An inscribed triangle with the hypotenuse being the diameter of the circle has angle a be 42 degrees. angle c is 90 degrees. the area of the triangle ack is 50cm^2. find the radius of the circle

The Radius of circle is 7.15 cm  

Step-by-step explanation:

Given as :

A Triangle is inscribed into circle

The center of circle is O

The Diameter of circle = AK = d

The hypotenuse of triangle being the diameter of circle

The Area of triangle = 50 cm²

Let The radius of circle = r cm

The Triangle is right angle at c

So , ∠ACK = 90°

∠CAK = 42°

∠AKC = 180° - (90° + 42°)

So, ∠AKC = 48°

Now, ∵ Area of triangle ACK = 50 cm²

So, × AC × CK = 50

Or, AC × CK = 50 × 2

i.e , AC × CK = 100        ..........1

From figure

Sin 48° =  

Or, 0.74 =  

∴ AC = 0.74 d             ..........2

Similarly

Sin 42° =  

Or, 0.66 =  

∴ CK = 0.66 d               .............3

Putting eq 2 and 3 value into eq 1

i.e  AC × CK = 100  

Or, 0.74 d  × 0.66 d = 100  

Or, 0.4884 × d² = 100  

∴ d² =

Or, d² = 204.75

Or, d =

Or, d = 14.30

So, The diameter of circle = d = 14.30 cm

Now, Radius of circle =

Or, r =  

i.e r = 7.15 cm

So, The Radius of circle =  r = 7.15 cm

Hence, The Radius of circle is 7.15 cm  Answer

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.