Geometry Help!?

Question

Please help me! These are some review problems for our final exam.

1. If the area of triangle GIC/the area of triangle AIM=2.25, find the ratio of the perimeters.

2. If a 12degree angle is inscribed on an arc of length 4pi, what's the area of the inscribed equilateral triangle?

3. Find the area of a circle that can be inscribed in an equilateral triangle with area 36root3 inches squared.

Thanks!

Answer 1

1. Take the square root of the ratio of the areas to find the linear ratios for any two similar figures (and perimeters are linear ratios)

2. The arc degree measure is twice the inscribed angle or 24 degrees. Using C for the circumference of the circle solve the proportion C/360 = (4 pi)/24

24C = 1440 (pi)

C = 60 (pi)

Find the radius of the circle

C = 2(pi)r
60(pi) = 2(pi)r
30 = r

Using a radius of 30, draw in three radii with 120 degree central angles and connect the endpoints to create an equilateral triangle

The area of each 120 triangle = (1/2)absin120 or 1/2(30)(30)(Sqrt3)/2

Multiply by 3 to get the area of the whole triangle

Area = 2700 SQRT(3) / 4 = 675 SQRT(3)

3. The length of the apothem of the equilateral triangle is the radius of the inscribed circle.

Using a for the apothem, then the area of the triangle
is 3a^2 SQRT(3)

3a^2 SQRT(3) = 36 SQRT(3)
3a^2 = 36
a^2 = 12
a = 2 SQRT (3)

The area of the circle is pi(a)^2 = 12pi.

Answer 2

1) Area is based on two dimensions, while perimeter is based on one. (It's the difference, say, between square inches and inches.)
If the ratio of areas is 2.25 : 1, then the ratio of perimeters is
√(2.25) : √1 = 1.5 : 1, or 3 : 2.

2) An inscribed angle measures half of the arc. This means you've a 24° arc with length 4π. (24° / 360° = 1 / 15 of the circumference.) The circumeference of the circle is 15(4π) = 60π. This makes the radius 30. Drawing an equilateral triangle and subsequent 30-60-90, you'll find the triangle to have side lengths of 30√3, and the height is 45. The area = ½(b)(h) = ½(30√3)(45) = 675√3.

Sorry I have to run for now, but maybe someone else can help with problem 3 for you. Good luck!

Answer 3

The answers given so far only work if triangle GIC is similar to triangle AIM. Otherwise, all bets are off.