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Area:

\(= \sqrt {\frac {3a} {2} \left( \frac {3a} {2} - a \right)^3} = \sqrt {\frac {3a^4} {2^4}} = \frac {a^2} {4} \sqrt {3}\)

Circumradius:

\(= \frac {a^3} {\frac {a^2} {4} \sqrt {3}} = \frac {4a} {\sqrt {3}} = \frac {4\sqrt{3} a} {3}\)

The in radius would be  \(a\sqrt3/6\)

         P                 Q

                   4

                   X

                   8

S                                   R

Triangle  PXQ  similar to triangle RXS

Ratio of sides of RXS  to PXQ   = sqrt [ 8/ 4 ]    =  sqrt (2) 

Note that sin PXS = sin PXQ  = sin QXR

Area of triangle  PXS  =     (1/2) PX * SX sin PXS =  (1/2) PX * sqrt (2) QX  *sin PXS  =

 sqrt 2  [ (1/2) PX * QX * sin PXS ] =   sqrt (2) [ (1/2) PX * QX * sin PXQ[ = sqrt (2) [ area PXQ]

Similarly

Area of triangle QXR = (1/2) QX * XR * sin QXR  =  (1/2) QX * sqrt (2) PX * sin QXR  =

sqrt (2)  [ (1/2) QX * PX * sin QXR ] =   sqrt (2) [ (1/2) QX * PX * sin (PXQ) ]  =  sqrt (2) [ area PXQ]

So  [ PQRS ] =   [ PQX] + [ QXR ]  + 2 sqrt (2) [ PXQ]  =  4 + 8 +  2sqrt (2) [ 4 ] = 

12 + 8 sqrt (2)  =   4 [ 3 + 2sqrt (2) ]  =  4  [ 3 + sqrt (8) ] 

https://web2.0calc.com/questions/angles_103

See your other question

                             A

                                              12

        12   Z                        Y

                                                    6

B                                                    C

Since BY is a bisector

AB / AY  = BC / CY

12 / 12  = BC / 6

BC = 6

And since CZ is a bisector  let BZ = x   and AZ   = 12  - x

BZ / BC = AZ / AY

x / 6  = (12 - x) / 12

12 x  = 6 (12 - x)

12x = 72  - 6x

18x  = 72

x = 72/ 18  = 4 =  AZ

x = AC  / sqrt 2  =  2 / sqrt 2  =  sqrt 2

https://web2.0calc.com/questions/sequence_48052

4 h 36 min =   4 + 36/60  =  4.6 hrs

Let the speed of the plane in still air = S

Let the speed of the plane against the wind = S  - 40

Let the speed of the plane with the wind = S + 40

Let the distance one way = D.... so round trip =  = 2D

Distance / Rate  =  Time

We have (with no wind)

2D / S  = 4.6

2D = 4.6S

D = 2.3S

With wind

D / (S + 40)  + D / (S -40)  = 5

D (S-40) + D (S + 40) = 5 (S + 40) (S - 40)

2DS  =  5 ( S^2  - 1600)

2 (2.3)S * S  =  5S^2  -  8000

4.6S^2  = 5S^2  - 8000

.4S^2  = 8000

S^2  =  (8000) / .4 =   80000/4  =  20000

S  =  sqrt (20000) =  141.42

D =  2.3 (141.42) ≈  325.3 miles 

If three of the history books are together

They can occupy  positions 123 and  be arranged in 3! ways.....the other history book  can occupy positions 5, 6 or7

And the other three math books can  be arranged in 3! ways

So   3! * 3 * 3!   =  108 ways

They can occupy  positions  234 and  be  arranged in 3! ways.....the oher history book can occupy positions 6 or 7

And the three math books can  be arranged in 3! ways

So   3! * 2 * 3!  =  72 ways

And occupying positions  345 or 456   also  produce 2 ( 72)  arrangements   =144

And occupying  567   also produces another 108  arrangements

So  if three of the history books are together we have  2(108) +3(72)  = 432 arrangements

If all 4 of the history books  are together they can occupy positions 

1 - 4

2 - 5

3 - 6

4 - 7

For each of these, they can  be arranged in 4! ways  and the other 3 math books can be arranged in 3! ways

So....if all are together, they can  be arranged in  4 * 4! * 3!   =  3456 ways

So....the total number of arrangements =  432 + 3456  =  3888 total arrangements

7!   =  5040  ways

  30    A

                     X

90     B       12             C

Angle B = 60

AB =  12sqrt 3

AC  = 24

Since BX is a bisector...let  XC  = 24-m   and AX = m

And we have that

m / AB  = (24-m) / BC

m / (sqrt 3)  = (24 -m)

m = (24 - m) sqrt 3

m = 24sqrt 3 - msqrt 3

m+ msqrt 3 = 24sqrt 3

m ( 1 + sqrt 3)  =24sqrt 3

m = 24sqrt 3 / ( 1 + sqrt 3)  =  12sqrt3 ( sqrt 3 - 1)  = 12 [ 3 - sqrt 3 ]

[ BXA ]  = (1/2) AX * AB * sin 30°  = (1/4) AX * AB  =  12 [3 -sqrt 3 ] * [12 sqrt 3] / 4 =

3 [ 3 -sqrt 3 ] [ 12sqrt 3 ]  =  108 (sqrt 3 - 1)

Properties of Equilateral Triangle:

Since triangle DEF is equilateral, all three sides (DE, DF, and EF) are equal in length. Additionally, all three angles are also equal to 60 degrees.

Right Triangle ACB:

Given that angle ACB = 90 degrees and CD = 5 and CE = 4, we recognize this as a Pythagorean triple (3, 4, 5). Therefore, AC = 3.

Isosceles Triangle DEC:

Since DE = DC + CE and all sides of triangle DEF are equal, triangle DEC is isosceles with DE = 9.

Dropping an Altitude from A:

Draw an altitude from point A to side DE, intersecting DE at point H. Since triangle ACB is a right triangle with a 90-degree angle at C,

line segment AH is also an altitude of triangle DEC, splitting it into two congruent right triangles (ADH and CHE).

Finding AH:

In right triangle ADH, we know the hypotenuse (AD = AC = 3) and want to find the altitude (AH).

Since angle A is a vertex of an equilateral triangle, angle HAD = 60 degrees (half of an equilateral triangle's angle).

We can use the trigonometric ratio tangent (tan) for this situation: tan(HAD) = AH/HD.

Knowing tan(60) = √3 and that HD = DE/2 (since AH bisects DE), we can set up the equation: tan(60) = √3 = AH / (DE/2)

Solving for AH: AH = √3 * (DE/2) = √3 * (9/2) = 3√3 / 2

Finding AE:

In right triangle AEH, we know the altitude (AH = 3√3 / 2) and the base (HE = CE - CH = 4 - AC = 4 - 3 = 1).

We can use the Pythagorean theorem: AE^2 = AH^2 + HE^2

Substitute the known values: AE^2 = (3√3 / 2)^2 + 1^2 = 27/4 + 1 = 31/4

Take the square root of both sides to find AE: AE = sqrt(31)/2

Therefore, the length of AE is equal to sqrt(31)/2.

\(83^{\circ} \: \text{is } \frac{83}{360} \text{of the circle} \\ {r}^{2}\pi*\frac{83}{360}=60\pi\\ \underline{\text{Solving, we get}\: \bf{\frac{60\sqrt{498}}{83}}} \text{.}\)

Because BAC = 60, ABC = 60, ACB should equal 60.

Therefore, ABC is equilateral.

Because AD is an angle bisector, then ABD, is 30, and so ADB = ADC = 90.

Therefore, set BD = x.

\({x}^{2}+{24}^{2}={(2x)}^{2}\).

We get \(x = 8\sqrt{3}\). (These triangle ratios are very special, see if you can find the ratios for a 30-60-90 triangle for yourself)

Therefore the area is \(\frac{8\sqrt{3}*24*2}{2}=192\sqrt{3}\).

Set the distance to the beach as x.

\(\text{time} = \frac{\text{distance}}{\text{rate}}\).

Driving to the beach at 40mph is regularly is \(\frac{x\; \text{mile}}{40 \: \frac{\text{mile}}{\text{hour}}}= \frac{x}{40}\text{hour}\)

Driving to the beach at 15mph is \(\frac{x \; \text{mile}}{15 \: \frac{\text{mile}}{\text{hour}}} = \frac{x}{15}\text{hour}\)

Because she arrived 45 minutes or 3/4 hours later, in the second case:

\(\frac{x}{40}=\frac{x}{15}-\frac{3}{4}\).

Solving, the solution is x = 18, or 18 miles.

Let the distance driven be d miles.

We know time = distance/rate.

Traveling at 40 mph:

Time taken (t1) = d / 40 hours

Traveling at 15 mph:

Time taken (t2) = d / 15 hours

We are given that t2 - t1 = 45 minutes (convert to hours: 45 minutes = 45/60 hours = 3/4 hours)

Setting up the equation based on the given information: d / 15 - d / 40 = 3/4

Simplifying the equation: 4d - d = 120 (common denominator of 60) 3d = 120 d = 40 miles

Therefore, you drove for 40 miles to reach the beach.

The prime factorization of 2000 is 2⁴*5³.

Catherine must have rolled 3 5s, to form the 5³.

The 2⁴ is more tricky. Catherine could've rolled 4's or 2's a certain number of times to form the 2⁴, but the number of rolls to acheive this must be 3, because she rolls a total of 6 rolls.

Consider all the cases:

#1: The rolls 2, 2 and 2² in some order.

#2: The rolls 1 2² and 2² in some order.

These are the only ways she coud've rolled a 4.

The number of ways to arrange 2, 2, 2², 3, 3, 3 is 60

The number of ways to arrange 1, 2², 2², 3, 3, 3 is also 60.

Therefore there could be 60+60 = 120 sequences. 

that is unfortantly incorrect

The diagram below consists of a small square, four equilateral triangles, and a large square.  Find the area of the large square.   

At the base of the triangles, extend the line on over to the right side of the square.  

That line is 2 & ¹/₂ triangle bases long.  The base of each triangle is one unit.  

So the width of the large square is 2 & ¹/₂ units.  

Two and a half squared is six and a quarter.  The area of the large square is 6.25 units².   

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                                     Q

                        M        

                                    O  

                                    1

           P                       N                        R

                                     1

If QN perp to PR  then QNR forms a right triangle  such that

ON^2 + (PR/2)^2  =  OR^2

1^2 +  (1/2)^2  = OR^2

1 + 1/4   =  OR^2

5/4  = OR^2

QR =  sqrt (5)  /  2

https://web2.0calc.com/questions/triangle_41241

PQ / FH  

Triangle HQN  similar to triangle FQG          Triangle MPF  similar to triangle GPH

HQ / HN  = FQ / FG                                       MF / PF = GH / PH                    

HQ / 2  =  FQ / 7                                            3 / PF = 8 / PH                                        

HQ/FQ = 2/7                                                  PF / PH= 3/8

HQ = (2/9)FH   

PF = ( 3/11)FH

HQ = (22/99)FH

FP  = (37/99)FH

PQ = [ 99 - 22 - 27 ] / 99 = 50/99 FH

PQ / FH   = 50 / 99

In a row of five squares, each square is to be colored either red, yellow, or blue, so that no two consecutive squares have the same color. How many ways are there to color the five squares, if there must be at least three yellow squares?  

In order not to have two yellows next to each other,  

the three yellows must occupy squares 1, 3, and 5. 

That leaves squares 2 and 4 to fill, and two colors  

to do it with. There are only four ways. 

Blue Blue     Red Red     Blue Red     Red Blue  

So, the answer is four ways.  

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