`A(3, 4),B(0,0) and c(3,0)` are vertices of `DeltaABC.` If 'P' is the point inside the `DeltaABC`, such that `d(P, BC)le` min. `{d(P, AB), d (P, AC)}`. Then the maximum of `d (P, BC)` is.(where `d(P, BC)` represent distance between `P and BC)`.

In DeltaABC, if D is any point on the side BC, show that (AB+BC+AC)gt 2AD.

Let O(0.0),A(6,0) and B(3,sqrt3) be the vertices of triangle OAB. Let R those points P inside triangle OAB which satisfy d(P, OA) minimum (d(P,OB),d(P,AB) where d(P, OA), d(P, OB) and d(P, AB) represent the distance of P from the sides OA, OB and AFB respectively. If the area of region R is 9 (a -sqrtb) where a and b are coprime, then find the value of (a+ b)

Consider a rectangle ABCD formed by the points A=(0,0), B= (6, 0), C =(6, 4) and D =(0,4), P (x, y) is a moving interior point of the rectangle, moving in such a way that d (P, AB)le min {d (P, BC), d (P, CD) and d (P, AD)} here d (P, AB), d (P, BC), d (P, CD) and d (P, AD) represent the distance of the point P from the sides AB, BC, CD and AD respectively. Area of the region representing all possible positions of the point P is equal to

A(4,3),B(6,5)and C(5,-2) are the vertices of a DeltaABC, if P is a point on BC such that BP:PC =2:3. Find the co-ordinates of P and then prove then ar (DeltaABP):ar(DeltaACP)=2:3.

Let O(0, 0), A(2,0) and B(1,(1)/(sqrt(3))) be the vertices of a triangle. Let R be the region consisting of all those points P inside Delta OAB satisfying. d(P,OA) lr min {d(P,OB), d(P,AB)} , where d denotes the distance from the point P to the corresponding line. Let M be peak of region R. The perimeter of region R is equal to

P and Q are the points on sides AB and AC respectively of a DeltaABC , such that PQ||BC .If AP|PB=2:3 and AQ=4 cm , then the length of AC is ….. Cm .

In the figure, DeltaABC ~ DeltaBPQ. If AB = BC and P is the midpoint of seg BC, then A(DeltaABC):A(DeltaBPQ) =