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Take a look at figure given below. Give it a try and when you are ready then watch the solution.
A triangle is inscribed in a semicircle and on both sides, a perpendicular line is drawn. From end of both lines two lines are drawn to In-center of triangle. You have to find area of small triangle formed in terms of (x,y) as shown in figure.
Step 1:
Mark angle θ and 90-θ on both ends of triangle.
Step 2:
Give dimensions to small triangle as a,b.
Step 3:
Mark naming to each point in the figure as shown in figure.
Step 4:
Construct lines BE, BO, AB as shown.
Note: BE will pass through Incenter.
Step 5:
AB and BC have same arc length so angle formed by them will be same.
∠AEB = ∠BEC = 1/2 ∠AEC = θ/2
Step 6:
∠BEC and ∠BAC are on same arc so they are same.
∠BEC = ∠BAC = θ/2
Step 7:
∠BOC must be twice of ∠BEC.
∠BOC = 2∠BEC = θ
Step 8:
Let radius of Semicircle be 'r'.
Step 9:
In triangle AGO,
AG = r sinθ
Step 10:
In triangle ACE:
CE = 2r cosθ
Step 11:
Now observe Triangles ABG and EFC:
∠BAG = ∠FEC = θ/2
∠AGB = ∠ECF = 90
⇒ Third angle also same.
⇒ Triangles ABG and EFC are similar.
⇒ BG/AG = FC/CE
⇒ x/r sinθ = a/2r cosθ
⇒ a = 2x cosθ / sinθ
Step 12:
As we can observe, this diagram is totally symmetrical,
⇒ a and b have similar values. Only θ will be replaced by 90-θ and x is replaced by y.
⇒ b = 2y sinθ / cosθ
Now area of Triangle will be,
Area = 1/2 * a * b
⇒ Area = 1/2 * 2x cosθ / sinθ * 2y sinθ / cosθ
⇒ Area = 2xy
