Math Analytical Geometry word problems

Question

Given:  Co-ordinates of garden: A(-2, -2), B(6, 2) and C (2, 4).

1.Verify that the garden is a non-right scalene triangle. 

2.Determine the height of the triangle from vertex C.

3.Determine the area of the garden if 1 unit = 2.3m 

Part 2:

1.Farmer Jack is interested in surrounding the garden with a fence and to add new soil inside the garden. If fencing costs $8.95/m and soil costs $4.45/m2, then how much will it cost to build his fence and to re sod the garden? 

2.Farmer Jack also would like to add an additional fence inside the garden to separate the vegetables from the berries. This fence is formed by the midsegment constructed using side lengths AC and AB.  

Show that this fence is parallel to BC and determine the cost of building this additional fence. 

3.Farmer Jack also wants to build another fence inside the garden. This fence is to be built on the median from vertex A.  How long will this fence be in metres?  

4.Two circles are centred at C(0,0). The equation of the first circle is x2 + y2 = 56.25. The smaller circle has a diameter of 12.4 units. How much bigger is the area of the larger circle compared to the area of the smaller circle? (Area=πr2)

Answer 1

AB (-2,-2) to (6,2) = sqr(64+16) = sqr80=2sqr20=4sqr5

BC (6,2) to (2,4) = sqr(16+4) =sqr20 = 2sqr5, AB=2BC

AC (-2,-2) to (2,4) = sqr(16+36) = sqr52 =2sqr13

all 3 sides are unequal, so it's a scalene triangle

if ABC were a right triangle, the longest side would be the hypotenuse, 4sqr5

and AB^2 would = BC^2 + AC^2

16(5) = ? 4(5)+ 4(13)

80 =? 20+ 52

80 does not = 72, 80>72 but it's close to a right triangle

use the law of cosines

AB^2 = BC^2 + AC^2 -2BC(AC)cosC

80=72-2(2sqr5)(2sqr13)cosC, to be true cosC<0 and C>90 degrees. it's an obtuse scalene triangle

vertex C is (2,4) side opposite C is AB=2sqr5

use Heron's formula to calculate the area,

divide the area by 2sqr5 to get the height or altitude from vertex C

4.

x^2 + y^2 = 56.25

diameter of smaller circle = 12.4 and radius = 6.2

radius of larger circle = sqr56.25

Area of smaller = pir^2 = pi(6.2)^2= 38.44pi

Area of larger = 56.25pi

difference = (56.25-38.44)pi = 17.81pi = how much larger the larger circle is compared to the smaller circle