A circle is inscribed into a rhombus ABCD with one angle 60º. 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 |2is equal toa)12b)11c)9d)NoneCorrect answer is option 'B'. Can you explain this answer?

Question

A circle is inscribed into a rhombus ABCD with one angle 60&ordm;. 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 |2is equal toa)12b)11c)9d)NoneCorrect answer is option 'B'. Can you explain this answer?

Answer

To solve this problem, we can start by drawing a diagram.

Let's label the points of the rhombus ABCD as follows:
A: top vertex
B: right vertex
C: bottom vertex
D: left vertex

Since we know that one angle of the rhombus is 60 degrees, we can label the other angles as follows:
A: 60 degrees
B: unknown
C: 120 degrees (opposite angle of A)
D: unknown

Now, let's draw a circle inscribed in the rhombus. The center of the circle will be the point where the diagonals of the rhombus intersect. Let's label this point O.

Since the diagonals of a rhombus are perpendicular bisectors of each other, we can draw the perpendicular bisectors of AB and BC. These bisectors will intersect at point O, which is the center of the inscribed circle.

Let's label the points where the perpendicular bisectors intersect AB and BC as E and F, respectively.

Now, we have a right triangle AOE, with angle AOE being 90 degrees. We also know that angle A is 60 degrees. Therefore, angle OAE is 180 - 90 - 60 = 30 degrees.

Since angle OAE is 30 degrees and angle OEA is 90 degrees, we can conclude that angle EOA is 180 - 30 - 90 = 60 degrees.

Since angle EOA is 60 degrees, this means that angle EOF is also 60 degrees (since they are opposite angles). And since angle EOF is 60 degrees, this means that angle OFC is 180 - 60 = 120 degrees.

Since angle OFC is 120 degrees and angle C is 120 degrees, this means that angle OFC and angle C are congruent. Therefore, triangle OFC is an isosceles triangle.

Now, let's consider the triangle OFC. Since triangle OFC is isosceles, this means that the sides OF and OC are congruent. And since point O is the center of the inscribed circle, this means that OF and OC are radii of the circle. Therefore, the circle is tangent to the sides OF and OC.

Similarly, the circle is also tangent to the sides OE and OA.

Therefore, we have shown that the circle is inscribed in the rhombus ABCD with one angle of 60 degrees.

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