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Trapezoid ABCD has vertices A(-1,0), B(0,2), C(m,4)  and D(k,0), with m > 0  and k > 0 . The line y = -x + 4 is perpendicular to the line containing side CD , and the area of trapezoid ABCD is 36 square units. What is the value of k?

Hello Guest!

\(m_{ab}=\frac{2-0}{0-(-1)}\)

\(m_{ab}=2\)

\(y_{dc}=2(x-k)+0\\ y_{dc}=2(x-m)+4\\ 2x-2k=2x-2m+4\\2k=2m-4\\ k=m-2\)

to be continued

  !

As follows:

In triangle ABC angle bisectors AD, BE, CF intersect at a point I. If DI = 3, BD = 4, and BI = 5, then compute the area of triangle ABC.

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Δ ABS is an isosceles triangle.

∠B = 2 [arctan(3 / 4)] = 73.73979529º

AD = tanB * BD = 13.71428571

Area = AD * BD = 54.85714286 u²

Since DI =3, BD = 4  and BI = 5, then triangle IDB  is right with angle IDB = ADB = 90°

And since  ID = 3  and BI = 5.....then sin (DBI)  = 3/5

And cos DBI = 4/5

But since BE is an angle bisector, then sin (ABC)  = sin (2 * DBI)  =  2 sin(DBI)cos(DBI) = 2 ( 3/5) (4/5) = 24/25

And ADB is also a right triangle with BD  = 4

And since sin ABC = 24/25  then  triangle ADB  is a 7 - 24- 25  Pythagorean right triangle

Then sin ABD = 24/25

So sin of DAB  = 7/25

Using the Law of Sines

AB/ sin 90 = BD / sin (DAB)

AB/ 1  = 4 / (7/25)  =  100/7

And

AD  / sin (ABC)  = BD / sin (DAB)

AD = sin (ABC) * BD / sin (DAB)

AD = (24/25) * 4  / ( 7/25)  = 96/7

And since  angle ADB is right, then so is angle ADC

And since AD is an  angle bisector then  sin (BAD)  = sin (DAC)  = 7/25

So sin (ACD)   = 24/25

So we can find DC using the Law of Sines

DC / sin (DAC)  = AD / sin (ACD)

DC = sin (DAC) * AD / sin (ACD)

DC = (7/25) * ( 96/7) / (24/25)

DC = (7/25) * (96/7) * (25/24)

DC = (96/24)  = 4

So  BC =  BD + DC  = 4 + 4  = 8

And  the area of triangle ABC =  (1/2) (BC) ( AB) * sin ( ABC)  = (1/2) ( 8) ( 100/7) ( 24/25)  = 

(4) ( 4) (24/7)  = 

384/7  units^2

See the figure below to  get a feel  for this :

∠(x) = arctan(10 / 8) = 51.34019175º

∠(y) = 90 - ∠x = 38.65980825º

AB = 2 [tan(y) * 8] = 12.8  

Points ABCDEFG form 7 equal arcs.

Arc AB = 360 / 7

Ange AEB = 1/2 (arcAB)

The sum of all 7 angles is 180 degrees. 

This seems trickier  than it really is

Look  at the following figure  :

Note  that   since the sides of the  square = 1....then DB = sqrt 2 = AC

E is the  center of the  square so  BE  = CE  =  sqrt (2)/2

And angle CEB = 90°

And the circle has a radius of 1/2

So....if we take  the area of 1/2 of this  circle  and  subtract  the  area of triangle CEB we will have the same area  as  the shaded region

Area of half-circle  with radius of 1/2  =  pi (1/2)^2  / 2 =  pi/8

Area of triangle CEB  =  area of an isosceles triangle  with equal sides of sqrt (2)/2  and an included angle of 90° =

(1/2) [ sqrt (2) / 2 ] ^2 *sin 90° =  (1/2) ( 2/4) * 1  =  2/8  = 1/4

So....the shaded area =   pi/8  - 1/4  =    [ pi - 2 ] / 8

The interior angles of a pentagon sum to 540°

So.....let  X  be  the measure of angle M , angle T  and angle H......and we  have that

X + X + X  +   58  =  540

3X  +  58  =  540      subtract  58  from both sides

3X   =  482   divide both sides  by  3

X =  482/3  =   ( 160 + 2/3  )°  =  measure of angle  H

(1,11)  and  (4,5)

Slope  =   [ 5 - 11] / [ 4 -1] =   -6/3  =  -2

Equation  of line

y = -2 (x -1)  + 11     simplify

y = -2x + 2 + 11

y = -2x + 13       add 2x to both sides

2x + y =  13        divide through  by 13

(2/13) x  +  (1/13) y =  1        note that we can write this as

  x               y

____    +  _____     =       1

(13/2)         13

This implies  that  a =   (13/2)

Oh, yeah, Melody is right. Thank you!

From point-slope form, a = 39/17 and b = 39/2.

You multipy

Note:  you have allowed that one or more child is able have no strawberry candy or no rhubarb candy or none of either.

Two sectors of a circle of radius 12 overlap as shown. Determine the area of the shaded region.

Hello Guest!
 

\(A=2\cdot (\frac{\pi r^2}{4}-\frac{r^2}{2})\\ A=2\cdot (\frac{\pi \cdot 12^2}{4}-\frac{12^2}{2})\)

\(A=82.195\ cm^2\)

  !

Chuck will have 3/4 of the area of a circle of radius 4        3/4 pi (4)^2 = 12 pi

    PLUS  one quarter circle of radius 2     1/4 pi(2)^2 = pi                

      PLUS   one quarter circle of radius 1    1/4 pi (1)^2 = pi/4                    total = (13 1/4) pi = 41.63 m²

The triangle shown is an equilateral triangle with side length 12 cm. A side of the triangle is the diameter of the circle.  What is the area of the shaded region?

Hello Guest1

\(A=A_▷-\frac{2}{4}A_▷-\frac{1}{6}\pi r^2\)

\(A=(\frac{12^2\sqrt{3}}{4}-\frac{1}{2}\cdot \frac{12^2\sqrt{3}}{4}-\frac{1}{6}\pi \cdot 6^2)\ cm^2\)

\(A=12.327\ cm^2\)

The area of the shaded region is \(12.327\ cm^2\)

  !

That is why I didn't try to confuse as you did with a L-O-N-G answer for a short problem !  

https://web2.0calc.com/questions/geometry_99280

https://web2.0calc.com/questions/help-is-needed_5

area = 13pi + pi/4 = 41.626 m²

37  prime number

add 4 students:

     41  prime number

Let ABC be a right triangle such that B is a right angle. A circle with a diameter of  BC meets side AC at D.  If the area of ABC is 150 and AC = 40 then what is BD?

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Area of Δ ABC:      A = 150 u²

The base of a triangle ABC:       AC = 40 units

BD = 2A / AC = 300 / 40          BD = 7.5 units  

Thank you!

X = 21  

Y = 7  

Pattern:  Triple the first one, then add the second one. 

_.

In triangle ABC , AB = 7 , BC = 8, and the length of median AM  is 4. Find AC.

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Δ ABC is the right-angled triangle.

AC = sqrt(BC² - AB²)