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Ah....thanks, Rom....I didn't think about that   !!!

Thanks x-squared :)

Of course, how silly of me, I should have thought of that!

Suppose we make a circular fence.

$C = 100 = 2\pi r\\ r=\dfrac{C}{2\pi}\\ A = \pi r^2 = \pi \left(\dfrac{C}{2\pi}\right)^2 = \\ \dfrac{100^2}{4\pi} = \dfrac{2500}{\pi} \approx 796~ft^2$

Call the lengths of the two legs x and y

We have this system of equations

x^2 + [(1/2)y]^2 = 25   ⇒  x^2 + (1/4)y^2 = 25    (1)

[(1/2)x]^2 + y^2 = 40    ⇒  (1/4)x^2 + y^2 = 40   (2)

Multiply equation (1)  through by -4 and we have this system

-4x^2   - y^2 =  -100

(1/4)x^2 + y^2 = 40       add these

-(15/4)x^2 = - 60

x^2 = 16

x = 4

And using (2) to find y, we have

(1/4)(4)^2 + y^2 = 40

4 + y^2 = 40

y^2 = 36

y = 6

So....the hypotenuse length is   sqrt (4^2 + 6^2 )  =  sqrt (52) = 2sqrt  (13)

And in a right triangle, the median drawn from the right angle is 1/2 the hypotenuse length = 

sqrt (13)

For any perimeter,P,  the largest area that can be enclosed is a square with a side of P/4

So....the max area is

(100/4)^2 =  25^2   =   625 ft^2

Floor area of 1 drum =

 pi * ( 1.25)^2  =    1.5625 pi  ft^2     (1)

Total Floor Area = 7 * 7.5 = 52.5 ft^2   (2)

Number of drums that can be placed =  (2) / (1) =  

52.5 / [ 1.5625 pi ]  ≈  10 drums

6. 

 Line NM is the median of trapezoid ABCD. If the area of NMCD=2* the area of ABMN what is CD/AB?

Area of NMCD = 2 * Area of ABMN = (1/2)h ( DC + NM) ⇒ Area of ABMN =(1/4)h (DC + MN)   (1)

Area of ABMN =  (1/2) h (AB + MN)    (2)

Equate (1), (2)

(1/4)h (DC + MN) = (1/2)h (AB + MN)

DC + MN =   2 (AB + MN)

DC + MN = 2AB + 2MN

DC +  ( AB + DC)/2 =  2AB + 2 [ (AB + DC)/ 2 ]

(3/2)DC + (1/2)AB = 2AB + AB + DC

(3/2)DC + (1/2)AB = 3AB + DC

(1/2)DC = (5/2)AB

DC = 5AB

CD / AB =  5

Hi Alan...great job thanks!

Prior to posting question and after long hours of problem solving I did obtain 36 deg and 108 degrees. My main issue was applying the standard formulas for the general angle...in this case cosine. Therefore I am now trying to see how I can reconcile those answers....hopefully I can solve !

All the best for Xmas of course and New Year....thanks again! 

$k \in \{1,2,3,4,5\}\\ 3x = k \pmod{6} \\ \text{I guess you are assuming }x \text{ is an integer as well?}$

$x \text{ even } \Rightarrow 3x \pmod{6} = 0\\ x \text{ odd } \Rightarrow 3x \pmod{6} = 3\\ \text{ so }k = 1,2,4,5\\ \text{has no solutions for }x \in \mathbb{Z}$

5. Trapezoid ABCD has bases line AB and line CD. The extensions of the two legs of the trapezoid intersect at P. If the area of ABD=3 and the area of BCD=7, then what is the area of PAB?

Triangles ABD and BDC are under the same heights

Area of BDC        7        (1/2) h *DC

__________  = ___ =   _________

Area of ABD       3        (1/2) h * AB

So.....the ratio of base DC to base AB is  7 : 3

But triangle PAB is similar to triangle PDC

So...Area of triangle PDC : Area of triangle PAB =  7^2 / 3^2 = 49 / 9

So area PDC - area of ABCD = area of triangle PAB

(49/9)[area of PAB] - 10 = area of triangle PAB

(49/9) [area of PAB] -area of PAB = 10

(40./9)[area of PAB] = 10

area of PAB =  10 (9/40) =   9/4 units^2 = 2.25 units^2

Please help me, why tan(90)=Infinity?

I typed this in calc: $tan(90)$

and it was equal to $infinity$.

$\text{The front surface is }15\times 4 = 60 ~ft^2\\ \text{The back surface is }15\times 10 = 150 ~ft^2\\ \text{The two sides are a }10\times 25~ft^2 \text{ rectangle minus a }\\ 6 \times 25 ~ft \text{ right triangle}\\ \text{Each side is thus }250 - 75 = 175~ft^2\\ \text{summing these we get }60+150+2\times 175 = 560~ft^2$

it is not 54

This is actually a pretty neat problem as we don't seem to be restricted to rectangles.

Do you know which planar shape has the largest area to perimeter ratio?

Think about it and when you figure it out the answer to this is pretty simple.

$\text{The triangle will be acute for }\\ 3 < x < \sqrt{3^2 + 6^2} = \\ 3 < x < 3\sqrt{5}\\ \text{That's a continuum of values so asking how many there are is sort of meaningless}$

Sorry...slight math error....see my corrected answer....!!!

3.  

ABCD is a trapezoid with bases line AB and line DC. We know that Angle C=60 degrees and Angle D=45 degrees. If AB=5-2sqrt(3) and BC=4, what is DC?

       A              B

D                                 C

Draw BE perpendicular to  DC

Angle BEC  = 90°

Angle EBC = 30°

Then, because BEC s a 30-60-90 right triangle, then EC = (1/2)BC = 2

And BE = 2sqrt(3) = height of trapezoid

Draw AF perpendicular to DC

AF = 2sqrt(3)

And angle DAF = 45°

So...we have a 45 - 45 - 90 right triangle with AF = DF = 2sqrt(3)

So...DC =  EC + EF + DF   ....and EF = AB....so...

DC =  [ 2]  + [ 5 - 2 sqrt (3) ] + 2sqrt(3) =   2 + 5 =   7

i don't think 108 is correct or 54 for number 4

4. 

The ratio of the four sides of a trapezoid is 2:1:1:1 The area of the trapezoid is 27sqrt(3) Find the length of the median of the trapezoid.

Let the bottom base be the longest side = 2S

Let the top base = S

Using the Pythagorean Theorem, the height, h, can be found as

sqrt [1 - (1/2)^2] = sqrt [ 3/4 ]  = sqrt(3)/2

So....using the formula for the area of a trapezoid, we have

27sqrt (3) = (1/2)h ( 2S + S)

27sqrt (3) =   (1/2) (sqrt(3)/2 ) (3S)

27 = (3/4)S

S = 36

The length of the median =   (sum of bases)/2 =  (3/2)S =  (3/2)*36  = 54  

Thanks!

1.

Angle EAB = Angle CBA = y

And because ABCE is a parallelogram, we have that

And EAB + DBA = 180

And DBA = y - 28

So

y + ( y - 28) = 180

2y - 28 = 180

2y = 208

y = 104°  =  EAB =  BAE

This is the "factorial" function

n!  =  the product of the first n positive integers

Example

3! =  1 * 2 * 3 =     6

y = 3(x-h)^2 + 8  and    y = mx + 17   

Since the point (1,11) is on both graphs, this implies that

11 = m(1) + 17

-6 = m

And also

11 = 3(1 - h)^2 + 8

3 = 3(1 - h)^2

1 = (1 - h)^2

Implies that h = 0    or    h = 2

A look at graph here, Drazil  confirms that  m = -6   and h = 2 are correct :

https://www.desmos.com/calculator/7iezpmqnrx