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Points S, O, M, and I are concyclic such that arc SO=arc IM. If the chord SM=IO are intersected at the points K. Prove that area of triangle SOK=area of triangle IMK and SM=IO

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Please find the file attached below (figure for the question)

To prove that area of the two triangles is equal we must know the formula for the area of a triangle and concept of concyclic points as given below:

Area of a Triangle:

Formula to find the area of a triangle is

  • [tex]A=\frac{1}{2}*(Base)(Height)[/tex]

Base can be any side of a triangle but height must be the side perpendicular to the base

Concyclic points:

Points which lie on the same circle having the same distance from the center of the circle.

Given Data:

Arc SO= Arc IM

chord SM= chord IO

Proof of Area of Triangle SOK=Area of Triangle IMK:

As the given points are concyclic points, so

  • SK=OK
  • KM=IM

Any of the above point is radius of the circle.

Thus,

  • SK=OK=KM=IM

Are of Triangle SOK:

  • [tex]Area\ of\ the\ Triangle\ SOK = \frac{1}{2}*(SK)(OK)[/tex]

where, SK is the Base for triangle SOK and OK is the Height for the triangle SOK

Area of Triangle IMK:

  • [tex]Area\ of\ Triangle\ IMK= \frac{1}{2}*(KM)(IK) \\[/tex]

Where, KM is Base of the triangle IMK and IK is Height of the triangle IMK

As we know

SK=OK=KM=IM

We can say directly that area of both the triangles is same

[tex]Area\ of\ the\ Triangle\ SOK = \frac{1}{2}*(SK)(OK)[/tex] =[tex]Area\ of\ Triangle\ IMK= \frac{1}{2}*(KM)(IK) \\[/tex]

OR

[tex]\frac{1}{2}*(SK)(SK)[/tex][tex]=\frac{1}{2}*(SK)(SK)[/tex]

thus proved

Proof of SM=IO:

As points are concyclic so they all have same distance from the center of the circle

i.e.SK=OK=KM=IM

thus SM=IO

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