_1,h_2,h_3` be the altitudes of the `DeltaABC` and let the inradius be r. Then show that `(h_1+r)/(h_1-r)+(h_2+r)/(h_2-r)+(h_3+r)/(h_3-r)ge 6.`

Let A_(1), A_(2), A_(3), A_(4) be the areas of the triangular faces of a tetrahedron, and h_(1), h_(2), h_(3), h_(4) be the corresponding altitudes of the tetrahedron. If the volume of tetrahedron is 1//6 cubic units, then find the minimum value of (A_(1) +A_(2) + A_(3) + A_(4))(h_(1)+ h_(2)+h_(3)+h_(4)) (in cubic units).

The value of the acceleration due to gravity at a height h from the surface of the earth is g_1 and that at a depth h below the earth's surface is g_2 .Show that (g_2)/(g_1)=(1+h/R) [Radius of earth Rgt gt h ]

A hail drop is falling freely due to gravity. It travels distance h_(1),h_(2) and h_(3) respectively in the first second and third seconds of motion. The relation among h_(1) , h_(2) and h_(3) is

Find the values of non-negative real number h_1, h_2, h_3, k_1, k_2, k_3 such that the algebraic sum of the perpendiculars drawn from the points (2,k_1),(3,k_2),*7,k_3),(h_1,4),(h_2,5),(h_3,-3) on a variable line passing through (2, 1) is zero.

In DeltaABC, R, r, r_(1), r_(2), r_(3) denote the circumradius, inradius, the exradii opposite to the vertices A,B, C respectively. Given that r_(1) :r_(2): r_(3) = 1: 2 : 3 The sides of the triangle are in the ratio

Assume that the earth is a uniform sphere of radius R. If the acceleration due to gravity at a height h above the earth's surface is g_1 and at the same depth below the surface of the earth is g_2(h lt R) , then show that, g_2=g_1(1-h/R)(1+h/R)^2 .

Lengths of the tangents from A,B and C to the incircle are in A.P., then (a) r_(1), r_(2), r_(3) are in H.P (b) r_(1), r_(2), r_(3) are in A.P (c)a, b, c are in A.P (d) cos A = (4c -3b)/(2c)

The volume of a right circular cone of radius of base r unit and of height h unit is (1)/(3)pi r^2h cubic-unit.

The ex-radii r_1,r_2,r_3 or Delta ABC are in H.P. Show that its sides a,b,c are in A.P.

The value of the acceleration due to gravity at a height h from the surface of the earth is g_1 and that at a depth h below the earth’s surface is g_2 . Show that g_2/g_1 = (1+h/R) (Radius of earth R >> h) I.