CPhill

If we have an isosceles right triangle....the legs will be of equal length

AC is one leg and AB  is the other

And they both equal 6cm

And they form the base and height of the triangle

And the area of a right triangle  =

  [ Base * Height} / 2  = [ Product of the leg lengths ]  / 2   =

[6 * 6 ] / 2  =

36 / 2  =

18 cm^2

P               Q

6      X

S      8         R

This is similar to your other problem

Triangle PSR  forms a  6-8-10  Pythagorean Right triangle with PR   = 10

So....PX  =  SX  =  (1/2) (10) =  5

Using the Law of Cosines

PS^2  = PX^2 + SX^2  - 2(PX * SX)cosPXS

6^2 = 5^2 + 5^2  - 2 (5 * 5)cosPXS

36 - 2(5)^2

_________  =  cos PXS

 -2(5)^2

36  - 50

______    =   cos PXS

  -50

50 - 36                           14            7

_____   =  cos PXS  =   ___ =     ___

   50                               50           25

Since  PXS is acute.....then its sine will be positive  and we can find it as

sin PXS  = √ [ 1 - cos^2 PXS]  =  √ [ 1 -  (7/25)^2 ]  = √ [ 1 - 49/625]  =  √ [625 - 49]/ 25  =

√576/ 25   =    24 / 25

P                    Q

10       A

S       24         R

We can use the Pythagorean Theorem to   find PR  =

√ [ SR^2 + PS^2]   = √ [ 576 + 100]  = √676  =  26

And  (1/2) of this  = PA  = SA  =  13

So.....using the Law of Cosines

PS^2  = PA^2 + SA^2  - 2(PA * SA) cos PAS

10^2 = 13^2 + 13^2 - 2(13 * 13) cos PAS

100 - 2(13)^2

___________  =   cos PAS

  -2 (13^2)

2(13)^2  - 100

___________   =   cos PAS

   2(13)^2

338 - 100

________    =    cos PAS

  338 

238                                119

___    =   cos PAS =     ____

338                                169

Let r be the common ratio

Let the first term  = a

So.....the third term is   ar^2

And the sixth term is  ar^2

So.....we have this ratio

ar^2          (4/3)   

____  =    _______

ar^5          (-32/27)

r^2           (4/3)

___  =     ______       and we can invert these fractions and write

r^5          (-32/27)

r^3   =  (-32/27)  / (4/3)

r^3  =  (-32/27) * ( 3/4)   =  (-8) / (9)

r =  (-8/9)^(1/3)  =   -2 / (9)^(1/3)   =   - 2 / (3^2)^(1/3)  =  - 2 / (3)^(2/3)

So....to find the second term, we have

(2nd term) * r  =  (3rd term)          divide both sides by r

(2nd term )  =    (3rd term ) / r

(2nd term)   =   (4/3)  / [  -2 / (3)^(2/3) ]  =  (4/3) * (3)^(2/3) / -2  = 

(4 / -2)  * ( 3^(2/3) / 3 )  =

-2 / (3)^1/3 

2x^2 -2x + 4 = 0       divide through by 2

x^2 - x + 2  = 0        subtract 2 from both sides

x^2 - x  =  - 2       complete the square on x

x^2 - x + 1/4  =  -2 + 1/4

(x - 1/2)^2  = -7/4       take both roots

x - 1/2  = ±i√7 / 2        add 1/2 to both sides

So

x = [ i√7 + 1] / 2     =  a

a^2  =  [ 1 + √7 i] [ 1 + √7 i ] / 4  =  [ 1 + 2√7 i - 7] / 4  =  [ √7 i - 3 ] / 2

Or

x = [ -i√7 + 1] / 2  =    b

b^2 =  [ 1 - √7 i ]  [ 1 - √7 i ] / 4 =  [ 1 -2√7 i  -  7 ] / 4  =  [ -√7 i - 3] / 2

a^2 + b^2   =   -6/ 2  =  -3

ab  =   [  1 +  i√7 ] / 2  *  [  1  - i√7 ] / 2   =  [ 1 + 7] / 4  =  2

So    ( x - a/b)  (x - b/a)  =

x^2 - (a/b)x   - (b/a)x +  (a/b)(b/a)  =

x^2 - ( a/b + b/a)x + 1   =

x^2 - ( a^2 + b^2) x / ab + 1  =

x^2 -  (-3)x / 2 + 1   =

x^2 + (3/2)x + 1

Let the side of the tetrahedron   =   s

Since  each face is an equilateral triangle......the slant height, h is given by

sin (60°)  = h / s ⇒     h = * sin (60°)  =  √3/2 * s = √3/2 

And he distance from M to the center of the triangle is given by

tan (30°) = C / [ (1/2)s ] ⇒  C = (1/2)s * (1/√3)  =  1 / [ 2√3]  =  √3/6 *s   (1)

And....using the Pythagorean Theorem, we can find the height of the triangle as

√ [  ( √3/2* s)^2  - (√3/6*s)^2 ]  =

s√ [ 3/4  - 3/36 ]  =

s√ [ 27 - 3] /36] =

s√[24/36] =

s√[2/3]     (2)

So....the tangent of AMB  =  (2) / (1)  =  s√[2/3]  / [ √3/6 *s] =

√2              6               6√2          

__    x      ___   =      _____  =       2√2     

√3             √3               3 

3a  = 30  ⇒  a  = 10

3b = 15 ⇒ b = 5

3c = 24 ⇒  c = 8

So

c + b x a   =   8 + ( 5 x 10)  = 8 + 50   =  58

Here's  12

Assuming that  we are dealing with positive integers....let the odd numbers  be

2n - 1   and 2n + 1

Adding these together we get   4n

Sometimes.....in these kinds of proofs, it is good to see if we can discover a pattern....

Let n = 2

And the sum of the odds =  8

And the difference of two squares that equals this  is 3^2 - 1^2 = 9 - 1 = 8

We can write this as  [ 2n - 1 ]^2 - [ 2n - 3]^2

Note that the numbers in red are just  (n - 1) and (n + 1), respectively

Now......Let  n = 3

And the sum of the odds = 12

And the difference of two squares that equals this is 4^2 - 2^2  = 16 - 4  = 12

Which we can write as [ 2n - 2 ]^2 - [ 2n - 4 ]^2 

Note that, once again, the numbers in red are just ( n - 1)  and  ( n + 1), respectively

So....it appears that we can write the larger  square  as   [2n - (n- 1)]^2   =  [ n + 1]^2

And it appears that we can write the other square as  [ 2n - (n + 1) ]^2  =  [ n - 1]^2

And their difference is

[ n + 1]^2  -   [ n - 1]^2   =

[ n^2 + 2n + 1 ]  - [ n^2 - 2n + 1]  =

4n

5. Square ABCD has side length 60. An ellipse E is circumscribed about the square and there is a point P on the ellipse such that PC = PD =50. What is the area of E?

This one is a little difficult !!!!

Let the circle and the ellipse be centered at the origin

We can let the vertices of the square be  A = (-30,30) , B =(30,30), C = (30, -30)  and D = ( -30, -30)

The mid point of the bottom of the square is the point (-30, 0)

Call this point, M

And  PC = PD = 50

So....we can form right triangle MPC  such that PC forms the hypotenuse = 50

And MC = 30

So....MP  is the other leg  =  √ [ PC^2 - MC^2] = √[ 50^2 - 30^2]  = √ [ 2500 - 900 ] =

√1600 = 40

So......the distance from the origin to M  = 30

And the distance between M an P  =  40

So....using symmetry.....we can let the vertical axis of the ellipse  = 2* (30 + 40) = 140  = 2b

So......in the equation

x^2            y^2

___   +     ___    =  1

 a^2          b^2

We know that one point on the ellipse is (30, 30)  = (x , y)

And "b"  is (1/2) * 140  = 70

So.....we can find "a" as follows

30^2           30^2

_____  +    _____   =      1

  a^2            70^2

900            900

____ +      _____  =    1

a^2            4900

900         9

___  +   ____  =    1

a^2         49

900                   9

____  =   1  -    ___ 

 a^2                   49

900             [ 49 - 9]

___  =       _________

a^2                 49

900                 40

____   =        ____

 a^2                49

So

40 a^2 =  900*49

40a^2 = 44100

a^2  =  44100 / 40

a^2 = 4410/4

a = √ [ 4410 / 4 ]

And  the area of the ellipse   =

pi* a * b  =

pi * √[ 4410/ 4} * 70  ≈  7301.9   units^2

Here's the graph : https://www.desmos.com/calculator/r9mwhqlg2e

4. A circle centered at the origin intersects the ellipse  x^2/9 + y^2/16 = 1   at the four vertices of a square. Find the area of that square.

Since the diagonals of a square are at right angles and the center of the elliipse is at (0,,0).....the lines y = x  and y = -x    will form diagonals that intersect at (0,0) and will be at right angles to each other

To find out where  y = x intersects the ellipse.....let us solve this

[ sub x for y  in the ellipse equation ]

x^2/9 + x^2/16  = 1       multiply through by 144

16x^2 + 9x^2  = 144

25x^2  = 144

x^2   =  144 / 25    take both roots

x = ±√ [144/25]   = ±12/5   =  ±2.4

And siince  y = x     and y = -x the coordinates  of the top  vertices of the square are

(2.4, 2.4) , ( -2.4, 2.4)

The distance between the points  is 4.8  =  the side of the square 

So....the area of the square  is  side^2   = 4.8^2   = 23.04 units^2

Here's a graph : https://www.desmos.com/calculator/uld9mzxaet