An isosceles triangle with equal sides of 5 inches and a base of 6 inches is inscribed in a circle. What is the radius, in inches, of the circle? Express your answer as a mixed number.

AdminMod2 Aug 18, 2017

#1**+3 **

An isosceles triangle with equal sides of 5 inches and a base of 6 inches is inscribed in a circle. What is the radius, in inches, of the circle? Express your answer as a mixed number.

Referring to my diagram.

\(\theta+\alpha =\pi\;\;radians\\ \alpha = \pi-\theta\\ so\\ sin\alpha=sin\theta\\ cos\alpha = -cos\theta\\ \text{I am going to use the identity }cos^2\theta+\sin^2\theta=1\\\)

\(\text{Using Cosine Rule}\\ 5^2=r^2+r^2-2*r*r*cos\theta\\ 25=2r^2-2r^2cos\theta\\ \frac{25-2r^2}{-2r^2}=cos\theta\\ cos\theta=\frac{25-2r^2}{-2r^2}\\ cos^2\theta=\frac{4r^4-100r^2+625}{4r^4}\\~\\ sin\alpha=\frac{3}{r}=sin\theta\\ sin^2\theta=\frac{9}{r^2}=\frac{36r^2}{4r^4}\\~\\ Sin^2\theta+cos^2\theta=\frac{36r^2+4r^4-100r^2+625}{4r^4}=1\\ 36r^2+4r^4-100r^2+625=4r^4\\ -64r^2+625=0\\ 64r^2=625\\ r=\frac{25}{8}=3\frac{1}{8} \)

The radius of the circle is 3 and 1/8 inches.

Melody Aug 19, 2017

#2**+2 **

Thanks, Melody.....here's another way that is non-trig based.....

Look at the following image :

Let's forget about the circle for a moment - we'll return to this in a second.....

Let us place the side with length 6 along the x axis such that its endpoints lie at (-3,0) and (3,0)...let's label these as A and C

Let the altitude of the triangle lie along the y axis......this will actually form two right triangles.....each with a hypotenuse of 5

So....using the Pythagorean Theorem the altitude of our triangle will be sqrt [ 5^2 - 3^2] = sqrt [ 16] = 4

And let us call the altitude BD

Now....the center of our circle will lie on AD....and let us call this point (0, y)

And.......the radius of our circle will pass through all three vertex points of triangle ABC.....this means that the distance from (0, y) to B will equal the distance from (0,y) to C...and B has the coordinates (0, 4) and C has the coordinates (3,0)

So...using the distance frommula, we have that

sqrt [ ( 0 - 0)^2 + ( y - 4)^2 ] = sqrt [ (0-3)^2 + (y - 0)^2] squaring both sides and simplifying, we have that

(y - 4)^2 = 9 + y^2

y^2 - 8y + 16 = 9 + y^2 subtract y^2 from both sides

-8y + 16 = 9 add 8y to both sides, subtract 9 from both sides

7 = 8y divide both sides by 8

7/8 = y.... so the center of our circle will lie at (0, 7/8)

And the distance from this point to D is the radius = 4 - 7/8 = 32/8 - 7/8 = 25 / 8 = 3.125 = 3 + 1/8

And the equation of the circle will be x^2 + (y - 7/8)^2 = (25/8)^2

CPhill Aug 19, 2017

#4**0 **

Here's another (trig based ) method.

The sine rule says that \(\displaystyle \frac{a}{\sin A} = \frac{b}{\sin B}=\frac{c}{\sin C} = 2R\),

where \(\displaystyle \text{R}\) is the radius of the circumscribing circle.

So, using the notation in Chris's diagram, \(\displaystyle \frac{a}{\sin A} = \frac{5}{4/5} = \frac{25}{4}=2R\), so \(\displaystyle R = \frac{25}{8} = 3\frac{1}{8}\).

Tiggsy.

Guest Aug 20, 2017

#5**0 **

It is really good to see you Tiggsy :)

I probably just have a brain freeze but where did the 2R come from?

I mean it is not a part of the sine rule :/

Melody
Aug 20, 2017

#6**0 **

Hi Melody

The full version of the sine rule is as I stated it.

It's actually derived using the circumscribing circle.

(Join the vertices of the triangle to the centre of the circle O. Then, angle BOC = 2*angle A, and from (half of) the isosceles triangle BOC, sin(A) = (a/2)/R, so a/sin(A) = 2R. Repeat for B and C.)

Tiggsy.

Guest Aug 20, 2017