Geometry: Triangle with a Circumscribed and Inscribed Circle

[Mathmaniac123]

Homework Statement

What is the area of a right triangle whose inscribed circle has radius 3 and whose circumscribed circle has a radius 8?

Homework Equations

The diameter must be the hypotenuse of the circle

The Attempt at a Solution

The answer is 57, but I do not know the steps to achieve it.

I tried making the triangle a 45° 45° 90°, but the area comes out to be 64.(since the incircle has a radius of 3, the triangle would not work)

I tried a 30° 60° 90° triangle, but the area comes out to be 55. The answer is closer, so that means the the angles are a little bit closer together.

Since the answer is 57, the base x height would be 114. The closest factors would be 6x19, but 19 is larger than the hypotenuse, which makes me think that either the answer is wrong, or the b, and/or h are decimals or fractions.

Good Luck

 

[haruspex]

You can approach it by algebra or geometry.
For a geometrical approach, notice that joining the incentre to each vertex and dropping perpendiculars from the incentre to each side cuts the triangle into 6 right angled triangles. Each of these has one side length r, the radius of the incircle. So they can be rearranged in pairs to form rectangles width r. What do the other sides of these rectangles add up to?

 

[Equilibrium]

You can approach it by algebra or geometry.
For a geometrical approach, notice that joining the incentre to each vertex and dropping perpendiculars from the incentre to each side cuts the triangle into 6 right angled triangles. Each of these has one side length r, the radius of the incircle. So they can be rearranged in pairs to form rectangles width r. What do the other sides of these rectangles add up to?

may i take the OP's place?
im not sure about this problem and I am just referring to the image found on google image


1. it seems that one side of the triangle is the diameter of the circumscribing center so AC = 8*2 = 16
let
L_{B} = Length of the center of the inscribed circle to the vertex B, or the incenter

2. the bottom left portion of the image seems to form a square with sides 3
so
{L_{B}}^{2}=3^{2}+3^{2}
L_{B}=4.243

3. the inscribing circle radius to side c seems to be aligned with L_{B} which makes it 3 + 4.243 = 7.243

4. Now i get to find side b with angle cBC equal to 45 degrees
\cos{45}=\frac{7.243}{b}
b=10.243

5. Angle BCA is 45 degrees then we can find area
A_{triangle}=\frac{1}{2}cb\sin{\theta}=\frac{1}{2}(18)(10.243)\sin{45}=65.186 \mbox{ square units}

The Answer is not 57 as stated in op though, so my approach is wrong

 

[haruspex]

2. the bottom left portion of the image seems to form a square with sides 3

I think you mean bottom right, but it isn't true even then. It only looks like that because of the way you've drawn it.

 

[SammyS]

I think you mean bottom right, but it isn't true even then. It only looks like that because of the way you've drawn it.

The "bottom right figure" formed by the two radii, and the portions of the legs of the triangle that intersect each other and the radii, do indeed form a square of 9 square unit area .

Thus \ \displaystyle L_B=3\sqrt{2}\ .\ (But you really don't need to know L_B.)

 

[haruspex]

The "bottom right figure" formed by the two radii, and the portions of the legs of the triangle that intersect each other and the radii, do indeed form a square of 9 square unit area .

You're right, sorry.

 

[Equilibrium]

I already know how

first form three triangles with altitude three

total area = 1/2 (ab) = 1/2(3)(a) + 1/2(3)(b) + 1/2(3)(16) eqn 1

this will simplify to
\frac{ab}{a+b+16}=3

Then using this circle theorem
"If two segments from the same exterior point
are tangent to a circle, then the segments are
congruent."

Let R = point where radius 3 intersects