In a right triangle, the difference of its hypotenuse and base is 50 cm and the difference of its hypotenuse and perpendicular is 1 cm. Find the area of the incircle of the triangle.
Give your answer upto 2 decimal places.
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To solve this, I had to assume that what is referred to as "base" and "perpendicular" are the legs of the right triangle in question.
Relationships between the hypotenuse h , base b , and perpendicular p are: h − b = 5 0 , h − p = 1 , and p 2 + b 2 = h 2 .
Solving for h we get two solutions: h = 6 1 and h = 4 1 . Calculating b from the second one of those will give us a negative number, so this is not geometrically reasonable.
Using the first solution, we have b = 1 1 , p = 6 0 , the area of the triangle A = 2 1 p b = 3 3 0 , and semiperimeter s = 2 1 ( b + p + h ) = 6 6 .
Radius of incircle R = s A = 5 . Area of incircle is therefore 2 5 π .
It would be a help to at least some of us if you would change the wording to a more standard terminology.
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The area of the incircle of a right triangle is given by 2 π ( h − b ) ( h − p ) , where h, p and b are the hypotenuse, height and base of the right triangle respectively.