CPhill

Area....

We have a quadrilateral with two parallel sides AD and BC...so....this is a trapezoid

The area is

(1/2) height of trapezoid * ( sum of the base lengths)

The bases are DC and AB.....we know these

The height is the tricky part....it is found as:

[ The y coordinate of B - The y coordinate of C ] =  [  1 - (-5) ] = [ 1 + 5] = 6 units

So.....the area is

(1/2) (6) ( 3 + 2√5 )   ≈    22.42 units^2

And that's it....!!!

One of my favorite tunes......!!!!!

AD is correct !!!

And you got BC correct.... we can simplify it as   3√5

Let's look at DC.....

√ [ ( -4 - (-1) )^2 + ( -5 - (-5) )^2 ] =  √ [ ( -3)^2 + ( 0 )^2] = √9 = 3

So.... the perimeter is

AB + AD + DC + BC =

2√5 + 2√5 + 3 + 3√5 =

7√5 + 3   units ≈  18.65 units

You did pretty well....!!!!

The area is a little tricky, RP.....I'll get back to it in just a bit.....I have a few more questions to get to....just hold on...!!!

We'll try, Toto.....!!!

Since SR parallel to PQ....they are equal

So....to find PS = QR we have

24 - 2SR = PS + QR

24 -2(5) = PS + QR

24 - 10 = PS + QR

14 = PS + QR

So  QR = 7 = PS

And the area of the parallelogram PQRS =  QR * ST

So

28 = 7 * ST    divide both sides by 7

4 = ST = OQ

And using the Pythagorean Theorem

RT =  sqrt [ RS^2 - ST^2] =  sqrt [ 5^2 - 4^2] = sqrt [ 25 - 16] = sqrt[9] = 3 = OP

So....the perimeter of QOST =

2 (OP + PS) + 2(ST) =

2 (3 + 7) + 2(4) =

2*10 + 2*4 =

28 units

a.) x^3 + 3x^2 - x = 3 

Subtract 3 from both sides

x^3 + 3x^2 - x - 3  = 0

Factor by grouping

x^2 (x + 3) - 1(x + 3) = 0

(x^2 - 1) ( x + 3) = 0

(x + 1) ( x - 1) ( x + 3) = 0

Setting each factor to 0 and solving for x produces

x = -1    or    x = 1    or     x = -3

b.) b^3 - 4b^2 -21b = 0

Factor as

b ( b^2  - 4b - 21) = 0

b (b - 7) ( b + 3) = 0

Setting each factor to 0  and solving for b produces

b = 0  or  b  = 7    or    b = -3

What is the area???

The perimeter is all solved using the distance formula, RP

The formula for the distance between two points is given by

√ [ (subtract the x coordinates of the two points)^2 + (subtract the y coordinates)^2 ]

So....for instance....the distance between A and B is

√ [ ( -2 - 2)^2 + ( -1 -1)^2 ] = √ [ (-4)^2 + (-2)^2] = √ [16 + 4] = √20 = 2√5

Now..using this example..compute the distances between

A and D

D and C

C and B

Add these four distances....this will be the perimeter

Do this part first....then....I will show you how to calculate the area

Let me know if you run into trouble......

I think you want the dimensions of the fence

Let the two sides perpendicular to the river be = x

The side parallel to the river is then [ 1000 - 2x] 

So, the area A is given by:

A = x [1000 - 2x]  =  -2x^2 + 1000x

The x value that maximizes this is =     - (-1000) / [ 2 * -2] = 1000/4 =  250 

So....the dimesions of the fence are    250 [1000 - 2*250] = 250 x 500

And the max area is  250 ft x 500 ft =   125000 ft^2

HAHAHA!!!!......me too, Synth......

[ This probably wasn't a "real" question ]