Home
Class 11
MATHS
A circle is inscribed ti.e. touches all ...

A circle is inscribed ti.e. touches all four sides ) into a rhombous ABCD with one angle 60°. The distance from the centre of the circle to the nearest vertex is equal to 1. If P is any point of the circle then `|PA|^2+|PB|^2+|PC|^2+|PD|^2` is equal to:

Promotional Banner

Similar Questions

Explore conceptually related problems

A circle is inscribed into a rhombous ABCD with one angle 60. The distance from the centre of the circle to the nearest vertex is equal to 1. If P is any point of the circle then |PA|^2+|PB|^2+|PC|^2+|PD|^2 is equal to:

P is a point inside the rectangle ABCD Prove that PA^2+ PC^2= PB^2 +PD^2 : .

Two equal line segments PA and PB are drawn form external point P of a circle . If the distance of PA from the centre of the circle be 3, cm then find the distance of PB form the centre .

An equilateral triangle ABC is inscribed in a circle of radius r if P be any point on the circle then find the value of PA^(2)+PB^(2)+PC^(2)

If tangents PA and PB from a point to a circle with centre O are inclined to each other at angle of 80 ∘ , then ∠POA is equal to ,