Question

The vertices of a right angled triangle are on a circle of radius R and the sides of the triangle are tangents to another circle of radius r If the
engths of the sides about the right angle are 16 and 30 determine the value of R+r​

Answer 1

Hi,

Answer: 23

Step-by-step explanation:

From the figure attached below, we can see that the vertices of the right-angled ∆ ABC are on the bigger circle such that the side AC forms the diameter of the circle, therefore,  

R = ½ * diameter = ½ * AC …… (i)  

Lengths of the sides of the right-angled ∆ ABC are given as 16 and 30, so let AB = 30 & BC = 16.

Step 1:

Applying Pythagoras theorem to right-angled ∆ ABC, we get

AC² = AB² + BC²

⇒ AC = √[30² + 16²]  

⇒ AC = √[1156]

⇒ AC = 34

Substituting the value of AC in eq. (i)

R = ½ * 34 = 17 ….. (ii)

Step 2:

It is also given to us that the sides of the right-angled ∆ ABC are tangents to another circle with radius “r”.

Here, at first, we will calculate the area of the right-angled ∆ ABC. Secondly, we will calculate the area of right-angled ∆ ABC for every side of the triangle considering radius (r) of the circle as height.  

∴ Area of ∆ ABC = ½ * AB * BC = ½ * 30 * 16 = 240 ….. (iii)

And,

∴ Area of ∆ ABC  

= area(∆ AGB) + area(∆ AGC) + area(∆ BGC)

= [½ * 30 * r] + [1/2 * 34 * r] + [1/2 * 16 * r]

= 40 r …….. (iv)

Equating eq. (iii) & (iv), we get

40 r = 240

⇒ r = 6 …… (v)

Step 3:

From eq. (ii) & (v)substituting the values of R & r, we get

The value of (R + r) is,

= 17 + 6

= 23

Thus, the value of R+r = 23.

Hope this is helpful!!!!!