9th grade Geometry:?

Question

9th grade Geometry:
In searching an ancient ruin an archeologist found a piece of the rim of an old wheel. In order to restore the wheel he needed the diameter. He marked three points, A, B, C, on the rim so that AB=AC. If AB=15 and BC=24 what the diameter of the wheel?

Answer 1

Drop a perpendicular from A to the midpoint of BC. Call that point M. Then

CM = MB = 24/2 = 12.

Now we have a right triangle AMC where AC is the hypotenuse. We have:

AM² = AC² - CM² = 15² - 12² = 225 - 144 = 81
AM = 9

Call the center of the circle O. Now we have another right triangle MOC where OC is the hypotenuse and the radius r.

OM = r - AM = r - 9

We have:

OC² = CM² + OM²
r² = 12² + (r - 9)²
r² = 144 + r² - 18r + 81
18r = 225
r = 225/18 = 25/2

The diameter d is 2r.

d = 2r = 2(25/2) = 25

Answer 2

Wheels are circles, so the points on what would have been the wheel's rim would also be the points on a circle's edge. "B" is in the middle with A on one side and C on the other. Here's how I did it.

Draw the following:
- Draw a cricle with the center labeled as "D"
- Mark points B, A, and C along the top edge of the circle
- Connect the dots to make BA, AC, BC, and three lines coming from the circle's center: DB, DA, DC
- Call "E" the point where BC and AD cross.

Note that triangle ABE is a right triangle. The hypotenuse is 15, and the bottom leg (BE) is half of the length of BC, so it's 12. Using Pythagoras, you can find that AE = 9.

Notice that BD and AD are both radii of the circle. Label BD as "r". That means DE is r - 9. Now notice that BED is also a right triangle, where the hypotenuse is r, the top leg is 12, and the side leg is r-9. So apply Pythagoras and do some simplification:

(r-9)^2 + 12^2 = r^2
(r^2 - 18r + 81) + 144 = r^2
-18r + 81 + 144 = 0
225 = 18r
r = 12.5, so the diameter of the wheel was 25 units.

Answer 3

You have a triangle with AB = AC = 15 and BC = 24. if D is the midpoint of BC then (AD)^2 = 225 - 144 = 81, or AD = 9 Because AD bisects BC perpendicularly, the center must lie on AD. The perpendicular bisector of AC intersects AD at some point O, which will be the center of the wheel. The triangle formed is similar to ACD, so that 7.5/9 = R/15. R = 15*7.5/9 = 12.5, and the diameter of the wheel is 25

Answer 4

You really need to draw this one out to see it. So I'll try to make it clear..

BAC forms a triangle with the two sides of length 15 around the perimeter of the circle and the base, at the interior of the circle being 24.

Drop a vertical between this triangle, from the apex to the midpoint of the base (distance of 12 from the perimeter, we'll call it M) and onwards through the circle.

This line splits the triangle into two mirrored right angled triangles and the line continues through the center of the circle.

Do a bit of trig on one of the of right angled triangles, say ACM, with theta being the angle at the corner at the apex of the triangle.

sin(theta) = o/h = 12/15 = 0.8

theta = sin^-1(0.8) = 53.13 degrees.

Now draw another line from the midpoint of the hypotenuse (the side of length 15 of the triangle) through the center of the circle. This will be perpendicular to the length 15 side.

If we call the center D. This forms another right angled triangle ACD (or ABD if you used that side). The new corner meets at the center of the circle.

Using the known angle theta in this triangle, the hypotenuse is the radius r and the adjacent side is length 15/2.

So the trig function that invokes a and h is cos as in
cos(theta) = a/h = (15/2)/r

r = (15/2) / cos(theta)
= 15/2 / cos(53.13) = 7.5 / 0.6 = 12.5

the diameter = 2*r = 25

Answer 5

it depends at ur school. At my school I took Algebra 1 in 9th grade but then took Geometry over the summer so now I'm taking Algebra 2 in 10th grade. At my school it is advanced.

Answer 6

(x-x1)^2 + (y-y1)^2 = r^2 >>>EQ. A (equation of a circle)
draw your triangle making AB & AC as your leg of the triangle... get the coordinate of your vertices, (0,0 ; 12,9 ; 24,0)

substitute each coordinate to EQ. A
result:
x^2 + y^2 = r^2 for 0,0
x^2 = r^2 - y^2 >>> EQ. 1

x^2-24x+144+y^2-18y+81=r^2 for 12,9 >>> EQ. 2

x^2-48x+576+y^2 for 24,9 >>> EQ. 3

substitute EQ.1 to EQ. 3 you'll get x=12

substitute EQ. 1 to EQ. 2 you;ll get 24x + 18y = 225 then substite x = 12: y = -3.5

substitute x=12 & y=-3.5 to x^2 + y2 = r^2
r = 12.5
therefore diameter = 25

to check the values, use AutoCAD (software):
draw your triangle, from the menu click draw>circle>3pts then click the 3 vertices of your triangle... measure the diameter...
use dimlinear command.....

Answer 7

Drawing it out, i get

D = 2(a/[2 sin(A)])
D = a/(sin(A))

a = AB
b = AC
c = BC

a^2 = b^2 + c^2 - 2bc(cos(A))
15^2 = 15^2 + 24^2 - 2(15 * 24)cos(A)
-576 = -720cos(A)
.8 = cos(A)
A = cos^-1(.8)

D = 15/sin(cos^-1(.8))
D = 25

The Diameter is 25